[{"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "In this video, I want to focus on a few more techniques for factoring polynomials. And in particular, I want to focus on quadratics that don't have a 1 as a leading coefficient. For example, if I wanted to factor 4x squared plus 25x minus 21. Everything we've factored so far, or all of the quadratics we've factored so far, had either a 1 or a negative 1 where this 4 is sitting. All of a sudden now we have this 4 here. So what I'm going to teach you is a technique called factoring by grouping. And it's a little bit more involved than what we've learned before, but it's a neat trick."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "Everything we've factored so far, or all of the quadratics we've factored so far, had either a 1 or a negative 1 where this 4 is sitting. All of a sudden now we have this 4 here. So what I'm going to teach you is a technique called factoring by grouping. And it's a little bit more involved than what we've learned before, but it's a neat trick. But to some degree, it'll become obsolete once you learn the quadratic formula, because frankly, the quadratic formula is a lot easier. But this is how it goes. I'll show you the technique, and then at the end of this video, I'll actually show you why it works."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "And it's a little bit more involved than what we've learned before, but it's a neat trick. But to some degree, it'll become obsolete once you learn the quadratic formula, because frankly, the quadratic formula is a lot easier. But this is how it goes. I'll show you the technique, and then at the end of this video, I'll actually show you why it works. So what we need to do here is we need to think of two numbers. We're going to think of two numbers, a and b, where a times b is equal to 4 times negative 21. So a times b is going to be equal to 4 times negative 21, which is equal to negative 84."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "I'll show you the technique, and then at the end of this video, I'll actually show you why it works. So what we need to do here is we need to think of two numbers. We're going to think of two numbers, a and b, where a times b is equal to 4 times negative 21. So a times b is going to be equal to 4 times negative 21, which is equal to negative 84. And those same two numbers, a and b, a plus b, need to be equal to 25. They need to be equal to 25. Let me be very clear."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "So a times b is going to be equal to 4 times negative 21, which is equal to negative 84. And those same two numbers, a and b, a plus b, need to be equal to 25. They need to be equal to 25. Let me be very clear. This is the 25, so they need to be equal to 25. This is where the 4 is, so they need to be equal to 4 times negative 21. That's a negative 21."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "Let me be very clear. This is the 25, so they need to be equal to 25. This is where the 4 is, so they need to be equal to 4 times negative 21. That's a negative 21. So what two numbers are there that would do this? Well, we have to look at the factors of negative 84. And once again, one of these are going to have to be positive."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "That's a negative 21. So what two numbers are there that would do this? Well, we have to look at the factors of negative 84. And once again, one of these are going to have to be positive. The other ones are going to have to be negative, because their product is negative. So let's think about the different factors that might work. 4 and negative 21 look tantalizing, but when you add them, you get negative 17."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "And once again, one of these are going to have to be positive. The other ones are going to have to be negative, because their product is negative. So let's think about the different factors that might work. 4 and negative 21 look tantalizing, but when you add them, you get negative 17. Or if you had negative 4 and 21, you'd get positive 17. It doesn't work. Let's try some other combinations."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "4 and negative 21 look tantalizing, but when you add them, you get negative 17. Or if you had negative 4 and 21, you'd get positive 17. It doesn't work. Let's try some other combinations. 1 and 84, too far apart when you take their difference, because that's essentially what you're going to do if one is negative and one is positive. Too far apart. Let's see."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "Let's try some other combinations. 1 and 84, too far apart when you take their difference, because that's essentially what you're going to do if one is negative and one is positive. Too far apart. Let's see. You could do 2 and 42. Once again, too far apart. Negative 2 plus 42 is 40."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "Let's see. You could do 2 and 42. Once again, too far apart. Negative 2 plus 42 is 40. 2 plus negative 2 is negative 40. Too far apart. 3 goes into 84."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "Negative 2 plus 42 is 40. 2 plus negative 2 is negative 40. Too far apart. 3 goes into 84. 3 goes into 8. 2 times 3 is 6. 8 minus 6 is 2."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "3 goes into 84. 3 goes into 8. 2 times 3 is 6. 8 minus 6 is 2. Bring down the 4. It goes exactly 8 times. So 3 and 28."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "8 minus 6 is 2. Bring down the 4. It goes exactly 8 times. So 3 and 28. This seems interesting. 3 and 28. And remember, one of these has to be negative."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "So 3 and 28. This seems interesting. 3 and 28. And remember, one of these has to be negative. So if we have negative 3 plus 28, that is equal to 25. Now, we found our two numbers, but it's not going to be quite as simple of an operation as what we did when this wasn't a 1 or when this was a 1 or a negative 1. What we're going to do now is split up this term right here."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "And remember, one of these has to be negative. So if we have negative 3 plus 28, that is equal to 25. Now, we found our two numbers, but it's not going to be quite as simple of an operation as what we did when this wasn't a 1 or when this was a 1 or a negative 1. What we're going to do now is split up this term right here. We're going to split this up into negative... We're going to split it up into positive 28x minus 3x. We're just going to split that term. That term is that term right there."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "What we're going to do now is split up this term right here. We're going to split this up into negative... We're going to split it up into positive 28x minus 3x. We're just going to split that term. That term is that term right there. And of course, you have your minus 21 there. And you have your 4x squared over here. Now, you might say, how did you pick the 28 to go here and the negative 3 to go there?"}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "That term is that term right there. And of course, you have your minus 21 there. And you have your 4x squared over here. Now, you might say, how did you pick the 28 to go here and the negative 3 to go there? And it actually does matter. The way I thought about it is 3 or negative 3 and 21 or negative 21, they have some common factors in particular. They have the factor 3 in common."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "Now, you might say, how did you pick the 28 to go here and the negative 3 to go there? And it actually does matter. The way I thought about it is 3 or negative 3 and 21 or negative 21, they have some common factors in particular. They have the factor 3 in common. And 28 and 4 have some common factors. So I kind of grouped the 28 on the side of the 4. And you're going to see what I mean in a second."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "They have the factor 3 in common. And 28 and 4 have some common factors. So I kind of grouped the 28 on the side of the 4. And you're going to see what I mean in a second. If we literally group these, so that term becomes 4x squared plus 28x. And then this side over here in pink, well, I could say it's plus negative 3x minus 21. Once again, I picked these."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "And you're going to see what I mean in a second. If we literally group these, so that term becomes 4x squared plus 28x. And then this side over here in pink, well, I could say it's plus negative 3x minus 21. Once again, I picked these. I grouped the negative 3 with the 21 or the negative 21 because they're both divisible by 3. And I grouped the 28 with the 4 because they're both divisible by 4. And now, in each of these groups, we factor as much out as we can."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "Once again, I picked these. I grouped the negative 3 with the 21 or the negative 21 because they're both divisible by 3. And I grouped the 28 with the 4 because they're both divisible by 4. And now, in each of these groups, we factor as much out as we can. So both of these terms are divisible by 4x. So this orange term is equal to 4x times x. 4x squared divided by 4x is just x."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "And now, in each of these groups, we factor as much out as we can. So both of these terms are divisible by 4x. So this orange term is equal to 4x times x. 4x squared divided by 4x is just x. Plus 28x divided by 4x is just 7. Now, this second term, remember, you factor out everything that you can factor out. Well, both of these terms are divisible by 3 or negative 3."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "4x squared divided by 4x is just x. Plus 28x divided by 4x is just 7. Now, this second term, remember, you factor out everything that you can factor out. Well, both of these terms are divisible by 3 or negative 3. So let's factor out a negative 3. And this becomes x plus 7. And now, something might pop out at you."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "Well, both of these terms are divisible by 3 or negative 3. So let's factor out a negative 3. And this becomes x plus 7. And now, something might pop out at you. We have x plus 7 times 4x plus x plus 7 times negative 3. So we can factor out an x plus 7. We can factor out an x plus 7."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "And now, something might pop out at you. We have x plus 7 times 4x plus x plus 7 times negative 3. So we can factor out an x plus 7. We can factor out an x plus 7. This might not be completely obvious. You're probably not used to factoring out an entire binomial, but you could view this as like a. Or if you have 4xa minus 3a, you would be able to factor out an a."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "We can factor out an x plus 7. This might not be completely obvious. You're probably not used to factoring out an entire binomial, but you could view this as like a. Or if you have 4xa minus 3a, you would be able to factor out an a. And I could just leave this as a minus sign. Let me delete this plus right here because it's just minus 3. It's just minus 3."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "Or if you have 4xa minus 3a, you would be able to factor out an a. And I could just leave this as a minus sign. Let me delete this plus right here because it's just minus 3. It's just minus 3. Plus negative 3, same thing as minus 3. So what can we do here? We have x plus 7 times 4x."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "It's just minus 3. Plus negative 3, same thing as minus 3. So what can we do here? We have x plus 7 times 4x. We have an x plus 7 times negative 3. Let's factor out the x plus 7. We get x plus 7 times 4x minus 3."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "We have x plus 7 times 4x. We have an x plus 7 times negative 3. Let's factor out the x plus 7. We get x plus 7 times 4x minus 3. Minus that 3 right there. And we've factored our binomial. Sorry, we've factored our quadratic by grouping, and we factored it into two binomials."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "We get x plus 7 times 4x minus 3. Minus that 3 right there. And we've factored our binomial. Sorry, we've factored our quadratic by grouping, and we factored it into two binomials. Let's do another example of that and it's a little bit involved, but once you get the hang of it, it's kind of fun. So let's say we want to factor 6x squared plus 7x plus 1. Same drill."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "Sorry, we've factored our quadratic by grouping, and we factored it into two binomials. Let's do another example of that and it's a little bit involved, but once you get the hang of it, it's kind of fun. So let's say we want to factor 6x squared plus 7x plus 1. Same drill. We want to find a times b that is equal to 1 times 6, and we want to find an a plus b needs to be equal to 7. This is a little bit more straightforward. The obvious one is 1 and 6."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "Same drill. We want to find a times b that is equal to 1 times 6, and we want to find an a plus b needs to be equal to 7. This is a little bit more straightforward. The obvious one is 1 and 6. 1 times 6 is 6. 1 plus 6 is 7. So we have a is equal to 1."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "The obvious one is 1 and 6. 1 times 6 is 6. 1 plus 6 is 7. So we have a is equal to 1. Let me not even assign them. The numbers here are 1 and 6. Now we want to split this into a 1x and a 6x, but we want to group it so it's on the side of something that it shares a factor with."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "So we have a is equal to 1. Let me not even assign them. The numbers here are 1 and 6. Now we want to split this into a 1x and a 6x, but we want to group it so it's on the side of something that it shares a factor with. So we're going to have a 6x squared here plus, and so I'm going to put the 6x first because 6 and 6 share a factor. Then we're going to have plus 1x. 6x plus 1x is 7x."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "Now we want to split this into a 1x and a 6x, but we want to group it so it's on the side of something that it shares a factor with. So we're going to have a 6x squared here plus, and so I'm going to put the 6x first because 6 and 6 share a factor. Then we're going to have plus 1x. 6x plus 1x is 7x. That was the whole point. They had to add up to 7. Then we have the final plus 1 there."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "6x plus 1x is 7x. That was the whole point. They had to add up to 7. Then we have the final plus 1 there. Now in each of these groups, we can factor out as much as we like. So in this first group, let's factor out a 6x. So this first group becomes 6x times 6x squared divided by 6x is just an x."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "Then we have the final plus 1 there. Now in each of these groups, we can factor out as much as we like. So in this first group, let's factor out a 6x. So this first group becomes 6x times 6x squared divided by 6x is just an x. 6x divided by 6x is just a 1. Then in the second group, we're going to have a plus here, but this second group, we just literally have an x plus 1, or we could even write a 1 times an x plus 1. You can imagine I just factored out a 1, so to speak."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "So this first group becomes 6x times 6x squared divided by 6x is just an x. 6x divided by 6x is just a 1. Then in the second group, we're going to have a plus here, but this second group, we just literally have an x plus 1, or we could even write a 1 times an x plus 1. You can imagine I just factored out a 1, so to speak. Now I have 6x times x plus 1 plus 1 times x plus 1. Well, I can factor out the x plus 1. If I factor out an x plus 1, that's equal to x plus 1 times 6x plus that 1."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "You can imagine I just factored out a 1, so to speak. Now I have 6x times x plus 1 plus 1 times x plus 1. Well, I can factor out the x plus 1. If I factor out an x plus 1, that's equal to x plus 1 times 6x plus that 1. I'm just doing the distributive property in reverse. So hopefully you didn't find that too bad. Now I'm going to actually explain why this little magical system actually works."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "If I factor out an x plus 1, that's equal to x plus 1 times 6x plus that 1. I'm just doing the distributive property in reverse. So hopefully you didn't find that too bad. Now I'm going to actually explain why this little magical system actually works. Why it actually works. Let me take an example. Let's say I have, well, I'll do it in very general terms."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "Now I'm going to actually explain why this little magical system actually works. Why it actually works. Let me take an example. Let's say I have, well, I'll do it in very general terms. Let's add ax plus b times cx. Actually, I don't want to use, well, I'm afraid to use the a's and the b's. I think that'll confuse you because I use a's and b's here, and they won't be the same thing."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "Let's say I have, well, I'll do it in very general terms. Let's add ax plus b times cx. Actually, I don't want to use, well, I'm afraid to use the a's and the b's. I think that'll confuse you because I use a's and b's here, and they won't be the same thing. Let me use completely different letters. Let's say I have fx plus g times hx plus, I'll use j instead of i. You'll learn in the future why I don't like using i as a variable."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "I think that'll confuse you because I use a's and b's here, and they won't be the same thing. Let me use completely different letters. Let's say I have fx plus g times hx plus, I'll use j instead of i. You'll learn in the future why I don't like using i as a variable. So what is this going to be equal to? Well, it's going to be fx times hx, which is fhx, and then fx times j, so plus fjx. Then we're going to have g times hx, so plus ghx, and then g times j, plus gj."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "You'll learn in the future why I don't like using i as a variable. So what is this going to be equal to? Well, it's going to be fx times hx, which is fhx, and then fx times j, so plus fjx. Then we're going to have g times hx, so plus ghx, and then g times j, plus gj. Or if we add these two middle terms, if you add the two middle terms, you have fh times x plus, add these two terms, fj plus ghx, plus gj. Now, what did I do here? Well, remember, in all of these problems where you have a non-1 or non-negative-1 coefficient, we look for two numbers that add up to this whose product is equal to the product of that times that."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "Then we're going to have g times hx, so plus ghx, and then g times j, plus gj. Or if we add these two middle terms, if you add the two middle terms, you have fh times x plus, add these two terms, fj plus ghx, plus gj. Now, what did I do here? Well, remember, in all of these problems where you have a non-1 or non-negative-1 coefficient, we look for two numbers that add up to this whose product is equal to the product of that times that. Well, here we have two numbers that add up. Let's say that a is equal to fj. Let's say that a is equal to fj."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "Well, remember, in all of these problems where you have a non-1 or non-negative-1 coefficient, we look for two numbers that add up to this whose product is equal to the product of that times that. Well, here we have two numbers that add up. Let's say that a is equal to fj. Let's say that a is equal to fj. That is a, and b is equal to gh. So a plus b is going to be equal to that middle coefficient. a plus b is going to be equal to that middle coefficient there."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "Let's say that a is equal to fj. That is a, and b is equal to gh. So a plus b is going to be equal to that middle coefficient. a plus b is going to be equal to that middle coefficient there. And then what is a times b? a times b is going to be equal to fj times gh, which we could just reorder these terms. We're just multiplying a bunch of terms, so that could be rewritten as f times h times g times j."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "a plus b is going to be equal to that middle coefficient there. And then what is a times b? a times b is going to be equal to fj times gh, which we could just reorder these terms. We're just multiplying a bunch of terms, so that could be rewritten as f times h times g times j. These are all the same things. Well, what is fh times gj? This is equal to fh times gj."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "We're just multiplying a bunch of terms, so that could be rewritten as f times h times g times j. These are all the same things. Well, what is fh times gj? This is equal to fh times gj. Well, this is equal to the first coefficient times the constant term. So if a plus b will be equal to the middle coefficient, then a times b will equal the first coefficient times the constant term. So that's why this whole factoring by grouping even works, or how we're able to figure out what a and b even are."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "This is equal to fh times gj. Well, this is equal to the first coefficient times the constant term. So if a plus b will be equal to the middle coefficient, then a times b will equal the first coefficient times the constant term. So that's why this whole factoring by grouping even works, or how we're able to figure out what a and b even are. Now I'm going to close up with something slightly different, but just to make sure that you have a well-rounded education in factoring things. What I want to do is teach you to factor things a little bit more completely. This is a little bit of an add-on."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "So that's why this whole factoring by grouping even works, or how we're able to figure out what a and b even are. Now I'm going to close up with something slightly different, but just to make sure that you have a well-rounded education in factoring things. What I want to do is teach you to factor things a little bit more completely. This is a little bit of an add-on. I was going to make a whole video on this, but I think on some level it might be a little obvious for you. So let's say we had 2... Let me get a good one here. Let's say we had negative x to the third plus 17x squared minus 70."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "This is a little bit of an add-on. I was going to make a whole video on this, but I think on some level it might be a little obvious for you. So let's say we had 2... Let me get a good one here. Let's say we had negative x to the third plus 17x squared minus 70. Now, I have 70x. Now, immediately you say, gee, this isn't even a quadratic. I don't know how to solve something like this."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "Let's say we had negative x to the third plus 17x squared minus 70. Now, I have 70x. Now, immediately you say, gee, this isn't even a quadratic. I don't know how to solve something like this. It has an x to the third power. The first thing you should realize is that every term here is divisible by x. So let's factor out an x, or even better, let's factor out a negative x."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "I don't know how to solve something like this. It has an x to the third power. The first thing you should realize is that every term here is divisible by x. So let's factor out an x, or even better, let's factor out a negative x. So if you factor out a negative x, this is equal to negative x times... Negative x to the third divided by negative x is x squared. 17x squared divided by negative x is negative 17x."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "So let's factor out an x, or even better, let's factor out a negative x. So if you factor out a negative x, this is equal to negative x times... Negative x to the third divided by negative x is x squared. 17x squared divided by negative x is negative 17x. Negative 70x divided by negative x is positive 70. These cancel out. And now you have something that might look a little bit familiar."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "17x squared divided by negative x is negative 17x. Negative 70x divided by negative x is positive 70. These cancel out. And now you have something that might look a little bit familiar. We have just a standard quadratic where the leading coefficient is a 1, so we just have to find 2 numbers whose product is 70 and that add up to negative 17. And the numbers that immediately jumped into my head are negative 10 and negative 7. You take their product, you get 70, you add them up, you get negative 17."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "And now you have something that might look a little bit familiar. We have just a standard quadratic where the leading coefficient is a 1, so we just have to find 2 numbers whose product is 70 and that add up to negative 17. And the numbers that immediately jumped into my head are negative 10 and negative 7. You take their product, you get 70, you add them up, you get negative 17. So this part right here is going to be x minus 10 times x minus 7. And of course, you have that leading negative x. The general idea here is just see if there's anything you can factor out and then it will get into a form that you might recognize."}, {"video_title": "More examples of factoring by grouping Algebra I Khan Academy.mp3", "Sentence": "You take their product, you get 70, you add them up, you get negative 17. So this part right here is going to be x minus 10 times x minus 7. And of course, you have that leading negative x. The general idea here is just see if there's anything you can factor out and then it will get into a form that you might recognize. Hopefully you found this helpful. Now, I want to reiterate what I showed you at the beginning of this video. I think it's a really cool trick, so to speak, to be able to factor things that have a non-1 or non-negative 1 leading coefficient."}, {"video_title": "Verifying inverse functions by composition Mathematics III High School Math Khan Academy.mp3", "Sentence": "And let's say that g of x, g of x, is equal to the cube root of x plus one, the cube root of x plus one minus seven. Now what I want to do now is evaluate f of, f of g of x, I want to evaluate f of g of x, and I also want to evaluate g of f of x, g of f of x, and see what I get. And I encourage you, like always, pause the video and try it out. All right, let's first evaluate f of g of x. So that means g of x, this expression, is going to be our input. So everywhere we see an x in the definition for f of x, we would replace it with all of g of x. So f of g of x is going to be equal to, so it's going to be equal to, well I see an x right over there, so I'd write all of g of x there."}, {"video_title": "Verifying inverse functions by composition Mathematics III High School Math Khan Academy.mp3", "Sentence": "All right, let's first evaluate f of g of x. So that means g of x, this expression, is going to be our input. So everywhere we see an x in the definition for f of x, we would replace it with all of g of x. So f of g of x is going to be equal to, so it's going to be equal to, well I see an x right over there, so I'd write all of g of x there. So that's the cube root of x plus one minus seven, and then I have plus seven, plus seven, to the third power minus one. Notice, wherever I saw the x, since I'm taking f of g of x, I replace it with what g of x is. And so that is the cube root of x plus one minus seven."}, {"video_title": "Verifying inverse functions by composition Mathematics III High School Math Khan Academy.mp3", "Sentence": "So f of g of x is going to be equal to, so it's going to be equal to, well I see an x right over there, so I'd write all of g of x there. So that's the cube root of x plus one minus seven, and then I have plus seven, plus seven, to the third power minus one. Notice, wherever I saw the x, since I'm taking f of g of x, I replace it with what g of x is. And so that is the cube root of x plus one minus seven. All right, now let's see if we can simplify this. Well we have a minus seven plus seven, so that simplifies nicely. So this just becomes, this is equal to, I can do it in a neutral color now, this is equal to the cube root of x plus one to the third power minus one."}, {"video_title": "Verifying inverse functions by composition Mathematics III High School Math Khan Academy.mp3", "Sentence": "And so that is the cube root of x plus one minus seven. All right, now let's see if we can simplify this. Well we have a minus seven plus seven, so that simplifies nicely. So this just becomes, this is equal to, I can do it in a neutral color now, this is equal to the cube root of x plus one to the third power minus one. Well if I take the cube root of x plus one and then I raise it to the third power, well that's just going to give me x plus one. So this part, this part just simplifies to x plus one, and then I subtract one. So it all simplified out to just being equal to x."}, {"video_title": "Verifying inverse functions by composition Mathematics III High School Math Khan Academy.mp3", "Sentence": "So this just becomes, this is equal to, I can do it in a neutral color now, this is equal to the cube root of x plus one to the third power minus one. Well if I take the cube root of x plus one and then I raise it to the third power, well that's just going to give me x plus one. So this part, this part just simplifies to x plus one, and then I subtract one. So it all simplified out to just being equal to x. So we're just left with an x. So f of g of x is just x. So now let's try what g of f of x is."}, {"video_title": "Verifying inverse functions by composition Mathematics III High School Math Khan Academy.mp3", "Sentence": "So it all simplified out to just being equal to x. So we're just left with an x. So f of g of x is just x. So now let's try what g of f of x is. So g of f of x is going to be equal to, I'll do it right over here, this is going to be equal to the cube root of, and actually let me write it out. Wherever I see an x, I can write f of x instead. I didn't do it that last time."}, {"video_title": "Verifying inverse functions by composition Mathematics III High School Math Khan Academy.mp3", "Sentence": "So now let's try what g of f of x is. So g of f of x is going to be equal to, I'll do it right over here, this is going to be equal to the cube root of, and actually let me write it out. Wherever I see an x, I can write f of x instead. I didn't do it that last time. I went directly and replaced with the definition of f of x. But just to make it clear what I'm doing. So everywhere I'm seeing an x, I replace it with an f of x."}, {"video_title": "Verifying inverse functions by composition Mathematics III High School Math Khan Academy.mp3", "Sentence": "I didn't do it that last time. I went directly and replaced with the definition of f of x. But just to make it clear what I'm doing. So everywhere I'm seeing an x, I replace it with an f of x. So the cube root of f of x plus one minus seven. Well that's going to be equal to the cube root of, cube root of f of x, which is all of this business over here. So that is x plus seven to the third power minus one, and then we add one."}, {"video_title": "Verifying inverse functions by composition Mathematics III High School Math Khan Academy.mp3", "Sentence": "So everywhere I'm seeing an x, I replace it with an f of x. So the cube root of f of x plus one minus seven. Well that's going to be equal to the cube root of, cube root of f of x, which is all of this business over here. So that is x plus seven to the third power minus one, and then we add one. And we add one, and then we subtract the seven. So lucky for us, this subtracting one and adding one, those cancel out. And so we're going to take the cube root of x plus seven to the third power."}, {"video_title": "Verifying inverse functions by composition Mathematics III High School Math Khan Academy.mp3", "Sentence": "So that is x plus seven to the third power minus one, and then we add one. And we add one, and then we subtract the seven. So lucky for us, this subtracting one and adding one, those cancel out. And so we're going to take the cube root of x plus seven to the third power. Well the cube root of x plus seven to the third power is just going to be x plus seven. So this is going to be x plus seven. So all of this business simplifies to x plus seven."}, {"video_title": "Verifying inverse functions by composition Mathematics III High School Math Khan Academy.mp3", "Sentence": "And so we're going to take the cube root of x plus seven to the third power. Well the cube root of x plus seven to the third power is just going to be x plus seven. So this is going to be x plus seven. So all of this business simplifies to x plus seven. And then we need to subtract seven. And these two cancel out, or they negate each other, and we are just left with x. So we see something very interesting."}, {"video_title": "Verifying inverse functions by composition Mathematics III High School Math Khan Academy.mp3", "Sentence": "So all of this business simplifies to x plus seven. And then we need to subtract seven. And these two cancel out, or they negate each other, and we are just left with x. So we see something very interesting. f of g of x is just x, and g of f of x is x. So if we, in this case, if we start with an x, if we start with an x, we input it into the function g, and we get g of x, we get g of x, and then we input that into the function f, and we put that into the function f, f of g of x gets us back to x. It gets us back to x."}, {"video_title": "Verifying inverse functions by composition Mathematics III High School Math Khan Academy.mp3", "Sentence": "So we see something very interesting. f of g of x is just x, and g of f of x is x. So if we, in this case, if we start with an x, if we start with an x, we input it into the function g, and we get g of x, we get g of x, and then we input that into the function f, and we put that into the function f, f of g of x gets us back to x. It gets us back to x. So we kind of did a round trip. And the same thing is happening over here. If I put x into f of x, if I put x into f of, sorry, if I put x into the function f, and I get f of x, the output is f of x, and then I input that into g, into the function g, into the function g, once again I do this round trip, and I get back to x."}, {"video_title": "Verifying inverse functions by composition Mathematics III High School Math Khan Academy.mp3", "Sentence": "It gets us back to x. So we kind of did a round trip. And the same thing is happening over here. If I put x into f of x, if I put x into f of, sorry, if I put x into the function f, and I get f of x, the output is f of x, and then I input that into g, into the function g, into the function g, once again I do this round trip, and I get back to x. Another way to think about it, another way to think about it, if you view this as, so these are both composite functions, but one way to think about it is, if these are the set of all possible inputs into either of these composite functions, and then these are the outputs, so you are starting with an x, you are starting with an x, I'll do this case first. So g is a mapping, let me write down, so g is going to be a mapping from x to g of x. So this is what g is doing."}, {"video_title": "Verifying inverse functions by composition Mathematics III High School Math Khan Academy.mp3", "Sentence": "If I put x into f of x, if I put x into f of, sorry, if I put x into the function f, and I get f of x, the output is f of x, and then I input that into g, into the function g, into the function g, once again I do this round trip, and I get back to x. Another way to think about it, another way to think about it, if you view this as, so these are both composite functions, but one way to think about it is, if these are the set of all possible inputs into either of these composite functions, and then these are the outputs, so you are starting with an x, you are starting with an x, I'll do this case first. So g is a mapping, let me write down, so g is going to be a mapping from x to g of x. So this is what g is doing. So the function g maps from x to some value g of x, g of x, and then if you were to apply f to this value right over here, if you apply f to this value, to g of x, you get all the way back to x. So that is f of g of x. And vice versa."}, {"video_title": "Verifying inverse functions by composition Mathematics III High School Math Khan Academy.mp3", "Sentence": "So this is what g is doing. So the function g maps from x to some value g of x, g of x, and then if you were to apply f to this value right over here, if you apply f to this value, to g of x, you get all the way back to x. So that is f of g of x. And vice versa. If you start with x, and you apply f of x first, so if you start with f, if you apply f of x first, let me do that. So if you apply f of x first, you say you get to this value, so that is f of x. So you applied the function f, but then you apply the function g to that, you apply the function g to that, you get back."}, {"video_title": "Verifying inverse functions by composition Mathematics III High School Math Khan Academy.mp3", "Sentence": "And vice versa. If you start with x, and you apply f of x first, so if you start with f, if you apply f of x first, let me do that. So if you apply f of x first, you say you get to this value, so that is f of x. So you applied the function f, but then you apply the function g to that, you apply the function g to that, you get back. So this is g of f of x, I should say, g of f, where we're applying the function g to the value f of x. And so since we get a round trip either way, we know that the functions g and f are inverses of each other. In fact, we can write, we can write that f of x is equal to the inverse of g of x, inverse of g of x, and vice versa."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping Khan Academy.mp3", "Sentence": "We need to factor negative 12 f squared minus 38f plus 22. So a good place to start is to see if, is there any common factor for all three of these terms? When you look at them, they're all even. And we don't like a negative number out here. So let's divide everything, or let's factor out a negative 2. So this expression right here is the same thing as negative 2 times negative 12 f squared divided by negative 2. So it's positive 6 f squared."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping Khan Academy.mp3", "Sentence": "And we don't like a negative number out here. So let's divide everything, or let's factor out a negative 2. So this expression right here is the same thing as negative 2 times negative 12 f squared divided by negative 2. So it's positive 6 f squared. Negative 38 divided by negative 2 is positive 19. So it'll be positive 19f. And then 22 divided by negative 2 is negative 11."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping Khan Academy.mp3", "Sentence": "So it's positive 6 f squared. Negative 38 divided by negative 2 is positive 19. So it'll be positive 19f. And then 22 divided by negative 2 is negative 11. So we've simplified it a bit. We have the 6 f squared plus 19f minus 11. We'll just focus on that part right now."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping Khan Academy.mp3", "Sentence": "And then 22 divided by negative 2 is negative 11. So we've simplified it a bit. We have the 6 f squared plus 19f minus 11. We'll just focus on that part right now. And the best way to factor this thing, since we don't have a 1 here as the coefficient on the f squared, is to factor it by grouping. So we need to look for two numbers whose products is 6 times negative 11. So two numbers."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping Khan Academy.mp3", "Sentence": "We'll just focus on that part right now. And the best way to factor this thing, since we don't have a 1 here as the coefficient on the f squared, is to factor it by grouping. So we need to look for two numbers whose products is 6 times negative 11. So two numbers. So a times b needs to be equal to 6 times negative 11, or negative 66. And a plus b needs to be equal to 19. So let's try a few numbers here."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping Khan Academy.mp3", "Sentence": "So two numbers. So a times b needs to be equal to 6 times negative 11, or negative 66. And a plus b needs to be equal to 19. So let's try a few numbers here. So let's see, 22, 22. I'm just thinking of numbers that are roughly 19 apart, because they're going to be of different signs. So 22 and 3, I think, will work."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping Khan Academy.mp3", "Sentence": "So let's try a few numbers here. So let's see, 22, 22. I'm just thinking of numbers that are roughly 19 apart, because they're going to be of different signs. So 22 and 3, I think, will work. Right. If we take 22 times negative 3, that is negative 66. And 22 plus negative 3 is equal to 19."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping Khan Academy.mp3", "Sentence": "So 22 and 3, I think, will work. Right. If we take 22 times negative 3, that is negative 66. And 22 plus negative 3 is equal to 19. And the way I kind of got pretty close to this number is, well, they're going to be of different signs. So the positive versions of them have to be about 19 apart. And that worked out."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping Khan Academy.mp3", "Sentence": "And 22 plus negative 3 is equal to 19. And the way I kind of got pretty close to this number is, well, they're going to be of different signs. So the positive versions of them have to be about 19 apart. And that worked out. So 22 and negative 3. So now we can rewrite this 19f right here as the sum of negative 3f and 22f. So it's negative 3f plus 22f."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping Khan Academy.mp3", "Sentence": "And that worked out. So 22 and negative 3. So now we can rewrite this 19f right here as the sum of negative 3f and 22f. So it's negative 3f plus 22f. That's the same thing as 19f. I just kind of broke it apart. And of course, we have the 6f squared."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping Khan Academy.mp3", "Sentence": "So it's negative 3f plus 22f. That's the same thing as 19f. I just kind of broke it apart. And of course, we have the 6f squared. And we have the minus 11 here. Now, you're probably saying, hey, Sal, why did you put the 22 here and the negative 3 there? Why didn't you do it the other way around?"}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping Khan Academy.mp3", "Sentence": "And of course, we have the 6f squared. And we have the minus 11 here. Now, you're probably saying, hey, Sal, why did you put the 22 here and the negative 3 there? Why didn't you do it the other way around? Why didn't you put the 22 and then the negative 3 there? And my main motivation for doing it, I like to put the negative 3 on the same side with the 6, because they have the common factor of the 3. I like to put the 22 with the negative 11."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping Khan Academy.mp3", "Sentence": "Why didn't you do it the other way around? Why didn't you put the 22 and then the negative 3 there? And my main motivation for doing it, I like to put the negative 3 on the same side with the 6, because they have the common factor of the 3. I like to put the 22 with the negative 11. They have the same factor of 11. So that's why I decided to do it that way. So now let's do the grouping."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping Khan Academy.mp3", "Sentence": "I like to put the 22 with the negative 11. They have the same factor of 11. So that's why I decided to do it that way. So now let's do the grouping. And of course, you can't forget this negative 2 that we have sitting out here the whole time. So let me put that negative 2 out there. But that'll just kind of hang out for a while."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping Khan Academy.mp3", "Sentence": "So now let's do the grouping. And of course, you can't forget this negative 2 that we have sitting out here the whole time. So let me put that negative 2 out there. But that'll just kind of hang out for a while. But let's do some grouping. So let's group these first two. And then we're going to group this."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping Khan Academy.mp3", "Sentence": "But that'll just kind of hang out for a while. But let's do some grouping. So let's group these first two. And then we're going to group this. Let me get a nice color here. And then we're going to group this second 2. That's almost an identical color."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping Khan Academy.mp3", "Sentence": "And then we're going to group this. Let me get a nice color here. And then we're going to group this second 2. That's almost an identical color. Let me do it in this purple color. And then we can group that second 2 right there. So these first two, we could factor out a negative 3f."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping Khan Academy.mp3", "Sentence": "That's almost an identical color. Let me do it in this purple color. And then we can group that second 2 right there. So these first two, we could factor out a negative 3f. So it's negative 3f times 6f squared divided by negative 3f is negative 2f. And the negative 3f divided by negative 3f is just positive f. Actually, a better way to start, instead of factoring out a negative 3f, let's just factor out 3f. So we don't have a negative out here."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping Khan Academy.mp3", "Sentence": "So these first two, we could factor out a negative 3f. So it's negative 3f times 6f squared divided by negative 3f is negative 2f. And the negative 3f divided by negative 3f is just positive f. Actually, a better way to start, instead of factoring out a negative 3f, let's just factor out 3f. So we don't have a negative out here. We could do it either way. But if we just factor out a 3f, 6f squared divided by 3f is 2f. And then negative 3f divided by 3f is negative 1."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping Khan Academy.mp3", "Sentence": "So we don't have a negative out here. We could do it either way. But if we just factor out a 3f, 6f squared divided by 3f is 2f. And then negative 3f divided by 3f is negative 1. So that's what that factors into. And then that second part, in that dark purple color, we can factor out an 11. So we factor out an 11."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping Khan Academy.mp3", "Sentence": "And then negative 3f divided by 3f is negative 1. So that's what that factors into. And then that second part, in that dark purple color, we can factor out an 11. So we factor out an 11. And if we factor that out, 22f divided by 11 is 2f. And negative 11 divided by 11 is negative 1. And of course, once again, you have that negative 2 hanging out there."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping Khan Academy.mp3", "Sentence": "So we factor out an 11. And if we factor that out, 22f divided by 11 is 2f. And negative 11 divided by 11 is negative 1. And of course, once again, you have that negative 2 hanging out there. You have that negative 2. Now, inside the parentheses, we have two terms, both of which have 2f minus 1 as a factor. So we can factor that out."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping Khan Academy.mp3", "Sentence": "And of course, once again, you have that negative 2 hanging out there. You have that negative 2. Now, inside the parentheses, we have two terms, both of which have 2f minus 1 as a factor. So we can factor that out. This whole thing is just an exercise in doing the reverse distributive property, if you will. So let's factor that out. So you have 2f minus 1 times this 3f, times that 3f, and then times that plus 11."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping Khan Academy.mp3", "Sentence": "So we can factor that out. This whole thing is just an exercise in doing the reverse distributive property, if you will. So let's factor that out. So you have 2f minus 1 times this 3f, times that 3f, and then times that plus 11. Let me do that in the same shade of purple right over there. And you can distribute, if you like. 2f minus 1 times 3f will give you this term."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping Khan Academy.mp3", "Sentence": "So you have 2f minus 1 times this 3f, times that 3f, and then times that plus 11. Let me do that in the same shade of purple right over there. And you can distribute, if you like. 2f minus 1 times 3f will give you this term. 2f minus 1 times 11 will give you that term. And we can't forget that we still have that negative 2 hanging out outside. I don't want to change the colors on it."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping Khan Academy.mp3", "Sentence": "2f minus 1 times 3f will give you this term. 2f minus 1 times 11 will give you that term. And we can't forget that we still have that negative 2 hanging out outside. I don't want to change the colors on it. We have the negative 2 hanging out, that same negative 2 over there. And we're done factoring it. Negative 12f squared minus 38."}, {"video_title": "Checking ordered pair solutions to equations example 2 Algebra I Khan Academy.mp3", "Sentence": "Four x minus one is equal to three y plus five. Now when we look at an ordered pair and we want to figure out whether it's a solution, we just have to remind ourselves that in these ordered pairs, the convention, the standard is, is that the first coordinate is the x coordinate and the second coordinate is the y coordinate. So they're gonna, if this is a solution, if this ordered pair is a solution, that means that if x is equal to three and y is equal to two, that that would satisfy this equation up here. So let's try that out. So we have four times x. Well, we're saying x needs to be equal to three minus one is going to be equal to three times y. Well, if this ordered pair is a solution, then y is going to be equal to two."}, {"video_title": "Checking ordered pair solutions to equations example 2 Algebra I Khan Academy.mp3", "Sentence": "So let's try that out. So we have four times x. Well, we're saying x needs to be equal to three minus one is going to be equal to three times y. Well, if this ordered pair is a solution, then y is going to be equal to two. So three times y, y is two, plus five. Notice all I did is wherever I saw the x, I substituted it with three, wherever I saw the y, I substituted it with two. Now let's see if this is true."}, {"video_title": "Checking ordered pair solutions to equations example 2 Algebra I Khan Academy.mp3", "Sentence": "Well, if this ordered pair is a solution, then y is going to be equal to two. So three times y, y is two, plus five. Notice all I did is wherever I saw the x, I substituted it with three, wherever I saw the y, I substituted it with two. Now let's see if this is true. Four times three is 12 minus one. Is this really the same thing as three times two, which is six plus five? See, 12 minus one is 11."}, {"video_title": "Checking ordered pair solutions to equations example 2 Algebra I Khan Academy.mp3", "Sentence": "Now let's see if this is true. Four times three is 12 minus one. Is this really the same thing as three times two, which is six plus five? See, 12 minus one is 11. Six plus five is also 11. This is true. 11 equals 11."}, {"video_title": "Checking ordered pair solutions to equations example 2 Algebra I Khan Academy.mp3", "Sentence": "See, 12 minus one is 11. Six plus five is also 11. This is true. 11 equals 11. This pair, three comma two, does satisfy this equation. Now let's see whether this one does. Two comma three."}, {"video_title": "Checking ordered pair solutions to equations example 2 Algebra I Khan Academy.mp3", "Sentence": "11 equals 11. This pair, three comma two, does satisfy this equation. Now let's see whether this one does. Two comma three. So this is saying when x is equal to two, y would be equal to three for this equation. Let's see if that's true. So four times x, we're now going to see if when x is two, y can be three."}, {"video_title": "Checking ordered pair solutions to equations example 2 Algebra I Khan Academy.mp3", "Sentence": "Two comma three. So this is saying when x is equal to two, y would be equal to three for this equation. Let's see if that's true. So four times x, we're now going to see if when x is two, y can be three. So four times x, four times two, minus one is equal to three times y. Now y we're testing to see if it can be three. Three times three plus five."}, {"video_title": "Checking ordered pair solutions to equations example 2 Algebra I Khan Academy.mp3", "Sentence": "So four times x, we're now going to see if when x is two, y can be three. So four times x, four times two, minus one is equal to three times y. Now y we're testing to see if it can be three. Three times three plus five. Let's see if this is true. Four times two is eight minus one. Is this equal to three times three?"}, {"video_title": "Checking ordered pair solutions to equations example 2 Algebra I Khan Academy.mp3", "Sentence": "Three times three plus five. Let's see if this is true. Four times two is eight minus one. Is this equal to three times three? So that's nine plus five. So is seven equal to 14? No, clearly seven is not equal to 14."}, {"video_title": "Checking ordered pair solutions to equations example 2 Algebra I Khan Academy.mp3", "Sentence": "Is this equal to three times three? So that's nine plus five. So is seven equal to 14? No, clearly seven is not equal to 14. So these things are not equal to each other. So this is not a solution. When x equals two, y cannot be equal to three and satisfy this equation."}, {"video_title": "Adding & subtracting rational expressions like denominators High School Math Khan Academy.mp3", "Sentence": "And like always, pause the video and try to work it out before I do. Well, when you look at this, we have these two rational expressions and we have the same denominator, two x squared minus seven. So you could say we have six two x squared minus sevenths and then we have negative three x minus eight two x squared minus sevenths is one way to think about it. So if you have the same denominator, this is going to be equal to, this is going to be equal to, our denominator's going to be two x squared minus seven, two x squared minus seven, and then we just add the numerators. So it's going to be six plus negative three x, negative three x minus eight. And so if we want to simplify this a little bit, we'd recognize that we could add these two constant terms, the six and the negative eight. Six plus negative eight is going to be negative two, so it's going to be negative two, and then adding a negative three x, that's the same thing as subtracting three x, so negative two minus three x, all of that over, all of that, do that same blue color, all of that over two x squared minus seven."}, {"video_title": "Adding & subtracting rational expressions like denominators High School Math Khan Academy.mp3", "Sentence": "So if you have the same denominator, this is going to be equal to, this is going to be equal to, our denominator's going to be two x squared minus seven, two x squared minus seven, and then we just add the numerators. So it's going to be six plus negative three x, negative three x minus eight. And so if we want to simplify this a little bit, we'd recognize that we could add these two constant terms, the six and the negative eight. Six plus negative eight is going to be negative two, so it's going to be negative two, and then adding a negative three x, that's the same thing as subtracting three x, so negative two minus three x, all of that over, all of that, do that same blue color, all of that over two x squared minus seven. And we're done, we've just added these two rational expressions. Let's do another example. So here, we want to subtract one rational expression from another."}, {"video_title": "Adding & subtracting rational expressions like denominators High School Math Khan Academy.mp3", "Sentence": "Six plus negative eight is going to be negative two, so it's going to be negative two, and then adding a negative three x, that's the same thing as subtracting three x, so negative two minus three x, all of that over, all of that, do that same blue color, all of that over two x squared minus seven. And we're done, we've just added these two rational expressions. Let's do another example. So here, we want to subtract one rational expression from another. So see if you can figure that out. Well, once again, both of these rational expressions have the exact same denominator. The denominator for both of them is 14 x squared minus nine, 14 x squared minus nine, so the denominator of the difference, I guess we can call it that, is going to be 14 x squared minus nine."}, {"video_title": "Adding & subtracting rational expressions like denominators High School Math Khan Academy.mp3", "Sentence": "So here, we want to subtract one rational expression from another. So see if you can figure that out. Well, once again, both of these rational expressions have the exact same denominator. The denominator for both of them is 14 x squared minus nine, 14 x squared minus nine, so the denominator of the difference, I guess we can call it that, is going to be 14 x squared minus nine. So 14 x squared minus nine. Did I say four x squared before? 14 x squared minus nine, that's the denominator of both of them, so that's going to be the denominator of our answer right over here."}, {"video_title": "Adding & subtracting rational expressions like denominators High School Math Khan Academy.mp3", "Sentence": "The denominator for both of them is 14 x squared minus nine, 14 x squared minus nine, so the denominator of the difference, I guess we can call it that, is going to be 14 x squared minus nine. So 14 x squared minus nine. Did I say four x squared before? 14 x squared minus nine, that's the denominator of both of them, so that's going to be the denominator of our answer right over here. And so, we can just subtract the numerators. So we're gonna have nine x squared plus three minus, minus all of this business, minus negative three x squared plus five, and so we can distribute the negative sign. This is going to be equal to, this is going to be equal to nine x squared plus three, and then if you distribute the negative sign, the negative of negative three x squared is going to be plus three x squared, and then the negative of positive five is going to be negative five, so we're gonna subtract five from that, and all of that is going to be over 14 x squared minus nine, 14 x squared minus nine."}, {"video_title": "Adding & subtracting rational expressions like denominators High School Math Khan Academy.mp3", "Sentence": "14 x squared minus nine, that's the denominator of both of them, so that's going to be the denominator of our answer right over here. And so, we can just subtract the numerators. So we're gonna have nine x squared plus three minus, minus all of this business, minus negative three x squared plus five, and so we can distribute the negative sign. This is going to be equal to, this is going to be equal to nine x squared plus three, and then if you distribute the negative sign, the negative of negative three x squared is going to be plus three x squared, and then the negative of positive five is going to be negative five, so we're gonna subtract five from that, and all of that is going to be over 14 x squared minus nine, 14 x squared minus nine. And so in the numerator, we can do some simplification. We have nine x squared plus three x squared, so that's going to be equal to 12 x squared, and then we have, we have three plus negative five, or we could say three minus five, so that's going to be negative two, and all of that is going to be over 14 x squared minus nine. Minus nine."}, {"video_title": "Adding & subtracting rational expressions like denominators High School Math Khan Academy.mp3", "Sentence": "This is going to be equal to, this is going to be equal to nine x squared plus three, and then if you distribute the negative sign, the negative of negative three x squared is going to be plus three x squared, and then the negative of positive five is going to be negative five, so we're gonna subtract five from that, and all of that is going to be over 14 x squared minus nine, 14 x squared minus nine. And so in the numerator, we can do some simplification. We have nine x squared plus three x squared, so that's going to be equal to 12 x squared, and then we have, we have three plus negative five, or we could say three minus five, so that's going to be negative two, and all of that is going to be over 14 x squared minus nine. Minus nine. 14 x squared minus nine. And we're all done. We have just subtracted, and we could think about is there any way to simplify this more?"}, {"video_title": "Does a vertical line represent a function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "In the following graph is y a function of x. So in order for y to be a function of x, for any x that you input into the function, any x for which the function is defined, so let's say we have y is equal to f of x, so we have our little function machine, it should spit out exactly one value of y. If it spits out multiple values of y, we don't know what f of x is going to be equal to. It could be equal to any of those possible values for y. So let's see if this, for this graph, whether for a given x it spits out exactly one y. Well, the function seems to be only defined, so the domain of this function is x is equal to negative two. That's the only place where we have a definition for it."}, {"video_title": "Does a vertical line represent a function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "It could be equal to any of those possible values for y. So let's see if this, for this graph, whether for a given x it spits out exactly one y. Well, the function seems to be only defined, so the domain of this function is x is equal to negative two. That's the only place where we have a definition for it. And if we try to input negative two into this little black box, what do we get? Do we get exactly one thing? No, we put in negative two here."}, {"video_title": "Does a vertical line represent a function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "That's the only place where we have a definition for it. And if we try to input negative two into this little black box, what do we get? Do we get exactly one thing? No, we put in negative two here. We could get anything. Negative two, the point negative two nine is on this relation. Negative two eight is on this relation."}, {"video_title": "Does a vertical line represent a function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "No, we put in negative two here. We could get anything. Negative two, the point negative two nine is on this relation. Negative two eight is on this relation. Negative two seven, negative two 7.5, negative two 3.14159, they're all on these. So if you put a negative two into this relation, you actually get, essentially, you actually get an infinite set of values. It could be nine, it could be 3.14, it could be eight, it could be negative eight."}, {"video_title": "Dividing complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "We're asked to divide, and we're dividing 6 plus 3i by 7 minus 5i. And in particular, when I divide this, I want to get another complex number. So I want to get something, you know, some real number plus some imaginary number. So some multiple of i. So let's think about how we can do this. Well, division is the same thing. We could rewrite this as 6 plus 3i over 7 minus 5i."}, {"video_title": "Dividing complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So some multiple of i. So let's think about how we can do this. Well, division is the same thing. We could rewrite this as 6 plus 3i over 7 minus 5i. These are clearly equivalent. Dividing by something is the same thing as a rational expression, where that something is in the denominator right over here. And so how do we simplify this?"}, {"video_title": "Dividing complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "We could rewrite this as 6 plus 3i over 7 minus 5i. These are clearly equivalent. Dividing by something is the same thing as a rational expression, where that something is in the denominator right over here. And so how do we simplify this? Well, we have a tool in our toolkit that can make sure that we don't have an imaginary or a complex number in the denominator. And that's the complex conjugate. If we multiply both the numerator and the denominator of this expression by the complex conjugate of the denominator, then we will get rid of, or we will have a real number in the denominator."}, {"video_title": "Dividing complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And so how do we simplify this? Well, we have a tool in our toolkit that can make sure that we don't have an imaginary or a complex number in the denominator. And that's the complex conjugate. If we multiply both the numerator and the denominator of this expression by the complex conjugate of the denominator, then we will get rid of, or we will have a real number in the denominator. So let's do that. Let's multiply the numerator and the denominator by the conjugate of this. So 7 plus 5i."}, {"video_title": "Dividing complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "If we multiply both the numerator and the denominator of this expression by the complex conjugate of the denominator, then we will get rid of, or we will have a real number in the denominator. So let's do that. Let's multiply the numerator and the denominator by the conjugate of this. So 7 plus 5i. Plus 7 plus 5i is the complex conjugate of 7 minus 5i. So we're going to multiply it by 7 plus 5i over 7 plus 5i. And anything divided by itself is going to be 1, assuming that you're not dealing with 0."}, {"video_title": "Dividing complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So 7 plus 5i. Plus 7 plus 5i is the complex conjugate of 7 minus 5i. So we're going to multiply it by 7 plus 5i over 7 plus 5i. And anything divided by itself is going to be 1, assuming that you're not dealing with 0. 0 over 0 is undefined. But 7 plus 5i over 7 plus 5i is 1. So we're not changing the value of this."}, {"video_title": "Dividing complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And anything divided by itself is going to be 1, assuming that you're not dealing with 0. 0 over 0 is undefined. But 7 plus 5i over 7 plus 5i is 1. So we're not changing the value of this. But what this does is it allows us to get rid of the imaginary part in the denominator. So let's multiply this out. Our numerator, we just have to multiply every part of this complex number times every part of this complex number."}, {"video_title": "Dividing complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So we're not changing the value of this. But what this does is it allows us to get rid of the imaginary part in the denominator. So let's multiply this out. Our numerator, we just have to multiply every part of this complex number times every part of this complex number. You can think of it as foil if you like. We're really just doing the distributive property twice. We have 6 times 6 is, or sorry, 6 times 7, which is 42."}, {"video_title": "Dividing complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Our numerator, we just have to multiply every part of this complex number times every part of this complex number. You can think of it as foil if you like. We're really just doing the distributive property twice. We have 6 times 6 is, or sorry, 6 times 7, which is 42. And then we have 6 times 5i, which is 30i plus 30i. And then we have 3i times 7. So that's plus 21i."}, {"video_title": "Dividing complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "We have 6 times 6 is, or sorry, 6 times 7, which is 42. And then we have 6 times 5i, which is 30i plus 30i. And then we have 3i times 7. So that's plus 21i. And then finally, we have 3i times 5i. 3 times 5 is 15. But we have i times i, or i squared, which is negative 1."}, {"video_title": "Dividing complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So that's plus 21i. And then finally, we have 3i times 5i. 3 times 5 is 15. But we have i times i, or i squared, which is negative 1. So it would be 15 times negative 1, or minus 15. So that's our numerator. And then our denominator is going to be, well, we have a plus b times a minus b."}, {"video_title": "Dividing complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "But we have i times i, or i squared, which is negative 1. So it would be 15 times negative 1, or minus 15. So that's our numerator. And then our denominator is going to be, well, we have a plus b times a minus b. You could think of it that way, or you could just do what we just did up here. Actually, let's just do what we did up here so you don't have to remember that difference of squares pattern and all of that. 7 times 7 is 49."}, {"video_title": "Dividing complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And then our denominator is going to be, well, we have a plus b times a minus b. You could think of it that way, or you could just do what we just did up here. Actually, let's just do what we did up here so you don't have to remember that difference of squares pattern and all of that. 7 times 7 is 49. Let's think of it in a foil way if that is helpful for you. So first, we did the 7 times a 7, then we could do the outer terms. 7 times 5i is plus 35i."}, {"video_title": "Dividing complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "7 times 7 is 49. Let's think of it in a foil way if that is helpful for you. So first, we did the 7 times a 7, then we could do the outer terms. 7 times 5i is plus 35i. Then we could do the inner terms. Negative 5i times 7 is minus 35i. These two are going to cancel out."}, {"video_title": "Dividing complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "7 times 5i is plus 35i. Then we could do the inner terms. Negative 5i times 7 is minus 35i. These two are going to cancel out. And then negative 5i times 5i is negative 25i squared. Negative 25i squared is the same thing as negative 25 times negative 1. So that is plus 25."}, {"video_title": "Dividing complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "These two are going to cancel out. And then negative 5i times 5i is negative 25i squared. Negative 25i squared is the same thing as negative 25 times negative 1. So that is plus 25. Now let's simplify them. These guys down here cancel out. Our denominator simplifies to 49 plus 25 is 74."}, {"video_title": "Dividing complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So that is plus 25. Now let's simplify them. These guys down here cancel out. Our denominator simplifies to 49 plus 25 is 74. And our numerator, we can add the real parts. So we have a 42 and a negative 15. So let's see, 42 minus 5 would be 37, minus another 10 would be 27."}, {"video_title": "Dividing complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Our denominator simplifies to 49 plus 25 is 74. And our numerator, we can add the real parts. So we have a 42 and a negative 15. So let's see, 42 minus 5 would be 37, minus another 10 would be 27. So that is 27. And then we're going to add our 30i plus the 21i. So 30 of something plus 21 of that same something is going to be 51 of that something."}, {"video_title": "Dividing complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So let's see, 42 minus 5 would be 37, minus another 10 would be 27. So that is 27. And then we're going to add our 30i plus the 21i. So 30 of something plus 21 of that same something is going to be 51 of that something. In this case, that something is the imaginary unit. It is i. Let me do that in magenta."}, {"video_title": "Dividing complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So 30 of something plus 21 of that same something is going to be 51 of that something. In this case, that something is the imaginary unit. It is i. Let me do that in magenta. I guess that's orange. So this is plus 51i. And I want to write it in the form of a plus bi, the traditional complex number form."}, {"video_title": "Dividing complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Let me do that in magenta. I guess that's orange. So this is plus 51i. And I want to write it in the form of a plus bi, the traditional complex number form. So this right over here is the same thing as 27 over 74 plus 51 over 74 times i. And we are done. I want to write that i in that same orange color."}, {"video_title": "Dividing complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And I want to write it in the form of a plus bi, the traditional complex number form. So this right over here is the same thing as 27 over 74 plus 51 over 74 times i. And we are done. I want to write that i in that same orange color. And we are done. We have a real part and we have an imaginary part. And if this last step just confuses you a little bit, remember, this is the same thing if it's helpful for you."}, {"video_title": "Dividing complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "I want to write that i in that same orange color. And we are done. We have a real part and we have an imaginary part. And if this last step just confuses you a little bit, remember, this is the same thing if it's helpful for you. We're essentially multiplying both of these terms times 1 over 74. We're dividing both of these terms by 74. We're distributing the 1 over 74 times both of these, I guess, is one way to think about it."}, {"video_title": "Special products of the form (x+a)(x-a) Algebra I High School Math Khan Academy.mp3", "Sentence": "Let's see if we can figure out what x plus three times x minus three is, and I encourage you to pause the video and see if you can work this out. Well, one way to tackle it is the way that we've always tackled when we multiply binomials is just apply the distributive property twice. So first, we could take this entire yellow x plus three and multiply it times each of these two terms. So first, we can multiply it times this x, so that's going to be x times x plus three, and then we are going to multiply it times, we could say this negative three. So we could write minus three times, now that's going to be multiplied by x plus three again, x plus three, and then we apply the distributive property one more time, where we take this magenta x and we distribute it across this x plus three, so x times x is x squared, x times three is three x, and then we do it on this side. Negative three times x is negative three x, and negative three times three is negative nine. And what does this simplify to?"}, {"video_title": "Special products of the form (x+a)(x-a) Algebra I High School Math Khan Academy.mp3", "Sentence": "So first, we can multiply it times this x, so that's going to be x times x plus three, and then we are going to multiply it times, we could say this negative three. So we could write minus three times, now that's going to be multiplied by x plus three again, x plus three, and then we apply the distributive property one more time, where we take this magenta x and we distribute it across this x plus three, so x times x is x squared, x times three is three x, and then we do it on this side. Negative three times x is negative three x, and negative three times three is negative nine. And what does this simplify to? Well, we're gonna get x squared, and we have three x minus three x, so these two characters cancel out, and we are just left with x squared minus nine. And you might see a little pattern here. Notice, I added three and then I subtracted three, and I got this, I got the x squared, and then if you take three and multiply it by negative three, you are going to get a negative nine."}, {"video_title": "Special products of the form (x+a)(x-a) Algebra I High School Math Khan Academy.mp3", "Sentence": "And what does this simplify to? Well, we're gonna get x squared, and we have three x minus three x, so these two characters cancel out, and we are just left with x squared minus nine. And you might see a little pattern here. Notice, I added three and then I subtracted three, and I got this, I got the x squared, and then if you take three and multiply it by negative three, you are going to get a negative nine. And notice, the middle terms canceled out, and one thing you might ask is, well, will that always be the case if we add a number and then we subtract that same number like that? And we could try it out. Let's talk in general terms."}, {"video_title": "Special products of the form (x+a)(x-a) Algebra I High School Math Khan Academy.mp3", "Sentence": "Notice, I added three and then I subtracted three, and I got this, I got the x squared, and then if you take three and multiply it by negative three, you are going to get a negative nine. And notice, the middle terms canceled out, and one thing you might ask is, well, will that always be the case if we add a number and then we subtract that same number like that? And we could try it out. Let's talk in general terms. So if we, instead of doing x plus three times x minus three, we could write the same thing as, instead of three, let's just say you have x plus, x plus a times x minus a, times x minus a. And I encourage you to pause this video and work it all out. Just assume a is some number, like three or some other number, and apply the distributive property twice and see what you get."}, {"video_title": "Special products of the form (x+a)(x-a) Algebra I High School Math Khan Academy.mp3", "Sentence": "Let's talk in general terms. So if we, instead of doing x plus three times x minus three, we could write the same thing as, instead of three, let's just say you have x plus, x plus a times x minus a, times x minus a. And I encourage you to pause this video and work it all out. Just assume a is some number, like three or some other number, and apply the distributive property twice and see what you get. Well, let's work through it. So first we can distribute this yellow x plus a onto the x and the negative a. So x plus a times x, or we could say x times x plus a."}, {"video_title": "Special products of the form (x+a)(x-a) Algebra I High School Math Khan Academy.mp3", "Sentence": "Just assume a is some number, like three or some other number, and apply the distributive property twice and see what you get. Well, let's work through it. So first we can distribute this yellow x plus a onto the x and the negative a. So x plus a times x, or we could say x times x plus a. So that's going to be, that's going to be x times x plus a, and then we're going to have minus a, or this negative a times x plus a. So minus, and then we're going to have this minus a times x plus a, times x plus a, times x plus a. Notice, all I did is I distributed this yellow, I just distributed this big chunk of this expression, I just distributed it onto the x and onto this negative a. I'm multiplying it times the x and I'm multiplying it by the negative a."}, {"video_title": "Special products of the form (x+a)(x-a) Algebra I High School Math Khan Academy.mp3", "Sentence": "So x plus a times x, or we could say x times x plus a. So that's going to be, that's going to be x times x plus a, and then we're going to have minus a, or this negative a times x plus a. So minus, and then we're going to have this minus a times x plus a, times x plus a, times x plus a. Notice, all I did is I distributed this yellow, I just distributed this big chunk of this expression, I just distributed it onto the x and onto this negative a. I'm multiplying it times the x and I'm multiplying it by the negative a. And now we can apply the distributive property again. X times x is x squared. X times a is ax."}, {"video_title": "Special products of the form (x+a)(x-a) Algebra I High School Math Khan Academy.mp3", "Sentence": "Notice, all I did is I distributed this yellow, I just distributed this big chunk of this expression, I just distributed it onto the x and onto this negative a. I'm multiplying it times the x and I'm multiplying it by the negative a. And now we can apply the distributive property again. X times x is x squared. X times a is ax. And then we get negative a times x is negative ax. And then negative a times a is negative a squared. And notice, regardless of my choice of a, I'm going to have ax and then minus ax."}, {"video_title": "Special products of the form (x+a)(x-a) Algebra I High School Math Khan Academy.mp3", "Sentence": "X times a is ax. And then we get negative a times x is negative ax. And then negative a times a is negative a squared. And notice, regardless of my choice of a, I'm going to have ax and then minus ax. So this is always going to cancel out. It didn't just work for the case when a was three. For any a, if I have a times x and then I subtract a times x, that's just going to cancel out."}, {"video_title": "Special products of the form (x+a)(x-a) Algebra I High School Math Khan Academy.mp3", "Sentence": "And notice, regardless of my choice of a, I'm going to have ax and then minus ax. So this is always going to cancel out. It didn't just work for the case when a was three. For any a, if I have a times x and then I subtract a times x, that's just going to cancel out. So this is just going to cancel out. And what are we going to be left with? We are going to be left with x squared minus a squared."}, {"video_title": "Special products of the form (x+a)(x-a) Algebra I High School Math Khan Academy.mp3", "Sentence": "For any a, if I have a times x and then I subtract a times x, that's just going to cancel out. So this is just going to cancel out. And what are we going to be left with? We are going to be left with x squared minus a squared. X squared minus a squared. And you could view this as a special case. When you have something x plus something times x minus that same something, it's going to be x squared minus that something squared."}, {"video_title": "Special products of the form (x+a)(x-a) Algebra I High School Math Khan Academy.mp3", "Sentence": "We are going to be left with x squared minus a squared. X squared minus a squared. And you could view this as a special case. When you have something x plus something times x minus that same something, it's going to be x squared minus that something squared. And this is a good one to know in general. This is a good one to know in general. And we could use it to quickly figure out the products of other binomials that fit this pattern here."}, {"video_title": "Special products of the form (x+a)(x-a) Algebra I High School Math Khan Academy.mp3", "Sentence": "When you have something x plus something times x minus that same something, it's going to be x squared minus that something squared. And this is a good one to know in general. This is a good one to know in general. And we could use it to quickly figure out the products of other binomials that fit this pattern here. So if I were to say, quick, what is x plus 10 times x minus 10? Well, you could say, all right, this fits the pattern. It's x plus a times x minus a."}, {"video_title": "Special products of the form (x+a)(x-a) Algebra I High School Math Khan Academy.mp3", "Sentence": "And we could use it to quickly figure out the products of other binomials that fit this pattern here. So if I were to say, quick, what is x plus 10 times x minus 10? Well, you could say, all right, this fits the pattern. It's x plus a times x minus a. So it's going to be x squared minus a squared. If a is 10, a squared is going to be 100. So you can do it really quick once you recognize the pattern."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "If you don't believe that one of these properties are true and you want them proved, I've made three or four videos that actually prove these properties. But what I'm going to do is I'm going to show you the properties and then show you how they can be used. It's going to be a little more hands-on. So let's just do a little bit of a review of just what a logarithm is. So if I say that a, oh, that's not the right, let's see, I want to change, there you go. Let's say I say that a, let me start over, a to the b is equal to c. a to the b to the power is equal to c. So another way to write this exact same relationship, instead of writing an exponent, is to write it as a logarithm. So we could say that the logarithm base a of c is equal to b."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "So let's just do a little bit of a review of just what a logarithm is. So if I say that a, oh, that's not the right, let's see, I want to change, there you go. Let's say I say that a, let me start over, a to the b is equal to c. a to the b to the power is equal to c. So another way to write this exact same relationship, instead of writing an exponent, is to write it as a logarithm. So we could say that the logarithm base a of c is equal to b. So these are essentially saying the same thing, they just have different kind of results. In one, you know a and b and you're kind of getting c. That's what exponentiation does for you. And the second one, you know a and you know that when you raise it to some power, you get c. And then you figure out what b is."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "So we could say that the logarithm base a of c is equal to b. So these are essentially saying the same thing, they just have different kind of results. In one, you know a and b and you're kind of getting c. That's what exponentiation does for you. And the second one, you know a and you know that when you raise it to some power, you get c. And then you figure out what b is. So the exact same relationship, just dated in a different way. Now I will introduce you to some interesting logarithm properties. And they actually just fall out of this relationship and the regular exponent rules."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "And the second one, you know a and you know that when you raise it to some power, you get c. And then you figure out what b is. So the exact same relationship, just dated in a different way. Now I will introduce you to some interesting logarithm properties. And they actually just fall out of this relationship and the regular exponent rules. So the first is that the logarithm, let me do a more cheerful color. The logarithm, let's say, of any base, so let's just call the base, let's say b for base, logarithm base b of a plus logarithm base b of c. And this only works if we have the same basis, so that's important to remember. That equals the logarithm of base b of a times c. Now what does this mean and how can we use it?"}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "And they actually just fall out of this relationship and the regular exponent rules. So the first is that the logarithm, let me do a more cheerful color. The logarithm, let's say, of any base, so let's just call the base, let's say b for base, logarithm base b of a plus logarithm base b of c. And this only works if we have the same basis, so that's important to remember. That equals the logarithm of base b of a times c. Now what does this mean and how can we use it? Or let's just even try it out with some, I don't know, examples. So this is saying that, I'll switch to another color. Let's make mauve my mauve."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "That equals the logarithm of base b of a times c. Now what does this mean and how can we use it? Or let's just even try it out with some, I don't know, examples. So this is saying that, I'll switch to another color. Let's make mauve my mauve. I don't know, I never know how to say that properly. Let's make that my example color. So let's say logarithm of base 2 of, I don't know, of 8 plus logarithm base 2 of, I don't know, let's say 32."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "Let's make mauve my mauve. I don't know, I never know how to say that properly. Let's make that my example color. So let's say logarithm of base 2 of, I don't know, of 8 plus logarithm base 2 of, I don't know, let's say 32. So in theory, this should equal, if we believe this property, this should equal logarithm base 2 of what? Well we say 8 times 32. So 8 times 32 is 240 plus 16, 256."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "So let's say logarithm of base 2 of, I don't know, of 8 plus logarithm base 2 of, I don't know, let's say 32. So in theory, this should equal, if we believe this property, this should equal logarithm base 2 of what? Well we say 8 times 32. So 8 times 32 is 240 plus 16, 256. Let's see if that's true just trying out this number. And this really isn't a proof, but it'll give you a little bit of an intuition, I think, for what's going on around here. So we just used our property, this little property that I presented to you, and let's see if it works out."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "So 8 times 32 is 240 plus 16, 256. Let's see if that's true just trying out this number. And this really isn't a proof, but it'll give you a little bit of an intuition, I think, for what's going on around here. So we just used our property, this little property that I presented to you, and let's see if it works out. So log base 2 of 8. 2 to what power is equal to 8? Well 2 to the third power is equal to 8, right?"}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "So we just used our property, this little property that I presented to you, and let's see if it works out. So log base 2 of 8. 2 to what power is equal to 8? Well 2 to the third power is equal to 8, right? 2 to the third power is equal to 8. So this term right here, that equals 3, right? Log base 2 of 8 is equal to 3."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well 2 to the third power is equal to 8, right? 2 to the third power is equal to 8. So this term right here, that equals 3, right? Log base 2 of 8 is equal to 3. 2 to what power is equal to 32? Let's see, 2 to the fourth power is 16, 2 to the fifth power is 32. So this right here is 5, right?"}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "Log base 2 of 8 is equal to 3. 2 to what power is equal to 32? Let's see, 2 to the fourth power is 16, 2 to the fifth power is 32. So this right here is 5, right? And 2 to the what power is equal to 256? Well, let's see. Well if you're a computer science major, you'll know that immediately."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "So this right here is 5, right? And 2 to the what power is equal to 256? Well, let's see. Well if you're a computer science major, you'll know that immediately. That a byte can have 256 values in it. So it's 2 to the eighth power. But if you don't know that, you could multiply it out yourself."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well if you're a computer science major, you'll know that immediately. That a byte can have 256 values in it. So it's 2 to the eighth power. But if you don't know that, you could multiply it out yourself. But this is 8. And I'm not doing it just because I knew that 3 plus 5 is equal to 8. I'm doing this independently."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "But if you don't know that, you could multiply it out yourself. But this is 8. And I'm not doing it just because I knew that 3 plus 5 is equal to 8. I'm doing this independently. So this is equal to 8. But it does turn out that 3 plus 5 is equal to 8. This may seem like magic to you, or it may seem obvious."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "I'm doing this independently. So this is equal to 8. But it does turn out that 3 plus 5 is equal to 8. This may seem like magic to you, or it may seem obvious. And for those of you who it might seem a little obvious, you're probably thinking, well, 2 to the third times 2 to the fifth is equal to 2 to the 3 plus 5, right? This is just an exponent rule. What do they call this?"}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "This may seem like magic to you, or it may seem obvious. And for those of you who it might seem a little obvious, you're probably thinking, well, 2 to the third times 2 to the fifth is equal to 2 to the 3 plus 5, right? This is just an exponent rule. What do they call this? The additive exponent? I don't know. I don't know the names of things."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "What do they call this? The additive exponent? I don't know. I don't know the names of things. And that equals 2 to the eighth. And that's exactly what we did here, right? On this side, we had 2 to the third times 2 to the fifth, essentially."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "I don't know the names of things. And that equals 2 to the eighth. And that's exactly what we did here, right? On this side, we had 2 to the third times 2 to the fifth, essentially. And on this side, you have them added to each other. And what makes logarithms interesting is, and why it's a little confusing at first, and you can watch the proofs if you really want a kind of a rigorous, not even my proofs aren't rigorous, but if you want kind of a better explanation of how this works. But this should hopefully give you an intuition for why this property holds, right?"}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "On this side, we had 2 to the third times 2 to the fifth, essentially. And on this side, you have them added to each other. And what makes logarithms interesting is, and why it's a little confusing at first, and you can watch the proofs if you really want a kind of a rigorous, not even my proofs aren't rigorous, but if you want kind of a better explanation of how this works. But this should hopefully give you an intuition for why this property holds, right? Because when you multiply two numbers of the same base, two exponential expressions of the same base, you can add their exponents. Similarly, when you have the log of two numbers multiplied by each other, that's equivalent to the log of each of the numbers added to each other. This is the same property."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "But this should hopefully give you an intuition for why this property holds, right? Because when you multiply two numbers of the same base, two exponential expressions of the same base, you can add their exponents. Similarly, when you have the log of two numbers multiplied by each other, that's equivalent to the log of each of the numbers added to each other. This is the same property. If you don't believe me, watch the proof videos. So let me show you another log property that's pretty much the same one. I almost view them the same."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "This is the same property. If you don't believe me, watch the proof videos. So let me show you another log property that's pretty much the same one. I almost view them the same. So this is log base b of a minus log base b of c is equal to log base b of a divided by c. That says a divided by c. And we can, once again, try it out with some numbers. I use 2 a lot, just because 2 is an easy number to figure out the powers. But let's use a different number."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "I almost view them the same. So this is log base b of a minus log base b of c is equal to log base b of a divided by c. That says a divided by c. And we can, once again, try it out with some numbers. I use 2 a lot, just because 2 is an easy number to figure out the powers. But let's use a different number. Let's say log base 3 of 1 ninth minus log base 3 of 81. So this property tells us that this is the same thing as, well, I'm ending up with a big number. Log base 3 of 1 ninth divided by 81."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "But let's use a different number. Let's say log base 3 of 1 ninth minus log base 3 of 81. So this property tells us that this is the same thing as, well, I'm ending up with a big number. Log base 3 of 1 ninth divided by 81. So that's the same thing as 1 ninth times 1 over 81. I used two large numbers for my example. But we'll move forward."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "Log base 3 of 1 ninth divided by 81. So that's the same thing as 1 ninth times 1 over 81. I used two large numbers for my example. But we'll move forward. So let's see. 9 times 8 is 720. Right?"}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "But we'll move forward. So let's see. 9 times 8 is 720. Right? 9 times 8 is 720. So this is 1 over 729. So this is log base 3 over 1 over 729."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "Right? 9 times 8 is 720. So this is 1 over 729. So this is log base 3 over 1 over 729. So 3 to what power is equal to 1 ninth? Well, 3 squared is equal to 9, right? 3 squared is equal to 9."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "So this is log base 3 over 1 over 729. So 3 to what power is equal to 1 ninth? Well, 3 squared is equal to 9, right? 3 squared is equal to 9. So we know that if 3 squared is equal to 9, then we know that 3 to the negative 2 is equal to 1 ninth, right? The negative just inverts it. So this is equal to negative 2."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "3 squared is equal to 9. So we know that if 3 squared is equal to 9, then we know that 3 to the negative 2 is equal to 1 ninth, right? The negative just inverts it. So this is equal to negative 2. Right? And then minus 3 to what power is equal to 81? Let's see."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "So this is equal to negative 2. Right? And then minus 3 to what power is equal to 81? Let's see. 3 to the third power is 27. So 3 to the fourth power. So we have minus 2 minus 4 is equal to, well, we could do it a couple of ways."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "Let's see. 3 to the third power is 27. So 3 to the fourth power. So we have minus 2 minus 4 is equal to, well, we could do it a couple of ways. Minus 2 minus 4 is equal to minus 6. And now we just have to confirm that 3 to the minus 6 power is equal to 1 over 729. So my question is, 3 to the minus 6 power, is that equal to 1 over 729?"}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "So we have minus 2 minus 4 is equal to, well, we could do it a couple of ways. Minus 2 minus 4 is equal to minus 6. And now we just have to confirm that 3 to the minus 6 power is equal to 1 over 729. So my question is, 3 to the minus 6 power, is that equal to 1 over 729? Well, that's the same thing as saying 3 to the 6th power is equal to 729, because that's all the negative exponent does, is inverts it. Let's see. We could multiply that out, but that should be the case."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "So my question is, 3 to the minus 6 power, is that equal to 1 over 729? Well, that's the same thing as saying 3 to the 6th power is equal to 729, because that's all the negative exponent does, is inverts it. Let's see. We could multiply that out, but that should be the case. Because, well, we could look here, but let's see. 3 to the third power. This would be 3 to the third power times 3 to the third power is equal to 27 times 27."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "We could multiply that out, but that should be the case. Because, well, we could look here, but let's see. 3 to the third power. This would be 3 to the third power times 3 to the third power is equal to 27 times 27. That looks pretty close. You can confirm it with a calculator if you don't believe me. Anyway, that's all the time I have in this video."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "This would be 3 to the third power times 3 to the third power is equal to 27 times 27. That looks pretty close. You can confirm it with a calculator if you don't believe me. Anyway, that's all the time I have in this video. In the next video, I'll introduce you to the last two logarithm properties. And if we have time, maybe I'll do examples with the leftover time. I'll see you soon."}, {"video_title": "Example 3 Subtracting polynomials Algebra I Khan Academy.mp3", "Sentence": "Or another way to think about it is you can distribute this negative sign along all of those terms. That's essentially what we're about to do here. We're just adding the negative of this entire thing. We're adding the opposite of it. So this first part, I'm not going to change it. That's still just 16x plus 14. But now I'm going to distribute the negative sign here."}, {"video_title": "Example 3 Subtracting polynomials Algebra I Khan Academy.mp3", "Sentence": "We're adding the opposite of it. So this first part, I'm not going to change it. That's still just 16x plus 14. But now I'm going to distribute the negative sign here. So negative 1 times 3x squared is negative 3x squared. Negative 1 times positive x is negative x, because that's positive 1x. Negative 1 times negative 9, remember you have to consider this negative right over there."}, {"video_title": "Example 3 Subtracting polynomials Algebra I Khan Academy.mp3", "Sentence": "But now I'm going to distribute the negative sign here. So negative 1 times 3x squared is negative 3x squared. Negative 1 times positive x is negative x, because that's positive 1x. Negative 1 times negative 9, remember you have to consider this negative right over there. That is part of the term. Negative 1 times negative 9 is positive 9. Negative times a negative is a positive."}, {"video_title": "Example 3 Subtracting polynomials Algebra I Khan Academy.mp3", "Sentence": "Negative 1 times negative 9, remember you have to consider this negative right over there. That is part of the term. Negative 1 times negative 9 is positive 9. Negative times a negative is a positive. So then we have positive 9. And now we just have to combine like terms. So what's our highest degree term here?"}, {"video_title": "Example 3 Subtracting polynomials Algebra I Khan Academy.mp3", "Sentence": "Negative times a negative is a positive. So then we have positive 9. And now we just have to combine like terms. So what's our highest degree term here? I like to write it in that order. We have only one x squared term, second degree term. We only have one of those, so let me write it over here."}, {"video_title": "Example 3 Subtracting polynomials Algebra I Khan Academy.mp3", "Sentence": "So what's our highest degree term here? I like to write it in that order. We have only one x squared term, second degree term. We only have one of those, so let me write it over here. Negative 3x squared. And then what do we have in terms of first degree terms? Just an x, x to the first power."}, {"video_title": "Example 3 Subtracting polynomials Algebra I Khan Academy.mp3", "Sentence": "We only have one of those, so let me write it over here. Negative 3x squared. And then what do we have in terms of first degree terms? Just an x, x to the first power. We have a 16x, and then from that we're going to subtract an x. Subtract 1x. So 16x minus 1x is 15x. If you have 16 of something and you subtract one of them away, you're going to have 15 of that something."}, {"video_title": "Example 3 Subtracting polynomials Algebra I Khan Academy.mp3", "Sentence": "Just an x, x to the first power. We have a 16x, and then from that we're going to subtract an x. Subtract 1x. So 16x minus 1x is 15x. If you have 16 of something and you subtract one of them away, you're going to have 15 of that something. And then finally, you have 14. You can view that as 14 times x to the 0, or just 14. 14 plus 9, they're both constant terms, or they're both being multiplied by x to the 0."}, {"video_title": "Example 3 Subtracting polynomials Algebra I Khan Academy.mp3", "Sentence": "If you have 16 of something and you subtract one of them away, you're going to have 15 of that something. And then finally, you have 14. You can view that as 14 times x to the 0, or just 14. 14 plus 9, they're both constant terms, or they're both being multiplied by x to the 0. 14 plus 9 is 23. And we're done. Negative 3x squared plus 15x plus 23."}, {"video_title": "Graph from slope-intercept equation example Algebra I Khan Academy.mp3", "Sentence": "And the general way of writing it is y is equal to mx plus b. Where m is the slope, and here in this case m is equal to 1 3rd, so let me write that down. And b is the y intercept. So in this case b is equal to negative 2. And you know that b is the y intercept because we know that the y intercept occurs when x is equal to 0. So if x is equal to 0 in either of these situations, this term just becomes 0 and y will be equal to b. So that's what we mean by b is the y intercept."}, {"video_title": "Graph from slope-intercept equation example Algebra I Khan Academy.mp3", "Sentence": "So in this case b is equal to negative 2. And you know that b is the y intercept because we know that the y intercept occurs when x is equal to 0. So if x is equal to 0 in either of these situations, this term just becomes 0 and y will be equal to b. So that's what we mean by b is the y intercept. So whenever you look at an equation in this form, it's actually fairly straightforward to graph this line. b is the y intercept, in this case it is negative 2. So that means that this line must intersect the y axis at y is equal to negative 2."}, {"video_title": "Graph from slope-intercept equation example Algebra I Khan Academy.mp3", "Sentence": "So that's what we mean by b is the y intercept. So whenever you look at an equation in this form, it's actually fairly straightforward to graph this line. b is the y intercept, in this case it is negative 2. So that means that this line must intersect the y axis at y is equal to negative 2. So it's this point right here. Negative 1, negative 2. This is the point 0, negative 2."}, {"video_title": "Graph from slope-intercept equation example Algebra I Khan Academy.mp3", "Sentence": "So that means that this line must intersect the y axis at y is equal to negative 2. So it's this point right here. Negative 1, negative 2. This is the point 0, negative 2. If you don't believe me, there's nothing magical about this. Try evaluating or try solving for y when x is equal to 0. When x is equal to 0, this term cancels out and you're just left with y is equal to negative 2."}, {"video_title": "Graph from slope-intercept equation example Algebra I Khan Academy.mp3", "Sentence": "This is the point 0, negative 2. If you don't believe me, there's nothing magical about this. Try evaluating or try solving for y when x is equal to 0. When x is equal to 0, this term cancels out and you're just left with y is equal to negative 2. So that's the y intercept right there. Now, this 1 3rd tells us the slope of the line. How much do we change in y for any change in x?"}, {"video_title": "Graph from slope-intercept equation example Algebra I Khan Academy.mp3", "Sentence": "When x is equal to 0, this term cancels out and you're just left with y is equal to negative 2. So that's the y intercept right there. Now, this 1 3rd tells us the slope of the line. How much do we change in y for any change in x? So this tells us that 1 3rd, so that right there is the slope. So it tells us that 1 3rd is equal to the change in y over the change in x. Or another way to think about it, if x changes by 3, then y will change by 1."}, {"video_title": "Graph from slope-intercept equation example Algebra I Khan Academy.mp3", "Sentence": "How much do we change in y for any change in x? So this tells us that 1 3rd, so that right there is the slope. So it tells us that 1 3rd is equal to the change in y over the change in x. Or another way to think about it, if x changes by 3, then y will change by 1. So let me graph that. So we know that this point is on the graph. That's the y intercept."}, {"video_title": "Graph from slope-intercept equation example Algebra I Khan Academy.mp3", "Sentence": "Or another way to think about it, if x changes by 3, then y will change by 1. So let me graph that. So we know that this point is on the graph. That's the y intercept. The slope tells us that if x changes by 3, so let me go 3 to the right, 1, 2, 3, that y will change by 1. So this must also be a point on the graph. And we could keep doing that."}, {"video_title": "Graph from slope-intercept equation example Algebra I Khan Academy.mp3", "Sentence": "That's the y intercept. The slope tells us that if x changes by 3, so let me go 3 to the right, 1, 2, 3, that y will change by 1. So this must also be a point on the graph. And we could keep doing that. If x changes by 3, y changes by 1. If x goes down by 3, y will go down by 1. If x goes down by 6, y will go down by 2."}, {"video_title": "Graph from slope-intercept equation example Algebra I Khan Academy.mp3", "Sentence": "And we could keep doing that. If x changes by 3, y changes by 1. If x goes down by 3, y will go down by 1. If x goes down by 6, y will go down by 2. It's that same ratio. So 1, 2, 3, 4, 5, 6, 1, 2. And you can see all of these points are on the line."}, {"video_title": "Graph from slope-intercept equation example Algebra I Khan Academy.mp3", "Sentence": "If x goes down by 6, y will go down by 2. It's that same ratio. So 1, 2, 3, 4, 5, 6, 1, 2. And you can see all of these points are on the line. And the line is the graph of this equation up here. So let me graph it. So it will look something like that."}, {"video_title": "Slope-intercept form Algebra I Khan Academy.mp3", "Sentence": "So for example, if you had the linear equation y is equal to two x plus three, that's one way to represent it, but I could represent this in an infinite number of ways. I could, let's see, I could subtract two x from both sides. I could write this as negative two x plus y is equal to three. I could manipulate it in ways where I get it to, and I'm not gonna do it right now, but this is another way of writing that same thing. Y minus five is equal to two times x minus one. You could actually simplify this and you could get either this equation here or that equation up on top. These are all equivalent."}, {"video_title": "Slope-intercept form Algebra I Khan Academy.mp3", "Sentence": "I could manipulate it in ways where I get it to, and I'm not gonna do it right now, but this is another way of writing that same thing. Y minus five is equal to two times x minus one. You could actually simplify this and you could get either this equation here or that equation up on top. These are all equivalent. You can get from one to the other with logical algebraic operations. So there's an infinite number of ways to represent a given linear equation, but what I wanna focus on in this video is this representation in particular because this one is a very useful representation of a linear equation, and we'll see in future videos this one and this one can also be useful depending on what you are looking for, but we're gonna focus on this one. And this one right over here, it's often called slope-intercept form."}, {"video_title": "Slope-intercept form Algebra I Khan Academy.mp3", "Sentence": "These are all equivalent. You can get from one to the other with logical algebraic operations. So there's an infinite number of ways to represent a given linear equation, but what I wanna focus on in this video is this representation in particular because this one is a very useful representation of a linear equation, and we'll see in future videos this one and this one can also be useful depending on what you are looking for, but we're gonna focus on this one. And this one right over here, it's often called slope-intercept form. Slope-intercept form. And hopefully in a few minutes it will be obvious why it is called slope-intercept form. And before I explain that to you, let's just try to graph this thing."}, {"video_title": "Slope-intercept form Algebra I Khan Academy.mp3", "Sentence": "And this one right over here, it's often called slope-intercept form. Slope-intercept form. And hopefully in a few minutes it will be obvious why it is called slope-intercept form. And before I explain that to you, let's just try to graph this thing. I'm gonna try to graph it. I'm just gonna plot some points here. So x comma y, and I'm gonna pick some x values where it's easy to calculate the y values."}, {"video_title": "Slope-intercept form Algebra I Khan Academy.mp3", "Sentence": "And before I explain that to you, let's just try to graph this thing. I'm gonna try to graph it. I'm just gonna plot some points here. So x comma y, and I'm gonna pick some x values where it's easy to calculate the y values. So maybe the easiest is if x is equal to zero. If x is equal to zero, then two times zero is zero. That term goes away, and you're only left with this term right over here, y is equal to three."}, {"video_title": "Slope-intercept form Algebra I Khan Academy.mp3", "Sentence": "So x comma y, and I'm gonna pick some x values where it's easy to calculate the y values. So maybe the easiest is if x is equal to zero. If x is equal to zero, then two times zero is zero. That term goes away, and you're only left with this term right over here, y is equal to three. Y is equal to three. And so if we were to plot this, actually let me start plotting it. So that is my y-axis."}, {"video_title": "Slope-intercept form Algebra I Khan Academy.mp3", "Sentence": "That term goes away, and you're only left with this term right over here, y is equal to three. Y is equal to three. And so if we were to plot this, actually let me start plotting it. So that is my y-axis. And let me do the x-axis. So that can be my x, oh that's not as straight as I would like it. So that looks pretty good."}, {"video_title": "Slope-intercept form Algebra I Khan Academy.mp3", "Sentence": "So that is my y-axis. And let me do the x-axis. So that can be my x, oh that's not as straight as I would like it. So that looks pretty good. All right, that is my x-axis. And let me mark off some hash marks here. So this is x equals one, x equals two, x equals three, this is y equals, let me do this, y equals one, y equals two, y equals three, and obviously I can keep going, I can keep going."}, {"video_title": "Slope-intercept form Algebra I Khan Academy.mp3", "Sentence": "So that looks pretty good. All right, that is my x-axis. And let me mark off some hash marks here. So this is x equals one, x equals two, x equals three, this is y equals, let me do this, y equals one, y equals two, y equals three, and obviously I can keep going, I can keep going. This would be y is equal to negative one. This would be x is equal to negative one, negative two, negative three, so on and so forth. So this point right over here, zero comma three, this is x is zero, y is three."}, {"video_title": "Slope-intercept form Algebra I Khan Academy.mp3", "Sentence": "So this is x equals one, x equals two, x equals three, this is y equals, let me do this, y equals one, y equals two, y equals three, and obviously I can keep going, I can keep going. This would be y is equal to negative one. This would be x is equal to negative one, negative two, negative three, so on and so forth. So this point right over here, zero comma three, this is x is zero, y is three. Well, the point that represents when x is equal to zero and y equals three, this is, we're right on the y-axis. If there of a line going through it and this line contains this point, this is going to be the y-intercept. So one way to think about it, the reason why this is called slope-intercept form, is it's very easy to calculate the y-intercept."}, {"video_title": "Slope-intercept form Algebra I Khan Academy.mp3", "Sentence": "So this point right over here, zero comma three, this is x is zero, y is three. Well, the point that represents when x is equal to zero and y equals three, this is, we're right on the y-axis. If there of a line going through it and this line contains this point, this is going to be the y-intercept. So one way to think about it, the reason why this is called slope-intercept form, is it's very easy to calculate the y-intercept. The y-intercept here is going to happen when it's written in this form, it's going to happen when x is equal to zero and y is equal to three. It's going to be this point right over here. So it's very easy to figure out the intercept, the y-intercept from this form."}, {"video_title": "Slope-intercept form Algebra I Khan Academy.mp3", "Sentence": "So one way to think about it, the reason why this is called slope-intercept form, is it's very easy to calculate the y-intercept. The y-intercept here is going to happen when it's written in this form, it's going to happen when x is equal to zero and y is equal to three. It's going to be this point right over here. So it's very easy to figure out the intercept, the y-intercept from this form. Now you might be saying, oh, well it's a slope-intercept form, it must also be easy to figure out the slope from this form. And if you made that conclusion, you would be correct, and we're about to see that in a few seconds. So let's plot some more points here, and I'm just going to keep increasing x by one."}, {"video_title": "Slope-intercept form Algebra I Khan Academy.mp3", "Sentence": "So it's very easy to figure out the intercept, the y-intercept from this form. Now you might be saying, oh, well it's a slope-intercept form, it must also be easy to figure out the slope from this form. And if you made that conclusion, you would be correct, and we're about to see that in a few seconds. So let's plot some more points here, and I'm just going to keep increasing x by one. So if you increase x by one, so we could write that our delta x, our change in x, delta Greek letter, this triangle's Greek letter delta represents change in, change in x here is one. We just increased x by one. What's going to be our corresponding change in y?"}, {"video_title": "Slope-intercept form Algebra I Khan Academy.mp3", "Sentence": "So let's plot some more points here, and I'm just going to keep increasing x by one. So if you increase x by one, so we could write that our delta x, our change in x, delta Greek letter, this triangle's Greek letter delta represents change in, change in x here is one. We just increased x by one. What's going to be our corresponding change in y? What's going to be our change in y? So let's see, when x is equal to one, you have two times one plus three is going to be five. So our change in y is going to be two."}, {"video_title": "Slope-intercept form Algebra I Khan Academy.mp3", "Sentence": "What's going to be our corresponding change in y? What's going to be our change in y? So let's see, when x is equal to one, you have two times one plus three is going to be five. So our change in y is going to be two. Let's do that again. Let's increase our x by one, change in x is equal to one. So then if we go from, if we're going to increase by one, we're going to go from x equals one to x equals two, what's our corresponding change in y?"}, {"video_title": "Slope-intercept form Algebra I Khan Academy.mp3", "Sentence": "So our change in y is going to be two. Let's do that again. Let's increase our x by one, change in x is equal to one. So then if we go from, if we're going to increase by one, we're going to go from x equals one to x equals two, what's our corresponding change in y? Well, when x is equal to two, two times two is four plus three is seven. Well, our change in y, our change in y is equal to two. We went from five, when x went from one to two, y went from five to seven."}, {"video_title": "Slope-intercept form Algebra I Khan Academy.mp3", "Sentence": "So then if we go from, if we're going to increase by one, we're going to go from x equals one to x equals two, what's our corresponding change in y? Well, when x is equal to two, two times two is four plus three is seven. Well, our change in y, our change in y is equal to two. We went from five, when x went from one to two, y went from five to seven. So for every one that we increase x, y is increasing by two. So for this linear equation, our change in y over change in x is always going to be, our change in y is two when our change in x is one, or it's equal to two. Or we could say that our slope is equal to two."}, {"video_title": "Slope-intercept form Algebra I Khan Academy.mp3", "Sentence": "We went from five, when x went from one to two, y went from five to seven. So for every one that we increase x, y is increasing by two. So for this linear equation, our change in y over change in x is always going to be, our change in y is two when our change in x is one, or it's equal to two. Or we could say that our slope is equal to two. And let's just graph this to make sure that we understand this. So when x equals one, y is equal to five. And actually we're going to have to graph five up here."}, {"video_title": "Slope-intercept form Algebra I Khan Academy.mp3", "Sentence": "Or we could say that our slope is equal to two. And let's just graph this to make sure that we understand this. So when x equals one, y is equal to five. And actually we're going to have to graph five up here. So when x is equal to one, y is equal to, and actually this is a little bit higher. Let me clean this up a little bit. So this one, let me erase that a little bit."}, {"video_title": "Slope-intercept form Algebra I Khan Academy.mp3", "Sentence": "And actually we're going to have to graph five up here. So when x is equal to one, y is equal to, and actually this is a little bit higher. Let me clean this up a little bit. So this one, let me erase that a little bit. Just like that. So that's y is equal to four, and this is y is equal to five. So when x is one, y is equal to five."}, {"video_title": "Slope-intercept form Algebra I Khan Academy.mp3", "Sentence": "So this one, let me erase that a little bit. Just like that. So that's y is equal to four, and this is y is equal to five. So when x is one, y is equal to five. So it's that point right over there. So our line is going to look, you only need two points to define a line. Our line is going to look like, let me do this in this color right over here."}, {"video_title": "Slope-intercept form Algebra I Khan Academy.mp3", "Sentence": "So when x is one, y is equal to five. So it's that point right over there. So our line is going to look, you only need two points to define a line. Our line is going to look like, let me do this in this color right over here. Our line is going to look like, is going to look, is going to look something like, is going to look, let me see if I can, I didn't draw it completely at scale, but it's going to look something like this. This is the line, this is the line, y is equal to two x plus three. And we already figured out that its slope is equal to two."}, {"video_title": "Slope-intercept form Algebra I Khan Academy.mp3", "Sentence": "Our line is going to look like, let me do this in this color right over here. Our line is going to look like, is going to look, is going to look something like, is going to look, let me see if I can, I didn't draw it completely at scale, but it's going to look something like this. This is the line, this is the line, y is equal to two x plus three. And we already figured out that its slope is equal to two. Our change, when our change in x is one, when our change in x is one, our change in y is two. If our change in x was negative one, if our change in x was negative one, our change in y is negative two. And you could see that."}, {"video_title": "Slope-intercept form Algebra I Khan Academy.mp3", "Sentence": "And we already figured out that its slope is equal to two. Our change, when our change in x is one, when our change in x is one, our change in y is two. If our change in x was negative one, if our change in x was negative one, our change in y is negative two. And you could see that. If from zero we went down one, if we went to negative one, then what's our y going to be? Two times negative one is negative two, plus three is one. So we see that the point one, or the point negative one comma one is on the line as well."}, {"video_title": "Slope-intercept form Algebra I Khan Academy.mp3", "Sentence": "And you could see that. If from zero we went down one, if we went to negative one, then what's our y going to be? Two times negative one is negative two, plus three is one. So we see that the point one, or the point negative one comma one is on the line as well. So the slope here, our change in y or change in x, if we're going from, between any two points on this line, is always going to be two. But where did you see two in this original equation? Well, you see the two right over here."}, {"video_title": "Slope-intercept form Algebra I Khan Academy.mp3", "Sentence": "So we see that the point one, or the point negative one comma one is on the line as well. So the slope here, our change in y or change in x, if we're going from, between any two points on this line, is always going to be two. But where did you see two in this original equation? Well, you see the two right over here. And when you write something in slope intercept form, where you explicitly solve for y, y is equal to some constant times x to the first power, plus some other constant, the second one is going to be your intercept, your y, or it's going to be a way to figure out the y intercept. The intercept itself is this point, the point at which the line intersects the y axis. And then this two is going to represent your slope."}, {"video_title": "Slope-intercept form Algebra I Khan Academy.mp3", "Sentence": "Well, you see the two right over here. And when you write something in slope intercept form, where you explicitly solve for y, y is equal to some constant times x to the first power, plus some other constant, the second one is going to be your intercept, your y, or it's going to be a way to figure out the y intercept. The intercept itself is this point, the point at which the line intersects the y axis. And then this two is going to represent your slope. And that makes sense, because every time you increase x by one, you're going to multiply that by two, so you're going to increase y by two. So this is just a kind of a, get your feet wet with the idea of slope intercept form, but you'll see, at least for me, this is the easiest form for me to think about what the graph of something looks like. Because if you were given another linear equation, let's say y is equal to negative x, negative x plus two."}, {"video_title": "Slope-intercept form Algebra I Khan Academy.mp3", "Sentence": "And then this two is going to represent your slope. And that makes sense, because every time you increase x by one, you're going to multiply that by two, so you're going to increase y by two. So this is just a kind of a, get your feet wet with the idea of slope intercept form, but you'll see, at least for me, this is the easiest form for me to think about what the graph of something looks like. Because if you were given another linear equation, let's say y is equal to negative x, negative x plus two. Well, immediately you say, okay, look, my y intercept is going to be the point zero comma two, so I'm going to intersect the y axis right at that point. And then I have a slope of, the coefficient here is really just negative one. So I have a slope of negative one."}, {"video_title": "Slope-intercept form Algebra I Khan Academy.mp3", "Sentence": "Because if you were given another linear equation, let's say y is equal to negative x, negative x plus two. Well, immediately you say, okay, look, my y intercept is going to be the point zero comma two, so I'm going to intersect the y axis right at that point. And then I have a slope of, the coefficient here is really just negative one. So I have a slope of negative one. So as we increase x by one, we're going to decrease y by one. Increase x by one, you're going to decrease y by one. If you increase x by two, you're going to decrease y by two."}, {"video_title": "Slope-intercept form Algebra I Khan Academy.mp3", "Sentence": "So I have a slope of negative one. So as we increase x by one, we're going to decrease y by one. Increase x by one, you're going to decrease y by one. If you increase x by two, you're going to decrease y by two. And so our line is going to look something like this. Let me see if I can draw it relatively neatly. It's going to look something, it's, let me, I can do it a little bit better than that."}, {"video_title": "Slope-intercept form Algebra I Khan Academy.mp3", "Sentence": "If you increase x by two, you're going to decrease y by two. And so our line is going to look something like this. Let me see if I can draw it relatively neatly. It's going to look something, it's, let me, I can do it a little bit better than that. It's because my graph paper is hand-drawn. It's not ideal. But I think you get, you get the point."}, {"video_title": "Slope-intercept form Algebra I Khan Academy.mp3", "Sentence": "It's going to look something, it's, let me, I can do it a little bit better than that. It's because my graph paper is hand-drawn. It's not ideal. But I think you get, you get the point. It's going to look something like that. So from slope-intercept form, very easy to figure out what the y intercept is, and very easy to figure out the slope. The slope here, slope here is negative one."}, {"video_title": "Slope-intercept form Algebra I Khan Academy.mp3", "Sentence": "But I think you get, you get the point. It's going to look something like that. So from slope-intercept form, very easy to figure out what the y intercept is, and very easy to figure out the slope. The slope here, slope here is negative one. That's this negative one right over here. And the y intercept, y intercept is the point zero comma two. Very easy to figure out, because essentially that gave you the information right there."}, {"video_title": "How to match function input to output given the graph (example) Algebra I Khan Academy.mp3", "Sentence": "So what they do over here is along the x axis, these are the inputs, and then the graph shows us what's the output. So when x is equal to seven, g of seven we see here is one. If x equals nine, g of nine here is two. If x equals six, g of six is equal to the y coordinate at this point, is equal to zero. So what is the input value for which g of x is equal to negative two? Well, this graph right over here, this is y equals g of x. So g of x equaling negative two means y is equal to negative two."}, {"video_title": "How to match function input to output given the graph (example) Algebra I Khan Academy.mp3", "Sentence": "If x equals six, g of six is equal to the y coordinate at this point, is equal to zero. So what is the input value for which g of x is equal to negative two? Well, this graph right over here, this is y equals g of x. So g of x equaling negative two means y is equal to negative two. And so when does y equal negative two? Well, when does y equal negative two? It looks like that happens right at this point."}, {"video_title": "How to match function input to output given the graph (example) Algebra I Khan Academy.mp3", "Sentence": "So g of x equaling negative two means y is equal to negative two. And so when does y equal negative two? Well, when does y equal negative two? It looks like that happens right at this point. And that happens when you input negative nine into g. G of negative nine is negative two. So this is going to be negative nine. And we're done."}, {"video_title": "Equations with rational expressions (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "All right, let's work through it together. Now, when I see things in the denominator like this, my instinct is to try to not have denominators like this. And so what we could do is, to get rid of this x minus one in the denominator on the left-hand side, we can multiply both sides of the equation times x minus one. x minus one. So we're gonna multiply both sides by x minus one. And once again, the whole point of doing that is so that we get rid of this x minus one in the denominator right over here. And then, to get rid of this x plus one in the denominator over here, we can multiply both sides of the equation times x plus one."}, {"video_title": "Equations with rational expressions (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "x minus one. So we're gonna multiply both sides by x minus one. And once again, the whole point of doing that is so that we get rid of this x minus one in the denominator right over here. And then, to get rid of this x plus one in the denominator over here, we can multiply both sides of the equation times x plus one. So, x plus one. Multiply both sides times x plus one. And so, what is that going to give us?"}, {"video_title": "Equations with rational expressions (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "And then, to get rid of this x plus one in the denominator over here, we can multiply both sides of the equation times x plus one. So, x plus one. Multiply both sides times x plus one. And so, what is that going to give us? Well, on the left-hand side, that is going to, x minus one divided by x minus one is just gonna be one for the x's where, for the x's where that's defined, for x not being equal to one. And so, we're gonna have x plus one times negative two x plus four. So let me write that down."}, {"video_title": "Equations with rational expressions (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "And so, what is that going to give us? Well, on the left-hand side, that is going to, x minus one divided by x minus one is just gonna be one for the x's where, for the x's where that's defined, for x not being equal to one. And so, we're gonna have x plus one times negative two x plus four. So let me write that down. So we have x plus, I think I'm gonna need some space, so let me make sure I don't write too big. x plus one times negative two x, negative two x plus four is going to be equal to, now, if we multiply both of these times three over x plus one, the x plus one's going to cancel with the x plus one and we're gonna be left with three times x minus one. So that is going to be three x minus three, three x minus three."}, {"video_title": "Equations with rational expressions (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "So let me write that down. So we have x plus, I think I'm gonna need some space, so let me make sure I don't write too big. x plus one times negative two x, negative two x plus four is going to be equal to, now, if we multiply both of these times three over x plus one, the x plus one's going to cancel with the x plus one and we're gonna be left with three times x minus one. So that is going to be three x minus three, three x minus three. And then, minus, one, times both of these. So, one times x minus one, times x plus one. So minus one times x minus one times x plus one."}, {"video_title": "Equations with rational expressions (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "So that is going to be three x minus three, three x minus three. And then, minus, one, times both of these. So, one times x minus one, times x plus one. So minus one times x minus one times x plus one. All I did is I multiplied, took the x minus one times x plus one, multiplied it times each of these terms. When I multiplied it times this first term, the x plus one and the x plus one canceled, so I just had to multiply it three times x minus one. And then, for the second term, I just multiplied it times both of these."}, {"video_title": "Equations with rational expressions (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "So minus one times x minus one times x plus one. All I did is I multiplied, took the x minus one times x plus one, multiplied it times each of these terms. When I multiplied it times this first term, the x plus one and the x plus one canceled, so I just had to multiply it three times x minus one. And then, for the second term, I just multiplied it times both of these. And now you might recognize this, if you have something x plus one times x minus one, that's going to be x squared minus one. So I could rewrite all of this right over here as being equal to, as being equal to x squared minus one. And once again, that's because this is the same thing as x squared minus one."}, {"video_title": "Equations with rational expressions (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "And then, for the second term, I just multiplied it times both of these. And now you might recognize this, if you have something x plus one times x minus one, that's going to be x squared minus one. So I could rewrite all of this right over here as being equal to, as being equal to x squared minus one. And once again, that's because this is the same thing as x squared minus one. And since I'm subtracting an x squared minus one, actually let me just, I don't wanna do too much on one step, so let's go to the next step. So I could multiply this out. So I could multiply x times negative two x, which would give us negative two x squared, x times four, which is going to give us plus four x, and I could multiply one times negative two x, so I'm gonna subtract two x, and then one times four, which is going to be plus four, and then that is going to be equal to, that is going to be equal to, we have three x minus three, and then we can distribute this negative sign, so we could say minus x squared plus one."}, {"video_title": "Equations with rational expressions (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "And once again, that's because this is the same thing as x squared minus one. And since I'm subtracting an x squared minus one, actually let me just, I don't wanna do too much on one step, so let's go to the next step. So I could multiply this out. So I could multiply x times negative two x, which would give us negative two x squared, x times four, which is going to give us plus four x, and I could multiply one times negative two x, so I'm gonna subtract two x, and then one times four, which is going to be plus four, and then that is going to be equal to, that is going to be equal to, we have three x minus three, and then we can distribute this negative sign, so we could say minus x squared plus one. And over here we can simplify it a little bit. This is going to be, that is four x minus two x is going to be, let me make sure I didn't write it over, yep, four x minus two x, so that would be two x. And so this is simplified to, let's see, well this is, we have a negative three and a one, so those two together are going to be equal to subtracting a two."}, {"video_title": "Equations with rational expressions (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "So I could multiply x times negative two x, which would give us negative two x squared, x times four, which is going to give us plus four x, and I could multiply one times negative two x, so I'm gonna subtract two x, and then one times four, which is going to be plus four, and then that is going to be equal to, that is going to be equal to, we have three x minus three, and then we can distribute this negative sign, so we could say minus x squared plus one. And over here we can simplify it a little bit. This is going to be, that is four x minus two x is going to be, let me make sure I didn't write it over, yep, four x minus two x, so that would be two x. And so this is simplified to, let's see, well this is, we have a negative three and a one, so those two together are going to be equal to subtracting a two. So we can rewrite everything as, we'll do it in a neutral color now, negative two x squared plus two x plus four is equal to negative x squared plus three x, plus three x minus two. Now we can try to get all of this business onto the right-hand side. So let's subtract it from both sides."}, {"video_title": "Equations with rational expressions (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "And so this is simplified to, let's see, well this is, we have a negative three and a one, so those two together are going to be equal to subtracting a two. So we can rewrite everything as, we'll do it in a neutral color now, negative two x squared plus two x plus four is equal to negative x squared plus three x, plus three x minus two. Now we can try to get all of this business onto the right-hand side. So let's subtract it from both sides. So we'll add x squared to both sides, add x squared, that gets rid of this white negative x squared. We subtract three x from both sides, subtract three x from both sides, add two to both sides, add two, and we will be left with, we are going to end up with, let's see, negative two x squared plus x squared is negative x squared. Two x minus three x is negative x."}, {"video_title": "Equations with rational expressions (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "So let's subtract it from both sides. So we'll add x squared to both sides, add x squared, that gets rid of this white negative x squared. We subtract three x from both sides, subtract three x from both sides, add two to both sides, add two, and we will be left with, we are going to end up with, let's see, negative two x squared plus x squared is negative x squared. Two x minus three x is negative x. And then four plus two is six, is going to be equal to, well that's going to cancel with that, that, that is equal to zero. I don't like having this negative on the x squared, so let's multiply both sides times negative one. And so if I do that, so if I just take the negative of both sides, so if I just multiply that times negative one, same thing as taking the negative of both sides, I'm going to get positive x squared plus x minus six is equal to zero."}, {"video_title": "Equations with rational expressions (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "Two x minus three x is negative x. And then four plus two is six, is going to be equal to, well that's going to cancel with that, that, that is equal to zero. I don't like having this negative on the x squared, so let's multiply both sides times negative one. And so if I do that, so if I just take the negative of both sides, so if I just multiply that times negative one, same thing as taking the negative of both sides, I'm going to get positive x squared plus x minus six is equal to zero. And we're making some good progress here. So we can factor this, and actually let me just do it right over here so that we can see the original problem. So if I were to factor this, what two numbers, their product is negative six, they're going to have different signs since their product is negative, and they add up to one, the coefficient on the first degree term."}, {"video_title": "Equations with rational expressions (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "And so if I do that, so if I just take the negative of both sides, so if I just multiply that times negative one, same thing as taking the negative of both sides, I'm going to get positive x squared plus x minus six is equal to zero. And we're making some good progress here. So we can factor this, and actually let me just do it right over here so that we can see the original problem. So if I were to factor this, what two numbers, their product is negative six, they're going to have different signs since their product is negative, and they add up to one, the coefficient on the first degree term. Well, positive three and negative two work, so I can rewrite this as x plus three times x minus two is equal to zero. Did I do that right? Yeah, three times negative two is negative six."}, {"video_title": "Equations with rational expressions (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "So if I were to factor this, what two numbers, their product is negative six, they're going to have different signs since their product is negative, and they add up to one, the coefficient on the first degree term. Well, positive three and negative two work, so I can rewrite this as x plus three times x minus two is equal to zero. Did I do that right? Yeah, three times negative two is negative six. Three x minus two x is positive x. All right, so I just factored, I just wrote this in this quadratic and factored form. And so the way that you get this equaling zero is if either one of those equals zero."}, {"video_title": "Equations with rational expressions (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "Yeah, three times negative two is negative six. Three x minus two x is positive x. All right, so I just factored, I just wrote this in this quadratic and factored form. And so the way that you get this equaling zero is if either one of those equals zero. x plus three equals zero, or x minus two is equal to zero. Well, this is going to happen if you subtract three from both sides, you get, that's going to happen if x is equal to negative three. Or over here, if you add two to both sides, x is equal to two."}, {"video_title": "Equations with rational expressions (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "And so the way that you get this equaling zero is if either one of those equals zero. x plus three equals zero, or x minus two is equal to zero. Well, this is going to happen if you subtract three from both sides, you get, that's going to happen if x is equal to negative three. Or over here, if you add two to both sides, x is equal to two. So either one of these will satisfy, but we want to be careful. We want to make sure that our original equation isn't going to be undefined for either one of these. And negative three does not make either of the denominators equal to zero, so that's cool."}, {"video_title": "Equations with rational expressions (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "Or over here, if you add two to both sides, x is equal to two. So either one of these will satisfy, but we want to be careful. We want to make sure that our original equation isn't going to be undefined for either one of these. And negative three does not make either of the denominators equal to zero, so that's cool. And positive two does not make either of the denominators equal to zero, so it looks like we're in good shape. There's two solutions to that equation. If one of them made any of the denominators equal to zero, then they would have been extraneous solutions."}, {"video_title": "Graphical relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Determine whether the points on this graph represent a function. Now, just as a refresher, a function is really just an association between members of a set that we call the domain and members of a set that we call a range. So if I take any member of the domain, let's call that x, and I give it to the function, the function should tell me what member of my range is that associated with it. So it should point to some other value. This is a function. It would not be a function if it says, well, it could point to y or it could point to z, or maybe it could point to e, or whatever else. This would not be a function because over here, so this right over here, not a function, because it's not clear if you input x what member of the range you're going to get."}, {"video_title": "Graphical relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So it should point to some other value. This is a function. It would not be a function if it says, well, it could point to y or it could point to z, or maybe it could point to e, or whatever else. This would not be a function because over here, so this right over here, not a function, because it's not clear if you input x what member of the range you're going to get. In order for it to be a function, it has to be very clear. For any input into the function, you have to be very clear that you're only going to get one output. Now, with that out of the way, let's think about this function that is defined graphically."}, {"video_title": "Graphical relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "This would not be a function because over here, so this right over here, not a function, because it's not clear if you input x what member of the range you're going to get. In order for it to be a function, it has to be very clear. For any input into the function, you have to be very clear that you're only going to get one output. Now, with that out of the way, let's think about this function that is defined graphically. So the ranges, or I should say the domains, the valid inputs, are the x values where this function is defined. So, for example, it tells us if x is equal to negative 1, if we assume that this over here is the x-axis and this is the y-axis, it tells us when x is equal to negative 1, we should output, or y is going to be equal to 3. So one way to write that mapping is you could say x, when you input it, or let me write it this way, negative 1, if you take negative 1 and you input it into our function, I'll put a little f box right over there, you will get the number 3."}, {"video_title": "Graphical relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Now, with that out of the way, let's think about this function that is defined graphically. So the ranges, or I should say the domains, the valid inputs, are the x values where this function is defined. So, for example, it tells us if x is equal to negative 1, if we assume that this over here is the x-axis and this is the y-axis, it tells us when x is equal to negative 1, we should output, or y is going to be equal to 3. So one way to write that mapping is you could say x, when you input it, or let me write it this way, negative 1, if you take negative 1 and you input it into our function, I'll put a little f box right over there, you will get the number 3. This is our x and this is our y. So that seems reasonable. Negative 1, very clear that you get to 3."}, {"video_title": "Graphical relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So one way to write that mapping is you could say x, when you input it, or let me write it this way, negative 1, if you take negative 1 and you input it into our function, I'll put a little f box right over there, you will get the number 3. This is our x and this is our y. So that seems reasonable. Negative 1, very clear that you get to 3. Let's see what happens when we go over here. If you put 2 into the function, when x is 2, y is negative 2. Once again, when x is 2, the function associates 2 for x, which is a member of the domain, it's defined for 2, it's not defined for 1."}, {"video_title": "Graphical relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Negative 1, very clear that you get to 3. Let's see what happens when we go over here. If you put 2 into the function, when x is 2, y is negative 2. Once again, when x is 2, the function associates 2 for x, which is a member of the domain, it's defined for 2, it's not defined for 1. We don't know what our function is equal to at once, so it's not defined there. So 1 isn't part of the domain, 2 is. It tells us when x is 2, then y is going to be equal to negative 2."}, {"video_title": "Graphical relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Once again, when x is 2, the function associates 2 for x, which is a member of the domain, it's defined for 2, it's not defined for 1. We don't know what our function is equal to at once, so it's not defined there. So 1 isn't part of the domain, 2 is. It tells us when x is 2, then y is going to be equal to negative 2. So it maps it or associates it with negative 2. That doesn't seem too troublesome just yet. Now let's look over here."}, {"video_title": "Graphical relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "It tells us when x is 2, then y is going to be equal to negative 2. So it maps it or associates it with negative 2. That doesn't seem too troublesome just yet. Now let's look over here. Our function is also defined at x is equal to 3. It associates 3, our function associates or maps 3 to the value y is equal to 2. y is equal to 2. That seems pretty straightforward."}, {"video_title": "Graphical relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Now let's look over here. Our function is also defined at x is equal to 3. It associates 3, our function associates or maps 3 to the value y is equal to 2. y is equal to 2. That seems pretty straightforward. Then we get to x is equal to 4, where it seems like this thing that could be a function, it is somewhat defined, it does try to associate 4 with things. What's interesting here is it tries to associate 4 with 2 different things. All of a sudden, in this thing that we think might have been a function, but it looks like it might not be, we don't know, do we associate 4 with 5?"}, {"video_title": "Graphical relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "That seems pretty straightforward. Then we get to x is equal to 4, where it seems like this thing that could be a function, it is somewhat defined, it does try to associate 4 with things. What's interesting here is it tries to associate 4 with 2 different things. All of a sudden, in this thing that we think might have been a function, but it looks like it might not be, we don't know, do we associate 4 with 5? Do we associate it with 5? Or do we associate it with negative 1? This thing right over here is actually a relation."}, {"video_title": "Graphical relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "All of a sudden, in this thing that we think might have been a function, but it looks like it might not be, we don't know, do we associate 4 with 5? Do we associate it with 5? Or do we associate it with negative 1? This thing right over here is actually a relation. You can have one member of the domain being related to multiple members of the range, but if you do have that, then you're not dealing with a function. Once again, because of this, this is not a function. It's not clear that when you input 4 into it, should you output 5 or should you output negative 1?"}, {"video_title": "Graphical relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "This thing right over here is actually a relation. You can have one member of the domain being related to multiple members of the range, but if you do have that, then you're not dealing with a function. Once again, because of this, this is not a function. It's not clear that when you input 4 into it, should you output 5 or should you output negative 1? Sometimes there's something called the vertical line test that tells you whether something is a function. When it's graphically defined like this, you literally say, when x is 4, if I draw a vertical line, do I intersect the function at 2 places or more? It could be 2 or more places."}, {"video_title": "Graphical relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "It's not clear that when you input 4 into it, should you output 5 or should you output negative 1? Sometimes there's something called the vertical line test that tells you whether something is a function. When it's graphically defined like this, you literally say, when x is 4, if I draw a vertical line, do I intersect the function at 2 places or more? It could be 2 or more places. If you do, that means that there's 2 or more values that are related to that value in the domain. There's 2 or more outputs for the input 4. If there are 2 or more outputs for that one input, then you're not dealing with a function."}, {"video_title": "Evaluating functions given their graph Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "The function f of x is graphed. Find f of negative 1. So this graph right over here is essentially a definition of our function. It tells us, given the allowed inputs into our function, what would the function output? So here they're saying, look, what gets output when we input x is equal to negative 1? So x equals negative 1 is right over here. x is equal to negative 1."}, {"video_title": "Evaluating functions given their graph Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "It tells us, given the allowed inputs into our function, what would the function output? So here they're saying, look, what gets output when we input x is equal to negative 1? So x equals negative 1 is right over here. x is equal to negative 1. And our function graph is right at 6 when f is equal to negative 1. So we can say that f of negative 1 is equal to 6. Let me write that over here."}, {"video_title": "Perpendicular lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "It has to be perpendicular to one of the other lines. You can't be just perpendicular by yourself. And perpendicular lines, just so you have a visualization for what perpendicular lines look like. Two lines are perpendicular if they intersect at right angles. So if this is one line right there, a perpendicular line will look like this. A perpendicular line will intersect it, but it won't just be any intersection. It will intersect at right angles."}, {"video_title": "Perpendicular lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "Two lines are perpendicular if they intersect at right angles. So if this is one line right there, a perpendicular line will look like this. A perpendicular line will intersect it, but it won't just be any intersection. It will intersect at right angles. It will intersect at right angles. So these two lines are perpendicular. Now, if two lines are perpendicular, if the slope of this orange line is m, so let's say its equation is y is equal to mx plus, let's say it's b1, so it's some y-intercept, then the equation of this yellow line, its slope is going to be the negative inverse of this guy."}, {"video_title": "Perpendicular lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "It will intersect at right angles. It will intersect at right angles. So these two lines are perpendicular. Now, if two lines are perpendicular, if the slope of this orange line is m, so let's say its equation is y is equal to mx plus, let's say it's b1, so it's some y-intercept, then the equation of this yellow line, its slope is going to be the negative inverse of this guy. This guy right here is going to be y is equal to negative 1 over mx plus some other y-intercept. Or another way to think about it is if two lines are perpendicular, the product of their slopes is going to be negative 1. And so you could write that there."}, {"video_title": "Perpendicular lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "Now, if two lines are perpendicular, if the slope of this orange line is m, so let's say its equation is y is equal to mx plus, let's say it's b1, so it's some y-intercept, then the equation of this yellow line, its slope is going to be the negative inverse of this guy. This guy right here is going to be y is equal to negative 1 over mx plus some other y-intercept. Or another way to think about it is if two lines are perpendicular, the product of their slopes is going to be negative 1. And so you could write that there. m times negative 1 over m. That's going to be, these two guys are going to cancel out, that's going to be equal to negative 1. So let's figure out the slopes of each of these lines and figure out if any of them are the negative inverse of any of the other ones. So line A, the slope is pretty easy to figure out."}, {"video_title": "Perpendicular lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "And so you could write that there. m times negative 1 over m. That's going to be, these two guys are going to cancel out, that's going to be equal to negative 1. So let's figure out the slopes of each of these lines and figure out if any of them are the negative inverse of any of the other ones. So line A, the slope is pretty easy to figure out. It's already in slope-intercept form. Its slope is 3. So line A has a slope of 3."}, {"video_title": "Perpendicular lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "So line A, the slope is pretty easy to figure out. It's already in slope-intercept form. Its slope is 3. So line A has a slope of 3. Line B, it's in standard form, not too hard to put it in slope-intercept form, so let's try to do it. So let's do line B over here. Line B, we have x plus 3y is equal to negative 21."}, {"video_title": "Perpendicular lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "So line A has a slope of 3. Line B, it's in standard form, not too hard to put it in slope-intercept form, so let's try to do it. So let's do line B over here. Line B, we have x plus 3y is equal to negative 21. Let's subtract x from both sides so that it ends up on the right-hand side. So this, we end up with 3y is equal to negative x minus 21. And now let's divide both sides of this equation by 3."}, {"video_title": "Perpendicular lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "Line B, we have x plus 3y is equal to negative 21. Let's subtract x from both sides so that it ends up on the right-hand side. So this, we end up with 3y is equal to negative x minus 21. And now let's divide both sides of this equation by 3. And we get y is equal to negative 1 third x minus 7. So this character's slope is negative 1 third. So here, m is equal to negative 1 third."}, {"video_title": "Perpendicular lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "And now let's divide both sides of this equation by 3. And we get y is equal to negative 1 third x minus 7. So this character's slope is negative 1 third. So here, m is equal to negative 1 third. So we already see they are the negative inverse of each other. You take the inverse of 3, it's 1 third, and then it's the negative of that. Or you take the inverse of negative 1 third, it's negative 3, and then this is the negative of that."}, {"video_title": "Perpendicular lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "So here, m is equal to negative 1 third. So we already see they are the negative inverse of each other. You take the inverse of 3, it's 1 third, and then it's the negative of that. Or you take the inverse of negative 1 third, it's negative 3, and then this is the negative of that. So these two lines are definitely perpendicular. Let's see this third line over here. So line C is 3x plus y is equal to 10."}, {"video_title": "Perpendicular lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "Or you take the inverse of negative 1 third, it's negative 3, and then this is the negative of that. So these two lines are definitely perpendicular. Let's see this third line over here. So line C is 3x plus y is equal to 10. If we subtract 3x from both sides, we get y is equal to negative 3x plus 10. So our slope in this case is negative 3. So our slope here is equal to negative 3."}, {"video_title": "Perpendicular lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "So line C is 3x plus y is equal to 10. If we subtract 3x from both sides, we get y is equal to negative 3x plus 10. So our slope in this case is negative 3. So our slope here is equal to negative 3. Now this guy is the negative of that guy. This guy's slope is the negative of that, but not the negative inverse. So it's not perpendicular."}, {"video_title": "Complex conjugates Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "I want to make a quick clarification and then add more tools in our complex number toolkit. In the first video, I said that if I had a complex number z and it's equal to a plus bi, I used a word. And I have to be careful about that word because I use it in kind of the everyday sense. But it also has a formal reality to it. So clearly, the real part of this complex number is a. Clearly, that is the real part. And clearly, this complex number is made up of a real number plus an imaginary number."}, {"video_title": "Complex conjugates Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "But it also has a formal reality to it. So clearly, the real part of this complex number is a. Clearly, that is the real part. And clearly, this complex number is made up of a real number plus an imaginary number. So I, just kind of talking in everyday terms, I called this the imaginary part. I called this imaginary number the imaginary part. But I want to just be careful there."}, {"video_title": "Complex conjugates Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And clearly, this complex number is made up of a real number plus an imaginary number. So I, just kind of talking in everyday terms, I called this the imaginary part. I called this imaginary number the imaginary part. But I want to just be careful there. I mean, I did make it clear that if you were to see the function, the real part of z, this would spit out the a. And the function, the imaginary part of z, this would spit out. And we talked about this in the first video."}, {"video_title": "Complex conjugates Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "But I want to just be careful there. I mean, I did make it clear that if you were to see the function, the real part of z, this would spit out the a. And the function, the imaginary part of z, this would spit out. And we talked about this in the first video. It would spit out the number that's scaling the i. So it would spit out the b. So if someone is talking in the formal sense about the imaginary part, they're really talking about the number that is scaling the i."}, {"video_title": "Complex conjugates Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And we talked about this in the first video. It would spit out the number that's scaling the i. So it would spit out the b. So if someone is talking in the formal sense about the imaginary part, they're really talking about the number that is scaling the i. But in my brain, when I think of a complex number, I think of it having a real number and an imaginary number. And if someone were to say, well, what part of that is the imaginary number, I would have given this whole thing. But if someone says, just what's the imaginary part, or they give you this function, just give them the b. Hopefully that clarifies things."}, {"video_title": "Complex conjugates Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So if someone is talking in the formal sense about the imaginary part, they're really talking about the number that is scaling the i. But in my brain, when I think of a complex number, I think of it having a real number and an imaginary number. And if someone were to say, well, what part of that is the imaginary number, I would have given this whole thing. But if someone says, just what's the imaginary part, or they give you this function, just give them the b. Hopefully that clarifies things. Frankly, I think the word imaginary part is badly named, because clearly this whole thing is an imaginary number. This right here is not an imaginary number. It's just a real number."}, {"video_title": "Complex conjugates Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "But if someone says, just what's the imaginary part, or they give you this function, just give them the b. Hopefully that clarifies things. Frankly, I think the word imaginary part is badly named, because clearly this whole thing is an imaginary number. This right here is not an imaginary number. It's just a real number. It's the real number scaling the i. So they should call this the number scaling the imaginary part of z. Anyway, with that said, what I want to introduce you to is the idea of a complex numbers conjugate."}, {"video_title": "Complex conjugates Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "It's just a real number. It's the real number scaling the i. So they should call this the number scaling the imaginary part of z. Anyway, with that said, what I want to introduce you to is the idea of a complex numbers conjugate. So if this is z, the conjugate of z, it'd be denoted with z with a bar over it. Sometimes it's z with a little asterisk right over there. That would just be equal to a minus bi."}, {"video_title": "Complex conjugates Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Anyway, with that said, what I want to introduce you to is the idea of a complex numbers conjugate. So if this is z, the conjugate of z, it'd be denoted with z with a bar over it. Sometimes it's z with a little asterisk right over there. That would just be equal to a minus bi. So let's see how they look on an Argand diagram. So that's my real axis. And then that is my imaginary axis."}, {"video_title": "Complex conjugates Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "That would just be equal to a minus bi. So let's see how they look on an Argand diagram. So that's my real axis. And then that is my imaginary axis. And then if I have z, this is z over here. This height over here is b. This base, or this length right here, is a."}, {"video_title": "Complex conjugates Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And then that is my imaginary axis. And then if I have z, this is z over here. This height over here is b. This base, or this length right here, is a. That's z. The conjugate of z is a minus bi. So it comes out a on the real axis, but it has minus b as its imaginary part."}, {"video_title": "Complex conjugates Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "This base, or this length right here, is a. That's z. The conjugate of z is a minus bi. So it comes out a on the real axis, but it has minus b as its imaginary part. So just like this. So this is the conjugate of z. So just to visualize it, it really is the conjugate of a complex number is really the mirror image of that complex number reflected over the x-axis."}, {"video_title": "Complex conjugates Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So it comes out a on the real axis, but it has minus b as its imaginary part. So just like this. So this is the conjugate of z. So just to visualize it, it really is the conjugate of a complex number is really the mirror image of that complex number reflected over the x-axis. You can imagine if this was a pool of water, we're seeing its reflection over here. And so we can actually look at this to visually add the complex number and its conjugate. So we said these are just like position vectors."}, {"video_title": "Complex conjugates Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So just to visualize it, it really is the conjugate of a complex number is really the mirror image of that complex number reflected over the x-axis. You can imagine if this was a pool of water, we're seeing its reflection over here. And so we can actually look at this to visually add the complex number and its conjugate. So we said these are just like position vectors. So if we were to add z and its conjugate, we could essentially just take this vector, shift it up here, do heads to tails. So this right here, we are adding z to its conjugate. And so this point right here, or the vector that specifies that point, is z plus z's conjugate."}, {"video_title": "Complex conjugates Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So we said these are just like position vectors. So if we were to add z and its conjugate, we could essentially just take this vector, shift it up here, do heads to tails. So this right here, we are adding z to its conjugate. And so this point right here, or the vector that specifies that point, is z plus z's conjugate. And you can see right here, just visually, this is going to be 2a. This is going to be 2a. And we could do that algebraically."}, {"video_title": "Complex conjugates Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And so this point right here, or the vector that specifies that point, is z plus z's conjugate. And you can see right here, just visually, this is going to be 2a. This is going to be 2a. And we could do that algebraically. If we were to add z, that's a plus bi, and add that to the conjugate, we would get a plus bi, and add that to its conjugate. So plus a minus bi, what are we going to get? These two guys cancel out."}, {"video_title": "Complex conjugates Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And we could do that algebraically. If we were to add z, that's a plus bi, and add that to the conjugate, we would get a plus bi, and add that to its conjugate. So plus a minus bi, what are we going to get? These two guys cancel out. We're just going to have 2a. Or another way to think about it, and really we're just playing around with math, if I take any complex number and to it I add its conjugate, I'm going to get 2 times the real part of the complex number. And this is also going to be 2 times the real part of the conjugate, because they have the exact same real part."}, {"video_title": "Complex conjugates Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "These two guys cancel out. We're just going to have 2a. Or another way to think about it, and really we're just playing around with math, if I take any complex number and to it I add its conjugate, I'm going to get 2 times the real part of the complex number. And this is also going to be 2 times the real part of the conjugate, because they have the exact same real part. Now with that said, let's think about where the conjugate could be useful. So let's say I had something like 1 plus 2i divided by 4 minus 5i. So it's no real obvious way to simplify this expression."}, {"video_title": "Complex conjugates Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And this is also going to be 2 times the real part of the conjugate, because they have the exact same real part. Now with that said, let's think about where the conjugate could be useful. So let's say I had something like 1 plus 2i divided by 4 minus 5i. So it's no real obvious way to simplify this expression. Maybe I don't like having this i in the denominator. Maybe I just want to write this as one complex number. If I divide one complex number by another, I should get another complex number."}, {"video_title": "Complex conjugates Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So it's no real obvious way to simplify this expression. Maybe I don't like having this i in the denominator. Maybe I just want to write this as one complex number. If I divide one complex number by another, I should get another complex number. But how do I do that? Well, one thing to do is to multiply the numerator and the denominator by the conjugate of the denominator. So 4 plus 5i over 4 plus 5i."}, {"video_title": "Complex conjugates Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "If I divide one complex number by another, I should get another complex number. But how do I do that? Well, one thing to do is to multiply the numerator and the denominator by the conjugate of the denominator. So 4 plus 5i over 4 plus 5i. And clearly I'm just multiplying by 1, because this is the same number over the same number. But the reason why this is valuable is if I multiply a number times its conjugate, I'm going to get a real number. So let me just show you that here."}, {"video_title": "Complex conjugates Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So 4 plus 5i over 4 plus 5i. And clearly I'm just multiplying by 1, because this is the same number over the same number. But the reason why this is valuable is if I multiply a number times its conjugate, I'm going to get a real number. So let me just show you that here. So let's just multiply this out. So we're going to get 1 times 4 plus 5i is 4 plus 5i. And then 2i times 4 is plus 8i."}, {"video_title": "Complex conjugates Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So let me just show you that here. So let's just multiply this out. So we're going to get 1 times 4 plus 5i is 4 plus 5i. And then 2i times 4 is plus 8i. And then 2i times 5i, that would be 10i squared, or negative 10. And then that will be over. Now this has the form a minus b times a plus b."}, {"video_title": "Complex conjugates Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And then 2i times 4 is plus 8i. And then 2i times 5i, that would be 10i squared, or negative 10. And then that will be over. Now this has the form a minus b times a plus b. So that's the product of, well, a plus b times a minus b is a squared minus b squared. So it's going to be equal to 4 squared, which is 16, minus 4 squared. Oh, I didn't have 4 plus 4i here."}, {"video_title": "Complex conjugates Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Now this has the form a minus b times a plus b. So that's the product of, well, a plus b times a minus b is a squared minus b squared. So it's going to be equal to 4 squared, which is 16, minus 4 squared. Oh, I didn't have 4 plus 4i here. This would be 4 plus 5i. What am I doing? 4 plus 5i."}, {"video_title": "Complex conjugates Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Oh, I didn't have 4 plus 4i here. This would be 4 plus 5i. What am I doing? 4 plus 5i. The same number over the same number. This was a 10 right over there. This is the conjugate."}, {"video_title": "Complex conjugates Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "4 plus 5i. The same number over the same number. This was a 10 right over there. This is the conjugate. I don't know, my brain must have been thinking in 4s. So obviously I don't want to change the number 4 plus 5i over 4 plus 5i. So let's multiply."}, {"video_title": "Complex conjugates Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "This is the conjugate. I don't know, my brain must have been thinking in 4s. So obviously I don't want to change the number 4 plus 5i over 4 plus 5i. So let's multiply. This is a minus b times a plus b. So 4 times 4. So this is going to be 4 squared minus 5i squared."}, {"video_title": "Complex conjugates Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So let's multiply. This is a minus b times a plus b. So 4 times 4. So this is going to be 4 squared minus 5i squared. And so this is going to be equal to 4 minus 10. Let's add the real parts. 4 minus 10 is negative 6."}, {"video_title": "Complex conjugates Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So this is going to be 4 squared minus 5i squared. And so this is going to be equal to 4 minus 10. Let's add the real parts. 4 minus 10 is negative 6. 5i plus 8i is 13i. Add the imaginary parts. And then you have 16 minus 5i squared."}, {"video_title": "Complex conjugates Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "4 minus 10 is negative 6. 5i plus 8i is 13i. Add the imaginary parts. And then you have 16 minus 5i squared. Well, 5i squared is negative 1. 5 squared, so this would be negative 25. The negative and the negative cancel out."}, {"video_title": "Complex conjugates Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And then you have 16 minus 5i squared. Well, 5i squared is negative 1. 5 squared, so this would be negative 25. The negative and the negative cancel out. So you have 16 plus 25. So that is 41. So we can write this as a complex number."}, {"video_title": "Complex conjugates Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "The negative and the negative cancel out. So you have 16 plus 25. So that is 41. So we can write this as a complex number. This is negative 6 over 41 plus 13 over 41i. We were able to divide these two complex numbers. So the useful thing here is the property that if I take any complex number and I multiply it by its conjugate, and obviously the conjugate of the conjugate is the original number."}, {"video_title": "Complex conjugates Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So we can write this as a complex number. This is negative 6 over 41 plus 13 over 41i. We were able to divide these two complex numbers. So the useful thing here is the property that if I take any complex number and I multiply it by its conjugate, and obviously the conjugate of the conjugate is the original number. But I would take any complex number and I multiply it by its conjugate. So this would be a plus bi times a minus bi. I'm going to get a real number."}, {"video_title": "Complex conjugates Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So the useful thing here is the property that if I take any complex number and I multiply it by its conjugate, and obviously the conjugate of the conjugate is the original number. But I would take any complex number and I multiply it by its conjugate. So this would be a plus bi times a minus bi. I'm going to get a real number. It's going to be a squared minus bi squared. Difference of squares, which is equal to a squared. Now, this is going to be negative b squared, but we have a negative sign out here."}, {"video_title": "Complex conjugates Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "I'm going to get a real number. It's going to be a squared minus bi squared. Difference of squares, which is equal to a squared. Now, this is going to be negative b squared, but we have a negative sign out here. So they cancel out. a squared plus b squared. And just out of curiosity, this is the same thing as the magnitude of our complex number squared."}, {"video_title": "Complex conjugates Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Now, this is going to be negative b squared, but we have a negative sign out here. So they cancel out. a squared plus b squared. And just out of curiosity, this is the same thing as the magnitude of our complex number squared. So this is the neat property. This is what makes conjugates really useful, especially when you want to simplify division of complex numbers. Anyway, hopefully you found that useful."}, {"video_title": "Interpret quadratic models Vertex form Algebra I Khan Academy.mp3", "Sentence": "Taylor wants to know what the restaurant's lowest net value will be, let me underline that, and when it will reach that value. So let's break it down step by step. The function which describes how the value of the restaurant, the net value of the restaurant changes over time is right over here. If I were to graph it, I can see that the coefficient on the quadratic term is positive, so it's going to be some form of upward opening parabola. I don't know exactly what it looks like, we can think about that in a second. And so it's going to have some point right over here, which really is the vertex of this parabola, where it's going to hit its lowest net value, and that's going to happen at some time t, if you can imagine that this right over here is the t-axis. So my first question is, is there some form, is there some way that I can rewrite this function algebraically, so it becomes very easy to pick out this low point, which is essentially the vertex of this parabola."}, {"video_title": "Interpret quadratic models Vertex form Algebra I Khan Academy.mp3", "Sentence": "If I were to graph it, I can see that the coefficient on the quadratic term is positive, so it's going to be some form of upward opening parabola. I don't know exactly what it looks like, we can think about that in a second. And so it's going to have some point right over here, which really is the vertex of this parabola, where it's going to hit its lowest net value, and that's going to happen at some time t, if you can imagine that this right over here is the t-axis. So my first question is, is there some form, is there some way that I can rewrite this function algebraically, so it becomes very easy to pick out this low point, which is essentially the vertex of this parabola. Pause this video and think about that. All right, so you could imagine the form that I'm talking about is vertex form, where you can clearly spot the vertex, and the way we can do that is actually by completing the square. So the first thing I will do is actually let me factor out a two here, because two is a common factor of both of these terms."}, {"video_title": "Interpret quadratic models Vertex form Algebra I Khan Academy.mp3", "Sentence": "So my first question is, is there some form, is there some way that I can rewrite this function algebraically, so it becomes very easy to pick out this low point, which is essentially the vertex of this parabola. Pause this video and think about that. All right, so you could imagine the form that I'm talking about is vertex form, where you can clearly spot the vertex, and the way we can do that is actually by completing the square. So the first thing I will do is actually let me factor out a two here, because two is a common factor of both of these terms. So v of t would be equal to two times t squared minus 10t, and I'm going to leave some space, because completing the square, which gets us to vertex form, is all about adding and subtracting the same value on one side, so we're not actually changing the value of that side, but writing it in a way that so we have a perfect square expression, and then we're probably going to add or subtract some value out here. Now, how do we make this a perfect square expression? And if any of this business about completing the square looks unfamiliar to you, I encourage you to look up completing the square on Khan Academy and review that."}, {"video_title": "Interpret quadratic models Vertex form Algebra I Khan Academy.mp3", "Sentence": "So the first thing I will do is actually let me factor out a two here, because two is a common factor of both of these terms. So v of t would be equal to two times t squared minus 10t, and I'm going to leave some space, because completing the square, which gets us to vertex form, is all about adding and subtracting the same value on one side, so we're not actually changing the value of that side, but writing it in a way that so we have a perfect square expression, and then we're probably going to add or subtract some value out here. Now, how do we make this a perfect square expression? And if any of this business about completing the square looks unfamiliar to you, I encourage you to look up completing the square on Khan Academy and review that. But the way that we complete the square is we look at this first degree coefficient right over here, it's negative 10, and we say, all right, well, let's take half of that and square it. So half of negative 10 is negative five, and if we were to square it, that's 25. So if we add 25 right over here, then this is going to become a perfect square expression, and you can see that it would be equivalent to this entire thing if we add 25 like that, is going to be equivalent to t minus five squared, just this part right over here."}, {"video_title": "Interpret quadratic models Vertex form Algebra I Khan Academy.mp3", "Sentence": "And if any of this business about completing the square looks unfamiliar to you, I encourage you to look up completing the square on Khan Academy and review that. But the way that we complete the square is we look at this first degree coefficient right over here, it's negative 10, and we say, all right, well, let's take half of that and square it. So half of negative 10 is negative five, and if we were to square it, that's 25. So if we add 25 right over here, then this is going to become a perfect square expression, and you can see that it would be equivalent to this entire thing if we add 25 like that, is going to be equivalent to t minus five squared, just this part right over here. That's why we took half of this and we squared it. But as I alluded to a few seconds ago or a few minutes ago, you can't just willy-nilly add 25 to one side or to one side of an equation like this. That will make this equality no longer true."}, {"video_title": "Interpret quadratic models Vertex form Algebra I Khan Academy.mp3", "Sentence": "So if we add 25 right over here, then this is going to become a perfect square expression, and you can see that it would be equivalent to this entire thing if we add 25 like that, is going to be equivalent to t minus five squared, just this part right over here. That's why we took half of this and we squared it. But as I alluded to a few seconds ago or a few minutes ago, you can't just willy-nilly add 25 to one side or to one side of an equation like this. That will make this equality no longer true. And so in fact, we didn't just add 25. Remember, we have this two out here. We added two times 25."}, {"video_title": "Interpret quadratic models Vertex form Algebra I Khan Academy.mp3", "Sentence": "That will make this equality no longer true. And so in fact, we didn't just add 25. Remember, we have this two out here. We added two times 25. You can verify that if you redistribute the two, you'd get two t squared minus 20t plus 50, plus two times 25. So in order to make the equality or in order to allow it to continue to be true, we have to subtract 50. So just to be clear, this isn't some kind of, you know, strange thing I'm doing."}, {"video_title": "Interpret quadratic models Vertex form Algebra I Khan Academy.mp3", "Sentence": "We added two times 25. You can verify that if you redistribute the two, you'd get two t squared minus 20t plus 50, plus two times 25. So in order to make the equality or in order to allow it to continue to be true, we have to subtract 50. So just to be clear, this isn't some kind of, you know, strange thing I'm doing. All I did is add 50 and subtract 50. You're saying, wait, you added 25, not 50. No, look, when I added 25 here, it's in a parentheses, and that whole expression is multiplied by two."}, {"video_title": "Interpret quadratic models Vertex form Algebra I Khan Academy.mp3", "Sentence": "So just to be clear, this isn't some kind of, you know, strange thing I'm doing. All I did is add 50 and subtract 50. You're saying, wait, you added 25, not 50. No, look, when I added 25 here, it's in a parentheses, and that whole expression is multiplied by two. So I really did add 50 here, so then I subtract 50 here to get to what I originally had. And when you view it that way, now v of t is going to be equal to two times this business, which we already established as t minus five squared, and then we have the minus 50. Now, why is this form useful?"}, {"video_title": "Interpret quadratic models Vertex form Algebra I Khan Academy.mp3", "Sentence": "No, look, when I added 25 here, it's in a parentheses, and that whole expression is multiplied by two. So I really did add 50 here, so then I subtract 50 here to get to what I originally had. And when you view it that way, now v of t is going to be equal to two times this business, which we already established as t minus five squared, and then we have the minus 50. Now, why is this form useful? This is vertex form. It's very easy to pick out the vertex. It's very easy to pick out when the low point is."}, {"video_title": "Interpret quadratic models Vertex form Algebra I Khan Academy.mp3", "Sentence": "Now, why is this form useful? This is vertex form. It's very easy to pick out the vertex. It's very easy to pick out when the low point is. The low point here happens when this part is minimized, and this part is minimized. Think about it. You have two times something squared."}, {"video_title": "Interpret quadratic models Vertex form Algebra I Khan Academy.mp3", "Sentence": "It's very easy to pick out when the low point is. The low point here happens when this part is minimized, and this part is minimized. Think about it. You have two times something squared. So if you have something squared, it's going to hit its lowest point when this something is zero. Otherwise, it's going to be a positive value. And so this part right over here is going to be equal to zero when t is equal to five."}, {"video_title": "Interpret quadratic models Vertex form Algebra I Khan Academy.mp3", "Sentence": "You have two times something squared. So if you have something squared, it's going to hit its lowest point when this something is zero. Otherwise, it's going to be a positive value. And so this part right over here is going to be equal to zero when t is equal to five. So the lowest value is when t is equal to five. Let me do that in a different color. I don't wanna reuse the colors too much."}, {"video_title": "Interpret quadratic models Vertex form Algebra I Khan Academy.mp3", "Sentence": "And so this part right over here is going to be equal to zero when t is equal to five. So the lowest value is when t is equal to five. Let me do that in a different color. I don't wanna reuse the colors too much. So if we say v of five is going to be equal to two times five minus five, I'm trying to keep up with the colors, minus five squared minus 50. Notice this whole thing becomes zero right over here. So v of five is equal to negative 50."}, {"video_title": "Interpret quadratic models Vertex form Algebra I Khan Academy.mp3", "Sentence": "I don't wanna reuse the colors too much. So if we say v of five is going to be equal to two times five minus five, I'm trying to keep up with the colors, minus five squared minus 50. Notice this whole thing becomes zero right over here. So v of five is equal to negative 50. That is when we hit our low point in terms of the net value of the restaurant. So t represents months. So we hit our low point."}, {"video_title": "Interpret quadratic models Vertex form Algebra I Khan Academy.mp3", "Sentence": "So v of five is equal to negative 50. That is when we hit our low point in terms of the net value of the restaurant. So t represents months. So we hit our low point. We rewrote our function in vertex form so it's easy to pick out this value. And we see that this low point happens at t equals five, which is at time five months. And then what is that lowest net value?"}, {"video_title": "Interpret quadratic models Vertex form Algebra I Khan Academy.mp3", "Sentence": "So we hit our low point. We rewrote our function in vertex form so it's easy to pick out this value. And we see that this low point happens at t equals five, which is at time five months. And then what is that lowest net value? Well, it's negative 50. And remember, the function gives us the net value in thousands of dollars. So it's negative $50,000 is the lowest net value of the restaurant."}, {"video_title": "Sum of logarithms with same base Logarithms Algebra II Khan Academy.mp3", "Sentence": "we're asked to simplify log base three of 27x. And frankly, this is already quite simple, but I'm assuming they want us to use some logarithm properties and manipulate this in some way, maybe to actually make it a little bit more complicated. But let's give our best shot at it. So the logarithm property that jumps out at me, because this right over here, we're saying what power do we have to raise three to to get 27x? 27x is the same thing as 27 times x. And so the logarithm property, it seems like they want us to use, is log base, let me write it, log base b of a times c. I'll write it this way. Log base b of a times c. This is equal to the logarithm base b of a plus the logarithm base b of c. And this comes straight out of the exponent properties, that if you have two exponents with the same base, you can add the exponents."}, {"video_title": "Sum of logarithms with same base Logarithms Algebra II Khan Academy.mp3", "Sentence": "So the logarithm property that jumps out at me, because this right over here, we're saying what power do we have to raise three to to get 27x? 27x is the same thing as 27 times x. And so the logarithm property, it seems like they want us to use, is log base, let me write it, log base b of a times c. I'll write it this way. Log base b of a times c. This is equal to the logarithm base b of a plus the logarithm base b of c. And this comes straight out of the exponent properties, that if you have two exponents with the same base, you can add the exponents. So let me make that a little bit clearer to you. And this part is a little confusing. The important part for this example is that you know how to apply this."}, {"video_title": "Sum of logarithms with same base Logarithms Algebra II Khan Academy.mp3", "Sentence": "Log base b of a times c. This is equal to the logarithm base b of a plus the logarithm base b of c. And this comes straight out of the exponent properties, that if you have two exponents with the same base, you can add the exponents. So let me make that a little bit clearer to you. And this part is a little confusing. The important part for this example is that you know how to apply this. But it's even better if you know the intuition. So let's say that log base b of a times c is equal to x. So this thing right over here evaluates to x."}, {"video_title": "Sum of logarithms with same base Logarithms Algebra II Khan Academy.mp3", "Sentence": "The important part for this example is that you know how to apply this. But it's even better if you know the intuition. So let's say that log base b of a times c is equal to x. So this thing right over here evaluates to x. Let's say that this thing right over here evaluates to y. So log base b of a is equal to y. And let's say that this thing over here evaluates to z."}, {"video_title": "Sum of logarithms with same base Logarithms Algebra II Khan Academy.mp3", "Sentence": "So this thing right over here evaluates to x. Let's say that this thing right over here evaluates to y. So log base b of a is equal to y. And let's say that this thing over here evaluates to z. So log base b of c is equal to z. Now, what we know is this thing right over here, this thing right over here, or this thing right over here tells us, tells us that b to the x power is equal to a times c. Now, this right over here is telling us that b to the y power is equal to a. b to the y power is equal to a. And this over here is telling us that b to the z power is equal to c. Let me do that in that same green."}, {"video_title": "Sum of logarithms with same base Logarithms Algebra II Khan Academy.mp3", "Sentence": "And let's say that this thing over here evaluates to z. So log base b of c is equal to z. Now, what we know is this thing right over here, this thing right over here, or this thing right over here tells us, tells us that b to the x power is equal to a times c. Now, this right over here is telling us that b to the y power is equal to a. b to the y power is equal to a. And this over here is telling us that b to the z power is equal to c. Let me do that in that same green. So I'm just writing the same truth I'm writing as an exponential function or exponential equation instead of a logarithmic equation. So b to the z power is equal to c. This is the same statement. This is the same statement."}, {"video_title": "Sum of logarithms with same base Logarithms Algebra II Khan Academy.mp3", "Sentence": "And this over here is telling us that b to the z power is equal to c. Let me do that in that same green. So I'm just writing the same truth I'm writing as an exponential function or exponential equation instead of a logarithmic equation. So b to the z power is equal to c. This is the same statement. This is the same statement. Or the same truth said in a different way. And this is the same truth said in a different way. Well, if we know that a is equal to this, is equal to b to the y, we can, and c, and c is equal to bz, then we can write b to the x power is equal to b to the y power, b to the y power, that's what a is, we know that already, times b to the z power, times b to the z power."}, {"video_title": "Sum of logarithms with same base Logarithms Algebra II Khan Academy.mp3", "Sentence": "This is the same statement. Or the same truth said in a different way. And this is the same truth said in a different way. Well, if we know that a is equal to this, is equal to b to the y, we can, and c, and c is equal to bz, then we can write b to the x power is equal to b to the y power, b to the y power, that's what a is, we know that already, times b to the z power, times b to the z power. And we know from our exponent properties, we know from our exponent properties that if we take b to the y times b to the z, this is the same thing as b to the, I'll do it in a neutral color, b to the y plus z power. This comes straight out of our exponent properties. And so if b to the y plus z power is the same thing as b to the x power, that tells us that x must be equal to y plus z. X must be equal to y plus z."}, {"video_title": "Sum of logarithms with same base Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, if we know that a is equal to this, is equal to b to the y, we can, and c, and c is equal to bz, then we can write b to the x power is equal to b to the y power, b to the y power, that's what a is, we know that already, times b to the z power, times b to the z power. And we know from our exponent properties, we know from our exponent properties that if we take b to the y times b to the z, this is the same thing as b to the, I'll do it in a neutral color, b to the y plus z power. This comes straight out of our exponent properties. And so if b to the y plus z power is the same thing as b to the x power, that tells us that x must be equal to y plus z. X must be equal to y plus z. If this is confusing to you, don't worry about it too much. The important thing, or at least the first important thing is that you know how to apply it, and then you can think about this a little bit more, and you can even try it out with some numbers. You just have to realize that logarithms are really just exponents."}, {"video_title": "Sum of logarithms with same base Logarithms Algebra II Khan Academy.mp3", "Sentence": "And so if b to the y plus z power is the same thing as b to the x power, that tells us that x must be equal to y plus z. X must be equal to y plus z. If this is confusing to you, don't worry about it too much. The important thing, or at least the first important thing is that you know how to apply it, and then you can think about this a little bit more, and you can even try it out with some numbers. You just have to realize that logarithms are really just exponents. And I know when people first tell me that, I was like, well, what does that mean? But when you evaluate a logarithm, you're getting an exponent that you would have to raise b to to get to a times c. But let's just apply this property right over here. So if we apply it to this one, we know that log base three of 27 times x, I'll write it that way, is equal to log base three of 27 plus log base three of x."}, {"video_title": "Sum of logarithms with same base Logarithms Algebra II Khan Academy.mp3", "Sentence": "You just have to realize that logarithms are really just exponents. And I know when people first tell me that, I was like, well, what does that mean? But when you evaluate a logarithm, you're getting an exponent that you would have to raise b to to get to a times c. But let's just apply this property right over here. So if we apply it to this one, we know that log base three of 27 times x, I'll write it that way, is equal to log base three of 27 plus log base three of x. And then this right over here we can evaluate. This tells us what power do I have to raise three to to get to 27? You could view it as this way."}, {"video_title": "Sum of logarithms with same base Logarithms Algebra II Khan Academy.mp3", "Sentence": "So if we apply it to this one, we know that log base three of 27 times x, I'll write it that way, is equal to log base three of 27 plus log base three of x. And then this right over here we can evaluate. This tells us what power do I have to raise three to to get to 27? You could view it as this way. Three to the question mark is equal to 27. Well, three to the third power is equal to 27. Three times three is nine, times three is 27."}, {"video_title": "Sum of logarithms with same base Logarithms Algebra II Khan Academy.mp3", "Sentence": "You could view it as this way. Three to the question mark is equal to 27. Well, three to the third power is equal to 27. Three times three is nine, times three is 27. So this right over here evaluates to three. So if we were to simplify, or I guess I wouldn't even call it simplifying it, I would just call it expanding it out or using this property because we now have two terms where we started off with one term. Actually, if we started with this, I'd say that this is the more simple version of it."}, {"video_title": "Sum of logarithms with same base Logarithms Algebra II Khan Academy.mp3", "Sentence": "Three times three is nine, times three is 27. So this right over here evaluates to three. So if we were to simplify, or I guess I wouldn't even call it simplifying it, I would just call it expanding it out or using this property because we now have two terms where we started off with one term. Actually, if we started with this, I'd say that this is the more simple version of it. But when we rewrite it, this first term becomes three. So this first term becomes three. And then we're left with plus log base three of x."}, {"video_title": "Sum of logarithms with same base Logarithms Algebra II Khan Academy.mp3", "Sentence": "Actually, if we started with this, I'd say that this is the more simple version of it. But when we rewrite it, this first term becomes three. So this first term becomes three. And then we're left with plus log base three of x. So this is just an alternate way of writing this original statement. Log base three of 27x. So once again, not clear that this is simpler than this right over here."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "Sometimes a quadratic polynomial, or just a quadratic itself, or quadratic expression. But all it means is a second degree polynomial. So something that's going to have a variable raised to the second power. In this case, and all of the examples will do, it'll be x. So let's say I have the quadratic expression, x squared plus 10x plus 9. And I want to factor it into the product of two binomials. How do we do that?"}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "In this case, and all of the examples will do, it'll be x. So let's say I have the quadratic expression, x squared plus 10x plus 9. And I want to factor it into the product of two binomials. How do we do that? Well, let's just think about what happens if we were to take x plus a and multiply that by x plus b. If we were to multiply these two things, what happens? Well, we have a little bit of experience doing this."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "How do we do that? Well, let's just think about what happens if we were to take x plus a and multiply that by x plus b. If we were to multiply these two things, what happens? Well, we have a little bit of experience doing this. This will be x times x, which is x squared, plus x times b, which is bx, plus a times x, plus a times b, plus ab. Or if we want to add these two in the middle right here, because they have the same, they're both coefficients of x, we could write this as x squared plus, I could write it as b plus a, or a plus bx, plus ab. So in general, if we assume that this is the product of two binomials, we see that this middle coefficient on the x term, or you could say the first degree coefficient there, that's going to be the sum of our a and b."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "Well, we have a little bit of experience doing this. This will be x times x, which is x squared, plus x times b, which is bx, plus a times x, plus a times b, plus ab. Or if we want to add these two in the middle right here, because they have the same, they're both coefficients of x, we could write this as x squared plus, I could write it as b plus a, or a plus bx, plus ab. So in general, if we assume that this is the product of two binomials, we see that this middle coefficient on the x term, or you could say the first degree coefficient there, that's going to be the sum of our a and b. This is going to be the sum of our a and b. And then the constant term is going to be the product of our a and b. Notice, this would map to this, and this would map to this."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "So in general, if we assume that this is the product of two binomials, we see that this middle coefficient on the x term, or you could say the first degree coefficient there, that's going to be the sum of our a and b. This is going to be the sum of our a and b. And then the constant term is going to be the product of our a and b. Notice, this would map to this, and this would map to this. And of course, this is the same thing as this. So can we somehow pattern match this to that? Is there some a and b where a plus b is equal to 10, and a times b is equal to 9?"}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "Notice, this would map to this, and this would map to this. And of course, this is the same thing as this. So can we somehow pattern match this to that? Is there some a and b where a plus b is equal to 10, and a times b is equal to 9? Well, let's just think about it a little bit. What are the factors of 9? What are the things that a and b could be equal to?"}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "Is there some a and b where a plus b is equal to 10, and a times b is equal to 9? Well, let's just think about it a little bit. What are the factors of 9? What are the things that a and b could be equal to? And we're assuming that everything is an integer, and normally when we're factoring, especially when we're beginning to factor, we're dealing with integer numbers. So what are the factors of 9? They're 1, 3, and 9."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "What are the things that a and b could be equal to? And we're assuming that everything is an integer, and normally when we're factoring, especially when we're beginning to factor, we're dealing with integer numbers. So what are the factors of 9? They're 1, 3, and 9. So this could be a 3 and a 3, or it could be a 1 and a 9. Now, if it's a 3 and a 3, then you'll have 3 plus 3. That doesn't equal 10."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "They're 1, 3, and 9. So this could be a 3 and a 3, or it could be a 1 and a 9. Now, if it's a 3 and a 3, then you'll have 3 plus 3. That doesn't equal 10. But if it's a 1 and a 9, 1 times 9 is 9. 1 plus 9 is 10. So it does work."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "That doesn't equal 10. But if it's a 1 and a 9, 1 times 9 is 9. 1 plus 9 is 10. So it does work. So a could be equal to 1, and b could be equal to 9. So we could factor this as being x plus 1 times x plus 9. And if you multiply these two out using the skills we developed in the last few videos, you'll see that it is indeed x squared plus 10x plus 9."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "So it does work. So a could be equal to 1, and b could be equal to 9. So we could factor this as being x plus 1 times x plus 9. And if you multiply these two out using the skills we developed in the last few videos, you'll see that it is indeed x squared plus 10x plus 9. So when you see something like this, when the coefficient on the x squared term, or the leading coefficient on this quadratic, is a 1, you can just say, all right, what two numbers add up to this coefficient right here? And what two numbers add up? And those same two numbers, when you take their product, have to be equal to 9."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "And if you multiply these two out using the skills we developed in the last few videos, you'll see that it is indeed x squared plus 10x plus 9. So when you see something like this, when the coefficient on the x squared term, or the leading coefficient on this quadratic, is a 1, you can just say, all right, what two numbers add up to this coefficient right here? And what two numbers add up? And those same two numbers, when you take their product, have to be equal to 9. And of course, this has to be in standard form. Or if it's not in standard form, you should put it in that form so that you can always say, OK, whatever's on the first degree coefficient, my a and b have to add to that. Whatever's my constant term, my a times b, the product has to be that."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "And those same two numbers, when you take their product, have to be equal to 9. And of course, this has to be in standard form. Or if it's not in standard form, you should put it in that form so that you can always say, OK, whatever's on the first degree coefficient, my a and b have to add to that. Whatever's my constant term, my a times b, the product has to be that. Let's do several more examples. And I think the more examples we do, the more sense this will make. Let's say we had x squared plus 10x plus, well, I already did 10x."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "Whatever's my constant term, my a times b, the product has to be that. Let's do several more examples. And I think the more examples we do, the more sense this will make. Let's say we had x squared plus 10x plus, well, I already did 10x. Let's do a different number. x squared plus 15x plus 50. We want to factor this."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "Let's say we had x squared plus 10x plus, well, I already did 10x. Let's do a different number. x squared plus 15x plus 50. We want to factor this. Well, same drill. We have an x squared term, x squared term. We have a first degree term."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "We want to factor this. Well, same drill. We have an x squared term, x squared term. We have a first degree term. This should be, this right here, should be the sum of two numbers, and then this term, the constant term right here, should be the product of two numbers. So we need to think of two numbers that when I multiply them, I get 50, and when I add them, I get 15. And this is going to be a bit of an art that you're going to develop, but the more practice you do, you're going to see that it'll start to come naturally."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "We have a first degree term. This should be, this right here, should be the sum of two numbers, and then this term, the constant term right here, should be the product of two numbers. So we need to think of two numbers that when I multiply them, I get 50, and when I add them, I get 15. And this is going to be a bit of an art that you're going to develop, but the more practice you do, you're going to see that it'll start to come naturally. So what could a and b be? Let's think about the factors of 50. It could be 1 times 50, 2 times 25."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "And this is going to be a bit of an art that you're going to develop, but the more practice you do, you're going to see that it'll start to come naturally. So what could a and b be? Let's think about the factors of 50. It could be 1 times 50, 2 times 25. See, 4 doesn't go into 50. It could be 5 times 10. I think that's all of them."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "It could be 1 times 50, 2 times 25. See, 4 doesn't go into 50. It could be 5 times 10. I think that's all of them. Let's try out these numbers and see if any of these add up to 15. So 1 plus 50 does not add up to 15. 2 plus 25 does not add up to 15."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "I think that's all of them. Let's try out these numbers and see if any of these add up to 15. So 1 plus 50 does not add up to 15. 2 plus 25 does not add up to 15. But 5 plus 10 does add up to 15. So this could be 5 plus 10, and this could be 5 times 10. So if we were to factor this, this would be equal to x plus 5 times x plus 10."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "2 plus 25 does not add up to 15. But 5 plus 10 does add up to 15. So this could be 5 plus 10, and this could be 5 times 10. So if we were to factor this, this would be equal to x plus 5 times x plus 10. And multiply it out. I encourage you to multiply this out and see that this is indeed x squared plus 15x plus 10. In fact, let's do it."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "So if we were to factor this, this would be equal to x plus 5 times x plus 10. And multiply it out. I encourage you to multiply this out and see that this is indeed x squared plus 15x plus 10. In fact, let's do it. x times x, x squared. x times 10 plus 10x. 5 times x plus 5x."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "In fact, let's do it. x times x, x squared. x times 10 plus 10x. 5 times x plus 5x. 5 times 10 plus 50. Notice the 5 times 10 gave us the 50. The 5x plus the 10x is giving us the 15x in between."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "5 times x plus 5x. 5 times 10 plus 50. Notice the 5 times 10 gave us the 50. The 5x plus the 10x is giving us the 15x in between. So it's x squared plus 15x plus 50. x squared plus 15x plus 50. Let's up the stakes a little bit, introduce some negative signs in here. Let's say I had x squared minus 11x plus 24."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "The 5x plus the 10x is giving us the 15x in between. So it's x squared plus 15x plus 50. x squared plus 15x plus 50. Let's up the stakes a little bit, introduce some negative signs in here. Let's say I had x squared minus 11x plus 24. Now, it's the exact same principle. I need to think of two numbers that when I add them need to be equal to negative 11. a plus b need to be equal to negative 11. And a times b need to be equal to 24."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "Let's say I had x squared minus 11x plus 24. Now, it's the exact same principle. I need to think of two numbers that when I add them need to be equal to negative 11. a plus b need to be equal to negative 11. And a times b need to be equal to 24. Now, there's something for you to think about. If when I multiply both of these numbers, I'm getting a positive number. I'm getting a 24."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "And a times b need to be equal to 24. Now, there's something for you to think about. If when I multiply both of these numbers, I'm getting a positive number. I'm getting a 24. That means that both of these need to be positive or both of these need to be negative. That's the only way I'm going to get a positive number here. Now, if when I add them I get a negative number, if these were positive, there's no way I can add two positive numbers and get a negative number."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "I'm getting a 24. That means that both of these need to be positive or both of these need to be negative. That's the only way I'm going to get a positive number here. Now, if when I add them I get a negative number, if these were positive, there's no way I can add two positive numbers and get a negative number. So the fact that their sum is negative and the fact that their product is positive tells me that both a and b are negative. a and b have to be negative. Remember, one can't be negative and the other one can't be positive because then the product would be negative."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "Now, if when I add them I get a negative number, if these were positive, there's no way I can add two positive numbers and get a negative number. So the fact that their sum is negative and the fact that their product is positive tells me that both a and b are negative. a and b have to be negative. Remember, one can't be negative and the other one can't be positive because then the product would be negative. And they both can't be positive because then this, when you add them, it would get you a positive number. So let's just think about what a and b can be. So two negative numbers."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "Remember, one can't be negative and the other one can't be positive because then the product would be negative. And they both can't be positive because then this, when you add them, it would get you a positive number. So let's just think about what a and b can be. So two negative numbers. So let's think about the factors of 24. And we'll kind of have to think of the negative factors. But let me see, it could be 1 times 24, 2 times 11, 3 times 8, or 4 times 6."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "So two negative numbers. So let's think about the factors of 24. And we'll kind of have to think of the negative factors. But let me see, it could be 1 times 24, 2 times 11, 3 times 8, or 4 times 6. Now, which of these, when I multiply these, well, obviously when I multiply 1 times 24 I get 24. When I get 2 times 12 I get 24. So we know that all of these, the products are 24."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "But let me see, it could be 1 times 24, 2 times 11, 3 times 8, or 4 times 6. Now, which of these, when I multiply these, well, obviously when I multiply 1 times 24 I get 24. When I get 2 times 12 I get 24. So we know that all of these, the products are 24. But which two of these, which two factors, when I add them, should I get 11? And then we could say, let's take the negative of both of those. So when you look at these, 3 and 8 jump out."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "So we know that all of these, the products are 24. But which two of these, which two factors, when I add them, should I get 11? And then we could say, let's take the negative of both of those. So when you look at these, 3 and 8 jump out. 3 times 8 is equal to 24. 3 plus 8 is equal to 11. But that doesn't quite work out, right?"}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "So when you look at these, 3 and 8 jump out. 3 times 8 is equal to 24. 3 plus 8 is equal to 11. But that doesn't quite work out, right? Because we have a negative 11 here. But what if we did negative 3 and negative 8? Negative 3 times negative 8 is equal to positive 24."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "But that doesn't quite work out, right? Because we have a negative 11 here. But what if we did negative 3 and negative 8? Negative 3 times negative 8 is equal to positive 24. Negative 3 minus 11, or sorry, negative 3 plus negative 8 is equal to negative 11. So negative 3 and negative 8 work. So if we factor this, this is going to x squared minus 11x plus 24 is going to be equal to x minus 3 times x minus 8."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "Negative 3 times negative 8 is equal to positive 24. Negative 3 minus 11, or sorry, negative 3 plus negative 8 is equal to negative 11. So negative 3 and negative 8 work. So if we factor this, this is going to x squared minus 11x plus 24 is going to be equal to x minus 3 times x minus 8. Let's do another one like that. Actually, let's mix it up a little bit. Let's say I had x squared plus 5x minus 14."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "So if we factor this, this is going to x squared minus 11x plus 24 is going to be equal to x minus 3 times x minus 8. Let's do another one like that. Actually, let's mix it up a little bit. Let's say I had x squared plus 5x minus 14. So here we have a different situation. The product of my two numbers is negative. a times b is equal to negative 14."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "Let's say I had x squared plus 5x minus 14. So here we have a different situation. The product of my two numbers is negative. a times b is equal to negative 14. My product is negative. That tells me that one of them is positive and one of them is negative. And when I add the two, a plus b, I get it being equal to 5."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "a times b is equal to negative 14. My product is negative. That tells me that one of them is positive and one of them is negative. And when I add the two, a plus b, I get it being equal to 5. So let's think about the factors of 14 and what combinations of them, when I add them, if one is positive and one is negative, or I'm really kind of taking the difference of the two, do I get 5? So if I take 1 and 14, I'm just going to try out things. Negative 1 plus 14 is negative 13."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "And when I add the two, a plus b, I get it being equal to 5. So let's think about the factors of 14 and what combinations of them, when I add them, if one is positive and one is negative, or I'm really kind of taking the difference of the two, do I get 5? So if I take 1 and 14, I'm just going to try out things. Negative 1 plus 14 is negative 13. So let me write all of the combinations that I could do. And eventually your brain will just zone in on it. So you could have negative 1 plus 14 is equal to 13."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "Negative 1 plus 14 is negative 13. So let me write all of the combinations that I could do. And eventually your brain will just zone in on it. So you could have negative 1 plus 14 is equal to 13. And 1 plus negative 14 is equal to negative 13. So those don't work. That doesn't equal 5."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "So you could have negative 1 plus 14 is equal to 13. And 1 plus negative 14 is equal to negative 13. So those don't work. That doesn't equal 5. Now what about 2 and 7? If I do negative 2 plus 7, that is equal to 5. We're done."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "That doesn't equal 5. Now what about 2 and 7? If I do negative 2 plus 7, that is equal to 5. We're done. That worked. I mean, we could have tried 2 plus negative 7, but that would have equaled negative 5. So that wouldn't have worked."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "We're done. That worked. I mean, we could have tried 2 plus negative 7, but that would have equaled negative 5. So that wouldn't have worked. But negative 2 plus 7 works. And negative 2 times 7 is negative 14. So there we have it."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "So that wouldn't have worked. But negative 2 plus 7 works. And negative 2 times 7 is negative 14. So there we have it. We know it's x minus 2 times x plus 7. That's pretty neat. Negative 2 times 7 is negative 14."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "So there we have it. We know it's x minus 2 times x plus 7. That's pretty neat. Negative 2 times 7 is negative 14. Negative 2 plus 7 is positive 5. Let's do several more of these, just to really get well honed the skill. So let's say we have x squared minus x minus 56."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "Negative 2 times 7 is negative 14. Negative 2 plus 7 is positive 5. Let's do several more of these, just to really get well honed the skill. So let's say we have x squared minus x minus 56. So the product of the two numbers have to be minus 56, have to be negative 56. And their difference, because 1 is going to be positive and 1 is going to be negative, their difference has to be negative 1. And the numbers that immediately jump out in my brain, and I don't know if they jump out in your brain, we just learned this in the times tables, 56 is 8 times 7."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "So let's say we have x squared minus x minus 56. So the product of the two numbers have to be minus 56, have to be negative 56. And their difference, because 1 is going to be positive and 1 is going to be negative, their difference has to be negative 1. And the numbers that immediately jump out in my brain, and I don't know if they jump out in your brain, we just learned this in the times tables, 56 is 8 times 7. I mean, there's other numbers. It's also 28 times 2. It's all sorts of things."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "And the numbers that immediately jump out in my brain, and I don't know if they jump out in your brain, we just learned this in the times tables, 56 is 8 times 7. I mean, there's other numbers. It's also 28 times 2. It's all sorts of things. But 8 times 7 really jumped out into my brain because they're very close to each other. We need numbers that are very close to each other. And one of these has to be positive and one of these has to be negative."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "It's all sorts of things. But 8 times 7 really jumped out into my brain because they're very close to each other. We need numbers that are very close to each other. And one of these has to be positive and one of these has to be negative. Now, the fact that when their sum is negative tells me that the larger of these two should probably be negative. So if we take negative 8 times 7, that's equal to negative 56, and then if we take negative 8 plus 7, that is equal to negative 1, which is exactly the coefficient right there. So when I factor this, this is going to be x minus 8 times x plus 7."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "And one of these has to be positive and one of these has to be negative. Now, the fact that when their sum is negative tells me that the larger of these two should probably be negative. So if we take negative 8 times 7, that's equal to negative 56, and then if we take negative 8 plus 7, that is equal to negative 1, which is exactly the coefficient right there. So when I factor this, this is going to be x minus 8 times x plus 7. This is often one of the hardest concepts people learn in algebra because it is a bit of an art. You have to look at all of the factors here, play with the positive and negative signs, see which of those factors, when 1 is positive and 1 is negative, add up to the coefficient on the x term. But as you do more and more practice, you'll see that it'll become a bit of second nature."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "So when I factor this, this is going to be x minus 8 times x plus 7. This is often one of the hardest concepts people learn in algebra because it is a bit of an art. You have to look at all of the factors here, play with the positive and negative signs, see which of those factors, when 1 is positive and 1 is negative, add up to the coefficient on the x term. But as you do more and more practice, you'll see that it'll become a bit of second nature. Now let's step up the stakes a little bit more. Let's say we had negative x squared. Everything we've done so far had a positive coefficient, a positive 1 coefficient on the x squared term."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "But as you do more and more practice, you'll see that it'll become a bit of second nature. Now let's step up the stakes a little bit more. Let's say we had negative x squared. Everything we've done so far had a positive coefficient, a positive 1 coefficient on the x squared term. But let's say we had a negative x squared minus 5x plus 24. How do we do this? Well, the easiest way I can think of doing it is factor out a negative 1, and then it becomes just like the problems we've been doing before."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "Everything we've done so far had a positive coefficient, a positive 1 coefficient on the x squared term. But let's say we had a negative x squared minus 5x plus 24. How do we do this? Well, the easiest way I can think of doing it is factor out a negative 1, and then it becomes just like the problems we've been doing before. So this is the same thing as negative 1 times positive x squared plus 5x minus 24. I just factored a negative 1 out. You can multiply negative 1 times all of these, and you'll see it becomes this."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "Well, the easiest way I can think of doing it is factor out a negative 1, and then it becomes just like the problems we've been doing before. So this is the same thing as negative 1 times positive x squared plus 5x minus 24. I just factored a negative 1 out. You can multiply negative 1 times all of these, and you'll see it becomes this. Or you could factor the negative 1 out and divide all of these by negative 1, and you get that right there. Now, same game as before. I need two numbers that when I take their product, I get negative 24."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "You can multiply negative 1 times all of these, and you'll see it becomes this. Or you could factor the negative 1 out and divide all of these by negative 1, and you get that right there. Now, same game as before. I need two numbers that when I take their product, I get negative 24. So one will be positive, one will be negative. And when I take their sum, it's going to be 5. So let's think about 24."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "I need two numbers that when I take their product, I get negative 24. So one will be positive, one will be negative. And when I take their sum, it's going to be 5. So let's think about 24. Is 1 and 24. Let's see, if this is negative 1 and 24, it's negative 23. Otherwise, if it's negative 1 and 24, it would be positive 23."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "So let's think about 24. Is 1 and 24. Let's see, if this is negative 1 and 24, it's negative 23. Otherwise, if it's negative 1 and 24, it would be positive 23. If it was the other way around, it would be negative 23. It doesn't work. What about 2 and 12?"}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "Otherwise, if it's negative 1 and 24, it would be positive 23. If it was the other way around, it would be negative 23. It doesn't work. What about 2 and 12? Well, if this is negative, if the 2 is negative, remember, one of these have to be negative. If the 2 is negative, their sum would be 10. If the 12 is negative, their sum would be negative 10."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "What about 2 and 12? Well, if this is negative, if the 2 is negative, remember, one of these have to be negative. If the 2 is negative, their sum would be 10. If the 12 is negative, their sum would be negative 10. Still doesn't work. 3 and 8. If the 3 is negative, their sum will be 5."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "If the 12 is negative, their sum would be negative 10. Still doesn't work. 3 and 8. If the 3 is negative, their sum will be 5. So it works. So if we pick negative 3 and 8, negative 3 and 8 work. So if we use it because negative 3 plus 8 is 5, negative 3 times 8 is negative 24."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "If the 3 is negative, their sum will be 5. So it works. So if we pick negative 3 and 8, negative 3 and 8 work. So if we use it because negative 3 plus 8 is 5, negative 3 times 8 is negative 24. So this is going to be equal to, can't forget that negative 1 out front. And then we factor the inside. Negative 1 times x minus 3 times x plus 8."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "So if we use it because negative 3 plus 8 is 5, negative 3 times 8 is negative 24. So this is going to be equal to, can't forget that negative 1 out front. And then we factor the inside. Negative 1 times x minus 3 times x plus 8. And if you really wanted to, you could multiply the negative 1 times this. You would get 3 minus x if you did, or you don't have to. Let's do one more of these."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "Negative 1 times x minus 3 times x plus 8. And if you really wanted to, you could multiply the negative 1 times this. You would get 3 minus x if you did, or you don't have to. Let's do one more of these. The more practice, the better, I think. All right, let's say I had negative x squared plus 18x minus 72. So once again, I like to factor out the negative 1."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "Let's do one more of these. The more practice, the better, I think. All right, let's say I had negative x squared plus 18x minus 72. So once again, I like to factor out the negative 1. So this is equal to negative 1 times x squared minus 18x plus 72. Now we just have to think of two numbers that when I multiply them, I get positive 72. So they have to be the same sign."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "So once again, I like to factor out the negative 1. So this is equal to negative 1 times x squared minus 18x plus 72. Now we just have to think of two numbers that when I multiply them, I get positive 72. So they have to be the same sign. And that makes it easier in our head, at least in my head. When I multiply them, I get positive 72. When I add them, I get negative 18."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "So they have to be the same sign. And that makes it easier in our head, at least in my head. When I multiply them, I get positive 72. When I add them, I get negative 18. So if they're the same sign and their sum is a negative number, they both must be negative. So they're both negative. And we could go through all of the factors of 72, but the one that springs up, maybe you think of 8 times 9, but 8 times 9, or negative 8 minus 9, or negative 8 plus negative 9 doesn't work."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "When I add them, I get negative 18. So if they're the same sign and their sum is a negative number, they both must be negative. So they're both negative. And we could go through all of the factors of 72, but the one that springs up, maybe you think of 8 times 9, but 8 times 9, or negative 8 minus 9, or negative 8 plus negative 9 doesn't work. That turns into 17. That was close. Let me show you that."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "And we could go through all of the factors of 72, but the one that springs up, maybe you think of 8 times 9, but 8 times 9, or negative 8 minus 9, or negative 8 plus negative 9 doesn't work. That turns into 17. That was close. Let me show you that. Negative 9 plus negative 8, that is equal to negative 17. Close, but no cigar. So what other ones are?"}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "Let me show you that. Negative 9 plus negative 8, that is equal to negative 17. Close, but no cigar. So what other ones are? We have 6 and 12. That actually seems pretty good. If we have negative 6 plus negative 12, that is equal to negative 18."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1 Algebra II Khan Academy.mp3", "Sentence": "So what other ones are? We have 6 and 12. That actually seems pretty good. If we have negative 6 plus negative 12, that is equal to negative 18. Notice, it's a bit of an art. You have to try the different factors here. So this will become negative 1, don't want to forget that, times x minus 6 times x minus 12."}, {"video_title": "Exponential model word problem bacteria growth High School Math Khan Academy.mp3", "Sentence": "We see it's an exponential model here. How many bacteria will make up the culture after 120 minutes? So, really they just want to say, well, what is B of 120 going to be? And so it's going to be 10 times two to the 120 divided by 12th power. So this is going to be equal to 10 times two to the 120 divided by 12 is 10th power. So this is going to be equal to 10 times, two to the 10th power is 1,024. If you want to verify that, you could say, well, two to the fifth is equal to 32."}, {"video_title": "Exponential model word problem bacteria growth High School Math Khan Academy.mp3", "Sentence": "And so it's going to be 10 times two to the 120 divided by 12th power. So this is going to be equal to 10 times two to the 120 divided by 12 is 10th power. So this is going to be equal to 10 times, two to the 10th power is 1,024. If you want to verify that, you could say, well, two to the fifth is equal to 32. And so two to the 10th is going to be two to the fifth times two to the fifth. And 32 times 32 is, let's see, 64, 0, 6. So let's see, we're going to have six, sorry."}, {"video_title": "Exponential model word problem bacteria growth High School Math Khan Academy.mp3", "Sentence": "If you want to verify that, you could say, well, two to the fifth is equal to 32. And so two to the 10th is going to be two to the fifth times two to the fifth. And 32 times 32 is, let's see, 64, 0, 6. So let's see, we're going to have six, sorry. Three times 32 is 96. And so you have a four, 12, 10, 24. So this is going to be 1024."}, {"video_title": "Exponential model word problem bacteria growth High School Math Khan Academy.mp3", "Sentence": "So let's see, we're going to have six, sorry. Three times 32 is 96. And so you have a four, 12, 10, 24. So this is going to be 1024. 10 times that is going to be equal to 1,024, 0. So 10,200, 10,240 bacteria. And we're done."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "All right, now let's work through this together. And to start, let's just remind ourselves what an average rate of change even is. You can view it as the change in your value of the function for a given change in the underlying variable, for a given change in x. We could also view this as, if we wanna figure out the interval, we could say our x final minus our x initial. And in the numerator, it would be the value of our function at the x final minus the value of the function at our x initial. Now, they aren't asking us to calculate this for all of these different intervals. They're just asking us whether it is positive."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "We could also view this as, if we wanna figure out the interval, we could say our x final minus our x initial. And in the numerator, it would be the value of our function at the x final minus the value of the function at our x initial. Now, they aren't asking us to calculate this for all of these different intervals. They're just asking us whether it is positive. And if you look over here, as long as our x final is greater than x initial, in order to have a positive average rate of change, we just need to figure out whether h at x final is greater than h at x initial. If the value of the function at the higher endpoint is larger than the value of the function at the lower endpoint, then we have a positive average rate of change. So let's see if that's happening for any of these choices."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "They're just asking us whether it is positive. And if you look over here, as long as our x final is greater than x initial, in order to have a positive average rate of change, we just need to figure out whether h at x final is greater than h at x initial. If the value of the function at the higher endpoint is larger than the value of the function at the lower endpoint, then we have a positive average rate of change. So let's see if that's happening for any of these choices. So let's see, h of zero, this endpoint, is going to be equal to zero. If I just say 1 1 8 times zero minus zero. And h of two is equal to 1 1 8 times two to the third power is eight."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "So let's see if that's happening for any of these choices. So let's see, h of zero, this endpoint, is going to be equal to zero. If I just say 1 1 8 times zero minus zero. And h of two is equal to 1 1 8 times two to the third power is eight. So 1 1 8 times eight is one minus four. So that's going to be, this is negative three. And so we don't have a situation where h at our endpoint, at our higher endpoint, is actually larger."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "And h of two is equal to 1 1 8 times two to the third power is eight. So 1 1 8 times eight is one minus four. So that's going to be, this is negative three. And so we don't have a situation where h at our endpoint, at our higher endpoint, is actually larger. This is a negative average rate of change. So I'll rule this one out. And actually, just to help us visualize this, I did go to Desmos and graph this function."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "And so we don't have a situation where h at our endpoint, at our higher endpoint, is actually larger. This is a negative average rate of change. So I'll rule this one out. And actually, just to help us visualize this, I did go to Desmos and graph this function. And we can visually see that we have a negative average rate of change from x equals zero to x equals two. At x equals zero, this is where our function is. At x equals two, this is where our function is."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "And actually, just to help us visualize this, I did go to Desmos and graph this function. And we can visually see that we have a negative average rate of change from x equals zero to x equals two. At x equals zero, this is where our function is. At x equals two, this is where our function is. And so you can see, at x equals two, our function has a lower value. You could also think of the average rate of change as the slope of the line that connects the two endpoints on the function. And so you can see it has a negative slope."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "At x equals two, this is where our function is. And so you can see, at x equals two, our function has a lower value. You could also think of the average rate of change as the slope of the line that connects the two endpoints on the function. And so you can see it has a negative slope. So we have a negative average rate of change between those two points. Now what about between these two? So h of zero, we already calculated as zero."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "And so you can see it has a negative slope. So we have a negative average rate of change between those two points. Now what about between these two? So h of zero, we already calculated as zero. And what is h of eight? Well, let's see, that's 1 8th times eight to the third power. Well, if I do eight to the third power, but then divide by eight, that's the same thing as eight to the second power."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "So h of zero, we already calculated as zero. And what is h of eight? Well, let's see, that's 1 8th times eight to the third power. Well, if I do eight to the third power, but then divide by eight, that's the same thing as eight to the second power. So that's going to be 64. Minus eight to the second power, minus 64. So that's equal to zero."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "Well, if I do eight to the third power, but then divide by eight, that's the same thing as eight to the second power. So that's going to be 64. Minus eight to the second power, minus 64. So that's equal to zero. So here, we have a zero average rate of change, because this numerator's going to be zero, so we can rule that out. And you see it right over here. When x is equal to zero, our function's there."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "So that's equal to zero. So here, we have a zero average rate of change, because this numerator's going to be zero, so we can rule that out. And you see it right over here. When x is equal to zero, our function's there. When x is equal to eight, our function is there. And you can see that the slope of the line that connects those two points is zero. So you have zero average rate of change between those two points."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "When x is equal to zero, our function's there. When x is equal to eight, our function is there. And you can see that the slope of the line that connects those two points is zero. So you have zero average rate of change between those two points. Now what about choice C? So let's see, h of six is going to be equal to 1 8th times six to the third power. So let's see, 36 times six is 180 plus 36."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "So you have zero average rate of change between those two points. Now what about choice C? So let's see, h of six is going to be equal to 1 8th times six to the third power. So let's see, 36 times six is 180 plus 36. So that is going to be 216. 216 minus 36. 216 is six times six times six."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "So let's see, 36 times six is 180 plus 36. So that is going to be 216. 216 minus 36. 216 is six times six times six. And then if we divide that by eight, that is going to be the same thing as. This is 36. And then we have 6 8ths of 36."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "216 is six times six times six. And then if we divide that by eight, that is going to be the same thing as. This is 36. And then we have 6 8ths of 36. So this is going to be, this is going to simplify to 3 4ths times 36 minus 36, which is going to be equal to negative nine. You could have done it with a calculator or done some long division. But hopefully what I just did makes some sense."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "And then we have 6 8ths of 36. So this is going to be, this is going to simplify to 3 4ths times 36 minus 36, which is going to be equal to negative nine. You could have done it with a calculator or done some long division. But hopefully what I just did makes some sense. It's a little bit of arithmetic. And so h of six, we have our function is negative nine. And then h of eight, I'll draw a line here so we don't make it too messy."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "But hopefully what I just did makes some sense. It's a little bit of arithmetic. And so h of six, we have our function is negative nine. And then h of eight, I'll draw a line here so we don't make it too messy. H of eight, we already know, is equal to zero. So our function at this end point is higher than the value of our function at this end point. So we do have a positive average rate of change."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "And then h of eight, I'll draw a line here so we don't make it too messy. H of eight, we already know, is equal to zero. So our function at this end point is higher than the value of our function at this end point. So we do have a positive average rate of change. So I would pick that choice right over there. And you could see it visually. H of six, when x is equal to six, our value of our function is negative nine."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "So we do have a positive average rate of change. So I would pick that choice right over there. And you could see it visually. H of six, when x is equal to six, our value of our function is negative nine. And when x is equal to eight, our value of our function is zero. And so the line that connects those two points definitely has a positive slope. So we have a positive average rate of change over that interval."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "H of six, when x is equal to six, our value of our function is negative nine. And when x is equal to eight, our value of our function is zero. And so the line that connects those two points definitely has a positive slope. So we have a positive average rate of change over that interval. Now, if we were just doing this on our own, we'd be done. But we could just check this one right over here. If we compare h of zero, we already know is zero."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "So we have a positive average rate of change over that interval. Now, if we were just doing this on our own, we'd be done. But we could just check this one right over here. If we compare h of zero, we already know is zero. And h of six, we already know, is equal to negative nine. So this is a negative average rate of change because at the higher end point right over here, we have a lower value of our function. So we'd rule this out."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "If we compare h of zero, we already know is zero. And h of six, we already know, is equal to negative nine. So this is a negative average rate of change because at the higher end point right over here, we have a lower value of our function. So we'd rule this out. And you see it right over here. If you go from x equals zero where the function is to x equals six where the function is, it looks something, it looks something like that. Clearly, that line has a negative slope."}, {"video_title": "Solving square-root equations no solution Mathematics III High School Math Khan Academy.mp3", "Sentence": "I encourage you to pause the video and see if you can solve for x before we work through it together. All right, so one thing we could do is we could try to isolate each of the radicals on either side of the equation. So let's subtract this one from both sides so I can get it onto the right-hand side, or a version of it on the right-hand side. So I'm subtracting it from the left-hand side and from the right-hand side. And so this is going to get us, that is going to get us. On the left-hand side, I just have square root, these cancel out, so I'm just left with the square root of three x minus seven is going to be equal to this, the negative of the square root of two x minus one. So now we can square both sides."}, {"video_title": "Solving square-root equations no solution Mathematics III High School Math Khan Academy.mp3", "Sentence": "So I'm subtracting it from the left-hand side and from the right-hand side. And so this is going to get us, that is going to get us. On the left-hand side, I just have square root, these cancel out, so I'm just left with the square root of three x minus seven is going to be equal to this, the negative of the square root of two x minus one. So now we can square both sides. And we always have to be careful when we're doing that because whether we're squaring the positive or the negative square root here, we're going to get the same value. So the solution we might get might be the version when we're solving for the positive square root, not when we take the negative of it. So we have to test our solutions at the end to make sure that they're actually valid for our original equation."}, {"video_title": "Solving square-root equations no solution Mathematics III High School Math Khan Academy.mp3", "Sentence": "So now we can square both sides. And we always have to be careful when we're doing that because whether we're squaring the positive or the negative square root here, we're going to get the same value. So the solution we might get might be the version when we're solving for the positive square root, not when we take the negative of it. So we have to test our solutions at the end to make sure that they're actually valid for our original equation. But if you square both sides, on the left-hand side, we're going to get three x minus seven. And on the right-hand side, a negative square is just a positive and the square root of two x minus one squared is going to be two x minus one. And I'll see, we can subtract two x from both sides to get all of our x's on one side."}, {"video_title": "Solving square-root equations no solution Mathematics III High School Math Khan Academy.mp3", "Sentence": "So we have to test our solutions at the end to make sure that they're actually valid for our original equation. But if you square both sides, on the left-hand side, we're going to get three x minus seven. And on the right-hand side, a negative square is just a positive and the square root of two x minus one squared is going to be two x minus one. And I'll see, we can subtract two x from both sides to get all of our x's on one side. So I'm trying to get rid of this. And we can add seven to both sides because I'm trying to get rid of the negative seven. So add seven to both sides."}, {"video_title": "Solving square-root equations no solution Mathematics III High School Math Khan Academy.mp3", "Sentence": "And I'll see, we can subtract two x from both sides to get all of our x's on one side. So I'm trying to get rid of this. And we can add seven to both sides because I'm trying to get rid of the negative seven. So add seven to both sides. And we are going to get, we are going to get three x minus two x is x is equal to negative one plus seven. X is equal to six. Now let's verify that this actually works."}, {"video_title": "Solving square-root equations no solution Mathematics III High School Math Khan Academy.mp3", "Sentence": "So add seven to both sides. And we are going to get, we are going to get three x minus two x is x is equal to negative one plus seven. X is equal to six. Now let's verify that this actually works. So if we look at our original equation, the square root of three times six minus seven, minus seven, needs to plus, plus the square root of two times six minus one needs to be equal to zero. So does this actually work out? Three times six minus seven."}, {"video_title": "Solving square-root equations no solution Mathematics III High School Math Khan Academy.mp3", "Sentence": "Now let's verify that this actually works. So if we look at our original equation, the square root of three times six minus seven, minus seven, needs to plus, plus the square root of two times six minus one needs to be equal to zero. So does this actually work out? Three times six minus seven. So this is going to be the square root of 11 plus the square root of 11 needs to be equal to zero, which clearly is not going to be equal to zero. This is two square roots of 11, which does not equal zero. So this does not work."}, {"video_title": "Solving square-root equations no solution Mathematics III High School Math Khan Academy.mp3", "Sentence": "Three times six minus seven. So this is going to be the square root of 11 plus the square root of 11 needs to be equal to zero, which clearly is not going to be equal to zero. This is two square roots of 11, which does not equal zero. So this does not work. And you might say, wait, how did this happen? I did all of this nice, neat algebra. I didn't make any mistakes."}, {"video_title": "Solving square-root equations no solution Mathematics III High School Math Khan Academy.mp3", "Sentence": "So this does not work. And you might say, wait, how did this happen? I did all of this nice, neat algebra. I didn't make any mistakes. But I got something that doesn't work. Well, this right here is an extraneous solution. Why is it an extraneous solution?"}, {"video_title": "Solving square-root equations no solution Mathematics III High School Math Khan Academy.mp3", "Sentence": "I didn't make any mistakes. But I got something that doesn't work. Well, this right here is an extraneous solution. Why is it an extraneous solution? Because it's actually the solution to the equation. It's a solution to the equation, the square root of three x minus seven minus the square root of two x minus one is equal to zero. And you might say, well, if it's the solution to that, if it's the solution to this thing right over here, how did I get the answer when I'm trying to do algebraic steps there?"}, {"video_title": "Solving square-root equations no solution Mathematics III High School Math Khan Academy.mp3", "Sentence": "Why is it an extraneous solution? Because it's actually the solution to the equation. It's a solution to the equation, the square root of three x minus seven minus the square root of two x minus one is equal to zero. And you might say, well, if it's the solution to that, if it's the solution to this thing right over here, how did I get the answer when I'm trying to do algebraic steps there? Well, the key is if when we added, when we took this onto the right-hand side and squared it, well, it all boiled down to this, regardless of what starting point you started with. If you did the exact same thing, you would have gotten to this point right over here. So the solution to this ended up being the solution to this starting point versus the one that we originally started with."}, {"video_title": "Solving square-root equations no solution Mathematics III High School Math Khan Academy.mp3", "Sentence": "And you might say, well, if it's the solution to that, if it's the solution to this thing right over here, how did I get the answer when I'm trying to do algebraic steps there? Well, the key is if when we added, when we took this onto the right-hand side and squared it, well, it all boiled down to this, regardless of what starting point you started with. If you did the exact same thing, you would have gotten to this point right over here. So the solution to this ended up being the solution to this starting point versus the one that we originally started with. So this one, interestingly, has no solutions. And it would actually be fun to think about why it has no solutions. We've shown to a certain degree the only solution you got by taking reasonable algebraic steps is an extraneous one."}, {"video_title": "Simplifying square roots Exponents, radicals, and scientific notation Pre-Algebra Khan Academy.mp3", "Sentence": "Let's see if we can simplify 5 times the square root of 117. So 117 doesn't jump out at me as some type of a perfect square. So let's actually take its prime factorization and see if any of those prime factors show up more than once. So it's clearly an odd number. It's clearly not divisible by 2. To test whether it's divisible by 3, we can add up all of the digits. And we explained why this works in another place on Khan Academy."}, {"video_title": "Simplifying square roots Exponents, radicals, and scientific notation Pre-Algebra Khan Academy.mp3", "Sentence": "So it's clearly an odd number. It's clearly not divisible by 2. To test whether it's divisible by 3, we can add up all of the digits. And we explained why this works in another place on Khan Academy. But if you add up all the digits, you get a 9. And 9 is divisible by 3. So 117 is going to be divisible by 3."}, {"video_title": "Simplifying square roots Exponents, radicals, and scientific notation Pre-Algebra Khan Academy.mp3", "Sentence": "And we explained why this works in another place on Khan Academy. But if you add up all the digits, you get a 9. And 9 is divisible by 3. So 117 is going to be divisible by 3. Now let's do a little aside here and figure out what 117 divided by 3 actually is. So 3 doesn't go into 1. It does go into 11 3 times."}, {"video_title": "Simplifying square roots Exponents, radicals, and scientific notation Pre-Algebra Khan Academy.mp3", "Sentence": "So 117 is going to be divisible by 3. Now let's do a little aside here and figure out what 117 divided by 3 actually is. So 3 doesn't go into 1. It does go into 11 3 times. 3 times 3 is 9. Subtract. You got a remainder of 2."}, {"video_title": "Simplifying square roots Exponents, radicals, and scientific notation Pre-Algebra Khan Academy.mp3", "Sentence": "It does go into 11 3 times. 3 times 3 is 9. Subtract. You got a remainder of 2. Bring down a 7. 3 goes into 27 9 times. 9 times 3 is 27."}, {"video_title": "Simplifying square roots Exponents, radicals, and scientific notation Pre-Algebra Khan Academy.mp3", "Sentence": "You got a remainder of 2. Bring down a 7. 3 goes into 27 9 times. 9 times 3 is 27. Subtract, and you're done. It goes in perfectly. And so we can factor 117 as 3 times 39."}, {"video_title": "Simplifying square roots Exponents, radicals, and scientific notation Pre-Algebra Khan Academy.mp3", "Sentence": "9 times 3 is 27. Subtract, and you're done. It goes in perfectly. And so we can factor 117 as 3 times 39. Now 39 we can factor as. That jumps out more at us. That's divisible by 3."}, {"video_title": "Simplifying square roots Exponents, radicals, and scientific notation Pre-Algebra Khan Academy.mp3", "Sentence": "And so we can factor 117 as 3 times 39. Now 39 we can factor as. That jumps out more at us. That's divisible by 3. That's equivalent to 3 times 13. And then all of these are now prime numbers. So we could say that this thing is the same as 5 times the square root of 3 times 3 times 13."}, {"video_title": "Simplifying square roots Exponents, radicals, and scientific notation Pre-Algebra Khan Academy.mp3", "Sentence": "That's divisible by 3. That's equivalent to 3 times 13. And then all of these are now prime numbers. So we could say that this thing is the same as 5 times the square root of 3 times 3 times 13. And this is going to be the same thing as, and we know this from our exponent properties, as 5 times the square root of 3 times 3 times the square root of 13. Now what's the square root of 3 times 3? Well, that's the square root of 9."}, {"video_title": "Simplifying square roots Exponents, radicals, and scientific notation Pre-Algebra Khan Academy.mp3", "Sentence": "So we could say that this thing is the same as 5 times the square root of 3 times 3 times 13. And this is going to be the same thing as, and we know this from our exponent properties, as 5 times the square root of 3 times 3 times the square root of 13. Now what's the square root of 3 times 3? Well, that's the square root of 9. That's the square root of 3 squared. Any of those, well, that's just going to give you 3. So this is just going to simplify to 3."}, {"video_title": "Simplifying square roots Exponents, radicals, and scientific notation Pre-Algebra Khan Academy.mp3", "Sentence": "Well, that's the square root of 9. That's the square root of 3 squared. Any of those, well, that's just going to give you 3. So this is just going to simplify to 3. So this whole thing is 5 times 3 times the square root of 13. So this part right over here would give us 15 times the square root of 13. Let's do one more example here."}, {"video_title": "Simplifying square roots Exponents, radicals, and scientific notation Pre-Algebra Khan Academy.mp3", "Sentence": "So this is just going to simplify to 3. So this whole thing is 5 times 3 times the square root of 13. So this part right over here would give us 15 times the square root of 13. Let's do one more example here. So let's try to simplify 3 times the square root of 26. Actually, I'm going to put 26 in yellow like I did in the previous problem. 26."}, {"video_title": "Simplifying square roots Exponents, radicals, and scientific notation Pre-Algebra Khan Academy.mp3", "Sentence": "Let's do one more example here. So let's try to simplify 3 times the square root of 26. Actually, I'm going to put 26 in yellow like I did in the previous problem. 26. Well, 26 is clearly an even number. So it's going to be divisible by 2. We can rewrite it as 2 times 13."}, {"video_title": "Simplifying square roots Exponents, radicals, and scientific notation Pre-Algebra Khan Academy.mp3", "Sentence": "26. Well, 26 is clearly an even number. So it's going to be divisible by 2. We can rewrite it as 2 times 13. And then we're done. 13 is a prime number. We can't factor this anymore."}, {"video_title": "Simplifying square roots Exponents, radicals, and scientific notation Pre-Algebra Khan Academy.mp3", "Sentence": "We can rewrite it as 2 times 13. And then we're done. 13 is a prime number. We can't factor this anymore. And so 26 doesn't have any perfect squares in it. It's not like we can factor it out as a factor of some other numbers and some perfect squares like we did here. 117 is 13 times 9."}, {"video_title": "Simplifying square roots Exponents, radicals, and scientific notation Pre-Algebra Khan Academy.mp3", "Sentence": "We can't factor this anymore. And so 26 doesn't have any perfect squares in it. It's not like we can factor it out as a factor of some other numbers and some perfect squares like we did here. 117 is 13 times 9. It's the product of a perfect square and 13. 26 isn't. So we've simplified this about as much as we can."}, {"video_title": "Graphing a parabola in vertex form Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So let me get my scratch pad out so we can think about this. So it's y is equal to negative 2 times x minus 2 squared plus 5. So one thing when you see a quadratic or a parabola expressed in this way, the thing that might jump out at you is that this term right over here, this term right over here, is always going to be positive because it's some quantity squared. Or I should say it's always going to be non-negative. It could be equal to 0. So it's always going to be some quantity squared and then we're multiplying it by a negative. So this whole quantity right over here is going to be non-positive."}, {"video_title": "Graphing a parabola in vertex form Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Or I should say it's always going to be non-negative. It could be equal to 0. So it's always going to be some quantity squared and then we're multiplying it by a negative. So this whole quantity right over here is going to be non-positive. It's always going to be less than or equal to 0. So if this thing is always less than or equal to 0, the maximum value that y will take on is when this thing actually does equal 0. So the maximum value for y is at 5."}, {"video_title": "Graphing a parabola in vertex form Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So this whole quantity right over here is going to be non-positive. It's always going to be less than or equal to 0. So if this thing is always less than or equal to 0, the maximum value that y will take on is when this thing actually does equal 0. So the maximum value for y is at 5. So y, the maximum value for y is 5. And when does that happen? Well, y hits 5 when this whole thing is 0."}, {"video_title": "Graphing a parabola in vertex form Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So the maximum value for y is at 5. So y, the maximum value for y is 5. And when does that happen? Well, y hits 5 when this whole thing is 0. And when does this thing equal 0? Well, this whole thing equals 0 when x minus 2 is equal to 0 and x minus 2 is equal to 0 when x is equal to 2. So the point 2,5 is the maximum point for this parabola and it is actually going to be the vertex."}, {"video_title": "Graphing a parabola in vertex form Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Well, y hits 5 when this whole thing is 0. And when does this thing equal 0? Well, this whole thing equals 0 when x minus 2 is equal to 0 and x minus 2 is equal to 0 when x is equal to 2. So the point 2,5 is the maximum point for this parabola and it is actually going to be the vertex. So if we were to graph this, so the point 2,5. So that's my y-axis. This is my x-axis."}, {"video_title": "Graphing a parabola in vertex form Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So the point 2,5 is the maximum point for this parabola and it is actually going to be the vertex. So if we were to graph this, so the point 2,5. So that's my y-axis. This is my x-axis. So this is 1, 2, 1, 2, 3, 4, 5. So this right over here is a point 2,5. This is a maximum point."}, {"video_title": "Graphing a parabola in vertex form Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "This is my x-axis. So this is 1, 2, 1, 2, 3, 4, 5. So this right over here is a point 2,5. This is a maximum point. It's a maximum point for this parabola. And now I want to find two more points so that I can really determine the parabola. Three points determine, completely determine a parabola."}, {"video_title": "Graphing a parabola in vertex form Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "This is a maximum point. It's a maximum point for this parabola. And now I want to find two more points so that I can really determine the parabola. Three points determine, completely determine a parabola. So that's one, the vertex. That's interesting. Now what I'd like to do is just get two points that are equidistant from the vertex."}, {"video_title": "Graphing a parabola in vertex form Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Three points determine, completely determine a parabola. So that's one, the vertex. That's interesting. Now what I'd like to do is just get two points that are equidistant from the vertex. And the easiest way to do that is to maybe figure out what happens when x is equal to, when x is equal to 1 and when x is equal to 3. So I can make a table here actually. Let me do that."}, {"video_title": "Graphing a parabola in vertex form Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Now what I'd like to do is just get two points that are equidistant from the vertex. And the easiest way to do that is to maybe figure out what happens when x is equal to, when x is equal to 1 and when x is equal to 3. So I can make a table here actually. Let me do that. So I care about x being equal to 1, 2, and 3 and what the corresponding y is. We already know that when x is equal to 2, y is equal to 5. 2,5 is our vertex."}, {"video_title": "Graphing a parabola in vertex form Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Let me do that. So I care about x being equal to 1, 2, and 3 and what the corresponding y is. We already know that when x is equal to 2, y is equal to 5. 2,5 is our vertex. When x is equal to 1, 1 minus 2 is negative 1, squared is just 1. So this thing is going to be negative 2 plus 5. So it's going to be 3."}, {"video_title": "Graphing a parabola in vertex form Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "2,5 is our vertex. When x is equal to 1, 1 minus 2 is negative 1, squared is just 1. So this thing is going to be negative 2 plus 5. So it's going to be 3. And when x is equal to 3, this is 3 minus 2, which is 1, squared is 1, times negative 2 is negative 2 plus 5 is 3 as well. So we have three points. We have the point 1,3, 1,1,3, the point 2,5, and the point 3,3,3 for this parabola."}, {"video_title": "Example 2 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Use completing the square to find the roots of the quadratic equation right here. Now when anyone talks about roots, this just means find the x's where y is equal to 0. That's what a root is. A root is an x value that will make this quadratic function equal 0. That will make y equal 0. So to find the x's, let's just make y equal 0 and then solve for x. So we get 0 is equal to 4x squared plus 40x plus 280."}, {"video_title": "Example 2 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "A root is an x value that will make this quadratic function equal 0. That will make y equal 0. So to find the x's, let's just make y equal 0 and then solve for x. So we get 0 is equal to 4x squared plus 40x plus 280. Now the first step that we might want to do, just because it looks like all three of these terms are divisible by 4, is just divide both sides of this equation by 4. That will make our math a little bit simpler. So let's just divide everything by 4 here."}, {"video_title": "Example 2 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So we get 0 is equal to 4x squared plus 40x plus 280. Now the first step that we might want to do, just because it looks like all three of these terms are divisible by 4, is just divide both sides of this equation by 4. That will make our math a little bit simpler. So let's just divide everything by 4 here. If we just divide everything by 4, we get 0 is equal to x squared plus 10x plus 280 divided by 4 is 70. Now, they say use completing the square. Actually, let me write that 70 a little bit further out."}, {"video_title": "Example 2 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So let's just divide everything by 4 here. If we just divide everything by 4, we get 0 is equal to x squared plus 10x plus 280 divided by 4 is 70. Now, they say use completing the square. Actually, let me write that 70 a little bit further out. You'll see why I did that in a second. So let me just write a plus 70 over here. Just have kind of an awkward space here."}, {"video_title": "Example 2 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Actually, let me write that 70 a little bit further out. You'll see why I did that in a second. So let me just write a plus 70 over here. Just have kind of an awkward space here. You'll see what I'm about to do with this space. That has everything to do with completing the square. So they say use completing the square, which means turn this, if you can, into a perfect square."}, {"video_title": "Example 2 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Just have kind of an awkward space here. You'll see what I'm about to do with this space. That has everything to do with completing the square. So they say use completing the square, which means turn this, if you can, into a perfect square. Turn at least part of this expression into a perfect square. Then we can use that to actually solve for x. So how do we turn this into a perfect square?"}, {"video_title": "Example 2 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So they say use completing the square, which means turn this, if you can, into a perfect square. Turn at least part of this expression into a perfect square. Then we can use that to actually solve for x. So how do we turn this into a perfect square? Well, we have a 10x here. We know that we can turn this into a perfect square trinomial if we take half of the 10, which is 5, and then we square that. So half of 10 is 5."}, {"video_title": "Example 2 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So how do we turn this into a perfect square? Well, we have a 10x here. We know that we can turn this into a perfect square trinomial if we take half of the 10, which is 5, and then we square that. So half of 10 is 5. You square it. You add a 25. Now, you can't just willy-nilly add a 25 to one side of the equation without doing something to the other or without just subtracting the 25 right here."}, {"video_title": "Example 2 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So half of 10 is 5. You square it. You add a 25. Now, you can't just willy-nilly add a 25 to one side of the equation without doing something to the other or without just subtracting the 25 right here. Think about it. I have not changed the equation. I have added 25 and I have subtracted 25."}, {"video_title": "Example 2 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Now, you can't just willy-nilly add a 25 to one side of the equation without doing something to the other or without just subtracting the 25 right here. Think about it. I have not changed the equation. I have added 25 and I have subtracted 25. So I've added nothing to the right-hand side. I could add a billion and subtract a billion and not change the equation. So I have not changed the equation at all right here."}, {"video_title": "Example 2 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "I have added 25 and I have subtracted 25. So I've added nothing to the right-hand side. I could add a billion and subtract a billion and not change the equation. So I have not changed the equation at all right here. But what I have done is I've made it possible to express these three terms as a perfect square. That right there, 2 times 5 is 10. 5 squared is 25."}, {"video_title": "Example 2 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So I have not changed the equation at all right here. But what I have done is I've made it possible to express these three terms as a perfect square. That right there, 2 times 5 is 10. 5 squared is 25. So that is x plus 5 squared. And if you don't believe me, multiply it out. You're going to have x squared plus 5x plus 5x, which will give you 10x, plus 5 squared, which is 25."}, {"video_title": "Example 2 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "5 squared is 25. So that is x plus 5 squared. And if you don't believe me, multiply it out. You're going to have x squared plus 5x plus 5x, which will give you 10x, plus 5 squared, which is 25. So those first three terms become that. And then the second two terms right there, you just add them. Let's see, negative 25 plus 70 would be positive 50."}, {"video_title": "Example 2 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "You're going to have x squared plus 5x plus 5x, which will give you 10x, plus 5 squared, which is 25. So those first three terms become that. And then the second two terms right there, you just add them. Let's see, negative 25 plus 70 would be positive 50. And then you have another 5, so it's plus 45. So we've just algebraically manipulated this equation. And we get 0 is equal to x plus 5 squared plus 45."}, {"video_title": "Example 2 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Let's see, negative 25 plus 70 would be positive 50. And then you have another 5, so it's plus 45. So we've just algebraically manipulated this equation. And we get 0 is equal to x plus 5 squared plus 45. Now, we could have, from the beginning if we wanted, we could have tried to factor it. But what we're going to do here, this will always work. Even if you have crazy decimal numbers here, you can solve for x using the method we're doing here, completing the square."}, {"video_title": "Example 2 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "And we get 0 is equal to x plus 5 squared plus 45. Now, we could have, from the beginning if we wanted, we could have tried to factor it. But what we're going to do here, this will always work. Even if you have crazy decimal numbers here, you can solve for x using the method we're doing here, completing the square. So to solve for x, let's just subtract 45 from both sides of this equation. Let's subtract 45. And so the left-hand side of this equation becomes negative 45."}, {"video_title": "Example 2 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Even if you have crazy decimal numbers here, you can solve for x using the method we're doing here, completing the square. So to solve for x, let's just subtract 45 from both sides of this equation. Let's subtract 45. And so the left-hand side of this equation becomes negative 45. And the right-hand side will be just the x plus 5 squared. These guys right here cancel out. Now, normally if I look at something like this, I'll say, OK, let's just take the square root of both sides of this equation."}, {"video_title": "Example 2 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "And so the left-hand side of this equation becomes negative 45. And the right-hand side will be just the x plus 5 squared. These guys right here cancel out. Now, normally if I look at something like this, I'll say, OK, let's just take the square root of both sides of this equation. And so you might be tempted to take the square root of both sides of this equation. But immediately when you do that, you'll notice something strange. We're trying to take the square root of a negative number."}, {"video_title": "Example 2 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Now, normally if I look at something like this, I'll say, OK, let's just take the square root of both sides of this equation. And so you might be tempted to take the square root of both sides of this equation. But immediately when you do that, you'll notice something strange. We're trying to take the square root of a negative number. And if we're dealing with real numbers, which is everything we've dealt with so far, you can't take a square root of a negative number. There is no real number that if you square it will give you a negative number. So it is not possible."}, {"video_title": "Example 2 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "We're trying to take the square root of a negative number. And if we're dealing with real numbers, which is everything we've dealt with so far, you can't take a square root of a negative number. There is no real number that if you square it will give you a negative number. So it is not possible. I don't care what you make x. It is not possible to add x to 5 and square it and get a negative number. So there is no x that can satisfy, if we assume that x is a real number, that can satisfy this equation."}, {"video_title": "Example 2 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So it is not possible. I don't care what you make x. It is not possible to add x to 5 and square it and get a negative number. So there is no x that can satisfy, if we assume that x is a real number, that can satisfy this equation. Because I don't care what x you put here, what real x you put here, you add 5 to it and square it, there's no way you're going to get a negative number. So there's no x that can satisfy this equation. So there are, we could say there are no, and I'm using the word real because in Algebra 2 you'll learn that there are things called complex numbers."}, {"video_title": "Example 2 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So there is no x that can satisfy, if we assume that x is a real number, that can satisfy this equation. Because I don't care what x you put here, what real x you put here, you add 5 to it and square it, there's no way you're going to get a negative number. So there's no x that can satisfy this equation. So there are, we could say there are no, and I'm using the word real because in Algebra 2 you'll learn that there are things called complex numbers. Don't worry about them right now. But there are no real roots to the quadratic equation. And we're done."}, {"video_title": "Evaluating functions given their formula Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "The function f of x is defined as f of x is equal to 49 minus x squared. Find the value of f of 5. So whenever you're dealing with a function, you take your input, in this case our input is going to be our 5, we input it into our little function box, and we need to get our output. And they define the function box here. Whatever your input is, take that, square it, and then subtract it from 49. So f of 5, every time I see an x here, since f of x is equal to this, every time I see an x, I would replace it with the input. So f of 5 is going to be equal to 49 minus, instead of writing x squared, I would write 5 squared."}, {"video_title": "Slope from two ordered pairs example 2 Algebra I Khan Academy.mp3", "Sentence": "Find the slope of the line that goes through the ordered pairs 7, negative 1 and negative 3, negative 1. Let me just do a quick graph of these just so we can visualize what they look like. So let me draw a quick graph over here. So our first point is 7, negative 1. So 1, 2, 3, 4, 5, 6, 7. This is the x-axis. 7, negative 1."}, {"video_title": "Slope from two ordered pairs example 2 Algebra I Khan Academy.mp3", "Sentence": "So our first point is 7, negative 1. So 1, 2, 3, 4, 5, 6, 7. This is the x-axis. 7, negative 1. So 7, negative 1 is right over there. 7, negative 1. This of course is the y-axis."}, {"video_title": "Slope from two ordered pairs example 2 Algebra I Khan Academy.mp3", "Sentence": "7, negative 1. So 7, negative 1 is right over there. 7, negative 1. This of course is the y-axis. And then the next point is negative 3, negative 1. So we go back 3 in the horizontal direction. Negative 3."}, {"video_title": "Slope from two ordered pairs example 2 Algebra I Khan Academy.mp3", "Sentence": "This of course is the y-axis. And then the next point is negative 3, negative 1. So we go back 3 in the horizontal direction. Negative 3. But the y-coordinate is still negative 1. It's still negative 1. So the line that connects these two points will look like this."}, {"video_title": "Slope from two ordered pairs example 2 Algebra I Khan Academy.mp3", "Sentence": "Negative 3. But the y-coordinate is still negative 1. It's still negative 1. So the line that connects these two points will look like this. It will look like that. Now, they're asking us to find the slope of the line that goes through the ordered pairs. Find the slope of this line."}, {"video_title": "Slope from two ordered pairs example 2 Algebra I Khan Academy.mp3", "Sentence": "So the line that connects these two points will look like this. It will look like that. Now, they're asking us to find the slope of the line that goes through the ordered pairs. Find the slope of this line. And just to give a little bit of intuition here, slope is a measure of a line's inclination. And the way that it's defined, slope is defined as rise over run, or change in y over change in x, or sometimes you'll see it defined as the variable m, and then they'll define change in y as just being the second y-coordinate minus the first y-coordinate, and then the change in x as the second x-coordinate minus the first x-coordinate. These are all different variations in slope, but hopefully you appreciate that these are measuring inclination."}, {"video_title": "Slope from two ordered pairs example 2 Algebra I Khan Academy.mp3", "Sentence": "Find the slope of this line. And just to give a little bit of intuition here, slope is a measure of a line's inclination. And the way that it's defined, slope is defined as rise over run, or change in y over change in x, or sometimes you'll see it defined as the variable m, and then they'll define change in y as just being the second y-coordinate minus the first y-coordinate, and then the change in x as the second x-coordinate minus the first x-coordinate. These are all different variations in slope, but hopefully you appreciate that these are measuring inclination. If I rise a ton when I run a little bit, if I move a little bit in the x-direction and I rise a bunch, then I have a very steep line. I have a very steep upward sloping line. If I don't change at all when I run a bit, then I have a very low slope, and that's actually what's happening here."}, {"video_title": "Slope from two ordered pairs example 2 Algebra I Khan Academy.mp3", "Sentence": "These are all different variations in slope, but hopefully you appreciate that these are measuring inclination. If I rise a ton when I run a little bit, if I move a little bit in the x-direction and I rise a bunch, then I have a very steep line. I have a very steep upward sloping line. If I don't change at all when I run a bit, then I have a very low slope, and that's actually what's happening here. I'm going from, you could either view this as the starting point or view this as the starting point, but let's view this as the starting point. So this negative 3 comma 1. If I go from negative 3 comma negative 1 to 7 comma negative 1, I'm running a good bit."}, {"video_title": "Slope from two ordered pairs example 2 Algebra I Khan Academy.mp3", "Sentence": "If I don't change at all when I run a bit, then I have a very low slope, and that's actually what's happening here. I'm going from, you could either view this as the starting point or view this as the starting point, but let's view this as the starting point. So this negative 3 comma 1. If I go from negative 3 comma negative 1 to 7 comma negative 1, I'm running a good bit. I'm going from negative 3, my x value is negative 3 here, and it goes all the way to 7. So my change in x here is 10. To go from negative 3 to 7, I change my x value by 10."}, {"video_title": "Slope from two ordered pairs example 2 Algebra I Khan Academy.mp3", "Sentence": "If I go from negative 3 comma negative 1 to 7 comma negative 1, I'm running a good bit. I'm going from negative 3, my x value is negative 3 here, and it goes all the way to 7. So my change in x here is 10. To go from negative 3 to 7, I change my x value by 10. But what's my change in y? Well, my y value here is negative 1, and my y value over here is still negative 1. So my change in y is 0."}, {"video_title": "Slope from two ordered pairs example 2 Algebra I Khan Academy.mp3", "Sentence": "To go from negative 3 to 7, I change my x value by 10. But what's my change in y? Well, my y value here is negative 1, and my y value over here is still negative 1. So my change in y is 0. My change in y is going to be 0. My y value does not change no matter how much I change my x value. So the slope here is going to be, when we run 10, what was our rise?"}, {"video_title": "Slope from two ordered pairs example 2 Algebra I Khan Academy.mp3", "Sentence": "So my change in y is 0. My change in y is going to be 0. My y value does not change no matter how much I change my x value. So the slope here is going to be, when we run 10, what was our rise? How much did we change in y? Well, we didn't rise at all. We didn't go up or down."}, {"video_title": "Slope from two ordered pairs example 2 Algebra I Khan Academy.mp3", "Sentence": "So the slope here is going to be, when we run 10, what was our rise? How much did we change in y? Well, we didn't rise at all. We didn't go up or down. So the slope here is 0. Or another way to think about it is this line has no inclination. It's a completely flat, it's a completely horizontal line."}, {"video_title": "Slope from two ordered pairs example 2 Algebra I Khan Academy.mp3", "Sentence": "We didn't go up or down. So the slope here is 0. Or another way to think about it is this line has no inclination. It's a completely flat, it's a completely horizontal line. So this should make sense. This is a 0. The slope here is 0."}, {"video_title": "Slope from two ordered pairs example 2 Algebra I Khan Academy.mp3", "Sentence": "It's a completely flat, it's a completely horizontal line. So this should make sense. This is a 0. The slope here is 0. And just to make sure that this gels with all of these other formulas that you might know, but I want to make it very clear, these are all just telling you rise over run, or change in y over change in x, a way to measure inclination. But let's just apply them just so hopefully it all makes sense to you. So we could also say slope is change in y over change in x."}, {"video_title": "Slope from two ordered pairs example 2 Algebra I Khan Academy.mp3", "Sentence": "The slope here is 0. And just to make sure that this gels with all of these other formulas that you might know, but I want to make it very clear, these are all just telling you rise over run, or change in y over change in x, a way to measure inclination. But let's just apply them just so hopefully it all makes sense to you. So we could also say slope is change in y over change in x. If we take this to be our start, and if we take this to be our end point, then we would call this over here x1, and then this is over here, this is y1, and then we would call this x2, and we would call this y2. If this is our start point and that is our end point. And so the slope here, the change in y, y2 minus y1, so it's negative 1 minus negative 1."}, {"video_title": "Slope from two ordered pairs example 2 Algebra I Khan Academy.mp3", "Sentence": "So we could also say slope is change in y over change in x. If we take this to be our start, and if we take this to be our end point, then we would call this over here x1, and then this is over here, this is y1, and then we would call this x2, and we would call this y2. If this is our start point and that is our end point. And so the slope here, the change in y, y2 minus y1, so it's negative 1 minus negative 1. All of that over x2, negative 3 minus x1, minus 7. So the numerator, negative 1 minus negative 1, that's the same thing as negative 1 plus 1. And our denominator is negative 3 minus 7, which is negative 10."}, {"video_title": "Slope from two ordered pairs example 2 Algebra I Khan Academy.mp3", "Sentence": "And so the slope here, the change in y, y2 minus y1, so it's negative 1 minus negative 1. All of that over x2, negative 3 minus x1, minus 7. So the numerator, negative 1 minus negative 1, that's the same thing as negative 1 plus 1. And our denominator is negative 3 minus 7, which is negative 10. So once again, negative 1 plus 1 is 0 over negative 10. And this is still going to be 0. And the only reason why we have a negative 10 here and a positive 10 there is because we swapped the starting and the ending points."}, {"video_title": "Slope from two ordered pairs example 2 Algebra I Khan Academy.mp3", "Sentence": "And our denominator is negative 3 minus 7, which is negative 10. So once again, negative 1 plus 1 is 0 over negative 10. And this is still going to be 0. And the only reason why we have a negative 10 here and a positive 10 there is because we swapped the starting and the ending points. In this example right over here, we took this as the start point and made this coordinate over here as the end point. Over here we swapped them around. 7, negative 1 was our start point, and negative 3, negative 1 is our end point."}, {"video_title": "Slope from two ordered pairs example 2 Algebra I Khan Academy.mp3", "Sentence": "And the only reason why we have a negative 10 here and a positive 10 there is because we swapped the starting and the ending points. In this example right over here, we took this as the start point and made this coordinate over here as the end point. Over here we swapped them around. 7, negative 1 was our start point, and negative 3, negative 1 is our end point. So if we start over here, our change in x is going to be negative 10, but our change in y is still going to be 0. So regardless of how you do it, the slope of this line is 0. It's a horizontal line."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "And like always, pause the video and see if you can work through it on your own. All right, well, what I like to do is I like to focus on this minimum point because I think that's a very easy thing to look at because both of them have that minimum point right over there. And so we can think about how do we shift f, at least, especially this minimum point, how do we shift it to get to overlapping with g? Well, the first thing that might jump out at us is that we would want to shift to the left. And we'd want to shift to the left four. So let me do this in a new color. So I would want to shift to the left by four."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Well, the first thing that might jump out at us is that we would want to shift to the left. And we'd want to shift to the left four. So let me do this in a new color. So I would want to shift to the left by four. So we have shifted to the left by four, or you could say we shifted by negative four. Either way, you could think about it. And then we need to shift down."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So I would want to shift to the left by four. So we have shifted to the left by four, or you could say we shifted by negative four. Either way, you could think about it. And then we need to shift down. So we need to shift, we need to go from y equals two to y is equal to negative five. So let me do that. So let's shift down."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "And then we need to shift down. So we need to shift, we need to go from y equals two to y is equal to negative five. So let me do that. So let's shift down. So we shift down by seven, or you could say we have a negative seven shift. So how do you express g of x if it's a version of f of x that's shifted to the left by four and shifted down by seven? Or you could say it had a negative four horizontal shift and had a negative seven vertical shift."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So let's shift down. So we shift down by seven, or you could say we have a negative seven shift. So how do you express g of x if it's a version of f of x that's shifted to the left by four and shifted down by seven? Or you could say it had a negative four horizontal shift and had a negative seven vertical shift. Well, one way to think about it is g of x is going to be equal to f of, let me do it in a little darker color. It's going to be equal to f of x minus your horizontal shift. I'll write horizontal shift."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Or you could say it had a negative four horizontal shift and had a negative seven vertical shift. Well, one way to think about it is g of x is going to be equal to f of, let me do it in a little darker color. It's going to be equal to f of x minus your horizontal shift. I'll write horizontal shift. So x minus your horizontal shift plus your vertical shift. So plus your vertical shift. Well, what is our horizontal shift here?"}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "I'll write horizontal shift. So x minus your horizontal shift plus your vertical shift. So plus your vertical shift. Well, what is our horizontal shift here? Well, we're shifting to the left, so it was a negative shift. So our horizontal shift is negative four. Now what's our vertical shift?"}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Well, what is our horizontal shift here? Well, we're shifting to the left, so it was a negative shift. So our horizontal shift is negative four. Now what's our vertical shift? Well, we went down, so our vertical shift is negative seven. So it's negative seven. So there you have it."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Now what's our vertical shift? Well, we went down, so our vertical shift is negative seven. So it's negative seven. So there you have it. We get g of x, let me do it in that same color. We get g of x is equal to f of x minus negative four, or x plus four. And then we have plus negative seven, or you could just say minus seven."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So there you have it. We get g of x, let me do it in that same color. We get g of x is equal to f of x minus negative four, or x plus four. And then we have plus negative seven, or you could just say minus seven. And we're done. And when I look at things like this, the negative seven is more intuitive to me. It's that I shifted it down, it makes sense that I have a negative seven."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "And then we have plus negative seven, or you could just say minus seven. And we're done. And when I look at things like this, the negative seven is more intuitive to me. It's that I shifted it down, it makes sense that I have a negative seven. But at first when you work on these, you say, hey, wait, I shifted to the left. Why is it a plus four? And the way I think about it is in order to get the same value out of the function, instead of inputting, so if you want to get the value of f of zero, you now have to put x equals negative four in, and then you get that same value."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "It's that I shifted it down, it makes sense that I have a negative seven. But at first when you work on these, you say, hey, wait, I shifted to the left. Why is it a plus four? And the way I think about it is in order to get the same value out of the function, instead of inputting, so if you want to get the value of f of zero, you now have to put x equals negative four in, and then you get that same value. You still get to zero. So I don't know if that helps or hurts in terms of your understanding, but it often helps to try out some different values for x and seeing how it actually does shift the function. And if you're just trying to get your head around this piece, the horizontal shift, I recommend not even using this example."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "And the way I think about it is in order to get the same value out of the function, instead of inputting, so if you want to get the value of f of zero, you now have to put x equals negative four in, and then you get that same value. You still get to zero. So I don't know if that helps or hurts in terms of your understanding, but it often helps to try out some different values for x and seeing how it actually does shift the function. And if you're just trying to get your head around this piece, the horizontal shift, I recommend not even using this example. Use an example that only has a horizontal shift, and it'll become a little bit more intuitive. And we have many videos that go into much more depth that explain that. Let's do another example of this."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "And if you're just trying to get your head around this piece, the horizontal shift, I recommend not even using this example. Use an example that only has a horizontal shift, and it'll become a little bit more intuitive. And we have many videos that go into much more depth that explain that. Let's do another example of this. So here we have y is equal to g of x in purple, and y is equal to f of x in blue. And they say given that f of x is equal to square root of x plus four minus two, write an expression for g of x in terms of x. So first, let me just write an expression for g of x in terms of f of x."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Let's do another example of this. So here we have y is equal to g of x in purple, and y is equal to f of x in blue. And they say given that f of x is equal to square root of x plus four minus two, write an expression for g of x in terms of x. So first, let me just write an expression for g of x in terms of f of x. We can see once again, it's just a shifted version of f of x. And remember, and I'll just write in general, so g of x is going to be equal to f of x minus your horizontal shift plus your vertical shift. Vertical shift."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So first, let me just write an expression for g of x in terms of f of x. We can see once again, it's just a shifted version of f of x. And remember, and I'll just write in general, so g of x is going to be equal to f of x minus your horizontal shift plus your vertical shift. Vertical shift. And so to go from f to g, what is your horizontal shift? Well, your horizontal shift is, if you take this point right over here, which should map to that point once we shift everything, your horizontal shift is two to the left. Or you could say it's a negative two horizontal shift, so that should be negative two."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Vertical shift. And so to go from f to g, what is your horizontal shift? Well, your horizontal shift is, if you take this point right over here, which should map to that point once we shift everything, your horizontal shift is two to the left. Or you could say it's a negative two horizontal shift, so that should be negative two. And then what is our vertical shift? Well, our vertical shift is we move, we go from y equals negative two to y equals three. So we're shifting five up."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Or you could say it's a negative two horizontal shift, so that should be negative two. And then what is our vertical shift? Well, our vertical shift is we move, we go from y equals negative two to y equals three. So we're shifting five up. So this is a vertical shift of positive five. So your vertical shift is five. So if we just wanted to write g of x in terms of f of x, like we just did in the previous example, we could say g of x is going to be equal to f of x minus negative two, which is x plus two, and then we have plus five."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So we're shifting five up. So this is a vertical shift of positive five. So your vertical shift is five. So if we just wanted to write g of x in terms of f of x, like we just did in the previous example, we could say g of x is going to be equal to f of x minus negative two, which is x plus two, and then we have plus five. But that's not what they asked us to do. They asked us to write, whoops, they asked us to write an expression for g of x in terms of x. And so here we're actually going to use the definition of f of x."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So if we just wanted to write g of x in terms of f of x, like we just did in the previous example, we could say g of x is going to be equal to f of x minus negative two, which is x plus two, and then we have plus five. But that's not what they asked us to do. They asked us to write, whoops, they asked us to write an expression for g of x in terms of x. And so here we're actually going to use the definition of f of x. So let me make it clear. We know that f of x, f of x, is going to be equal to square root of x plus four minus two. So given that, given that, what is f of x plus two?"}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "And so here we're actually going to use the definition of f of x. So let me make it clear. We know that f of x, f of x, is going to be equal to square root of x plus four minus two. So given that, given that, what is f of x plus two? Well, f of x plus two is going to be equal to, everywhere we see an x, we're going to replace it with an x plus two. Square root of x plus two plus four minus two, which is equal to the square root of x plus six minus two. Well, that's fair enough."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So given that, given that, what is f of x plus two? Well, f of x plus two is going to be equal to, everywhere we see an x, we're going to replace it with an x plus two. Square root of x plus two plus four minus two, which is equal to the square root of x plus six minus two. Well, that's fair enough. That's just f of x plus two. Now what is f of x plus two plus five? So f of x plus two plus five is going to be this thing right over here, plus five."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Well, that's fair enough. That's just f of x plus two. Now what is f of x plus two plus five? So f of x plus two plus five is going to be this thing right over here, plus five. So it's going to be equal to square root of x plus six minus two, and now we're going to add five. Let me do it in a different color. So plus five, so plus five."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So f of x plus two plus five is going to be this thing right over here, plus five. So it's going to be equal to square root of x plus six minus two, and now we're going to add five. Let me do it in a different color. So plus five, so plus five. And so what we end up with is going to be square root of x plus six minus two plus five is going to be plus three. So that is equal to g of x. Just as a reminder, what did we do here?"}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So plus five, so plus five. And so what we end up with is going to be square root of x plus six minus two plus five is going to be plus three. So that is equal to g of x. Just as a reminder, what did we do here? First I expressed g of x in terms of f of x. I said, hey, to get from f of x to g of x, I shift two to the left, two to the left. It's a little counterintuitive that it's plus two makes it a shift of two to the left. If this was minus two, it would be a shift of two to the right, but like I just said in the previous example, it's good to try out some x's and to see why that makes sense."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Just as a reminder, what did we do here? First I expressed g of x in terms of f of x. I said, hey, to get from f of x to g of x, I shift two to the left, two to the left. It's a little counterintuitive that it's plus two makes it a shift of two to the left. If this was minus two, it would be a shift of two to the right, but like I just said in the previous example, it's good to try out some x's and to see why that makes sense. And then we shifted five up. So this was g of x in terms of f of x. But then they told us what f of x actually is in terms of x."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "If this was minus two, it would be a shift of two to the right, but like I just said in the previous example, it's good to try out some x's and to see why that makes sense. And then we shifted five up. So this was g of x in terms of f of x. But then they told us what f of x actually is in terms of x. So I said, okay, well, what is f of x plus two? F of x plus two, we substituted x plus two for x and we got this, but g of x is f of x plus two plus five. So we took what we figured out f of x plus two is and then we added five and that's what g of x is."}, {"video_title": "Positive and negative intervals examples Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Highlight an interval where f of x is less than 0. So f of x, which is really being plotted on the vertical axis right over here, x is the horizontal axis. f of x being less than 0 really means that the graph is below the x-axis. So the graph, the function, is negative in this interval right over here and this interval over here. So I could put this anywhere right over here, or I could stick it anywhere right over here. Let me stick it right over here. There we go."}, {"video_title": "Positive and negative intervals examples Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So the graph, the function, is negative in this interval right over here and this interval over here. So I could put this anywhere right over here, or I could stick it anywhere right over here. Let me stick it right over here. There we go. Got it right. Let's do a couple more. So the function is plotted below."}, {"video_title": "Positive and negative intervals examples Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "There we go. Got it right. Let's do a couple more. So the function is plotted below. Highlight an interval where f of x is greater than 0. So I could do this area right over here where the function is above the x-axis, or I could do this area right over here where the function goes way above the x-axis. Wow, it even goes off the page."}, {"video_title": "Positive and negative intervals examples Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So the function is plotted below. Highlight an interval where f of x is greater than 0. So I could do this area right over here where the function is above the x-axis, or I could do this area right over here where the function goes way above the x-axis. Wow, it even goes off the page. So let's stick it right over here. Let's do one more. So highlight an interval where f of x is greater than 0."}, {"video_title": "Positive and negative intervals examples Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Wow, it even goes off the page. So let's stick it right over here. Let's do one more. So highlight an interval where f of x is greater than 0. Once again, I can do this region right over here where the function is above the x-axis, or over here where it's above the x-axis. I'll do it here just for fun. There we go."}, {"video_title": "Shifting absolute value graphs Mathematics II High School Math Khan Academy.mp3", "Sentence": "Negative one, absolute value is one. Zero, absolute value is zero. One, absolute value is one. So on and so forth. What I want to do in this video is think about how the equation will change if we were to shift this graph. So in particular, we're gonna first think about what would be the equation of this graph if we shift, if we shift three to the right and then think about how that will change if not only do we shift three to the right, but we also shift four up. Shift four up."}, {"video_title": "Shifting absolute value graphs Mathematics II High School Math Khan Academy.mp3", "Sentence": "So on and so forth. What I want to do in this video is think about how the equation will change if we were to shift this graph. So in particular, we're gonna first think about what would be the equation of this graph if we shift, if we shift three to the right and then think about how that will change if not only do we shift three to the right, but we also shift four up. Shift four up. And so once again, pause this video like we always say and figure out what would the equation be if you shift three to the right and four up. Alright, now let's do this together. So let's just first shift three to the right and think about how that might change the equation."}, {"video_title": "Shifting absolute value graphs Mathematics II High School Math Khan Academy.mp3", "Sentence": "Shift four up. And so once again, pause this video like we always say and figure out what would the equation be if you shift three to the right and four up. Alright, now let's do this together. So let's just first shift three to the right and think about how that might change the equation. So let's just visualize what we're even talking about. So if we're gonna shift three to the right, it would look like, it would look like this. So that's what we first want to figure out the equation for."}, {"video_title": "Shifting absolute value graphs Mathematics II High School Math Khan Academy.mp3", "Sentence": "So let's just first shift three to the right and think about how that might change the equation. So let's just visualize what we're even talking about. So if we're gonna shift three to the right, it would look like, it would look like this. So that's what we first want to figure out the equation for. And so how would we think about it? Well, one way to think about it is, well, something interesting is happening right over here at x equals three. Before, that interesting thing was happening at x equals zero."}, {"video_title": "Shifting absolute value graphs Mathematics II High School Math Khan Academy.mp3", "Sentence": "So that's what we first want to figure out the equation for. And so how would we think about it? Well, one way to think about it is, well, something interesting is happening right over here at x equals three. Before, that interesting thing was happening at x equals zero. Now it's happening at x equals three. And the interesting thing that happens here is that you switch signs inside the absolute value. You're instead of taking an absolute value of a negative, you're now taking the absolute value as you cross this point of a positive."}, {"video_title": "Shifting absolute value graphs Mathematics II High School Math Khan Academy.mp3", "Sentence": "Before, that interesting thing was happening at x equals zero. Now it's happening at x equals three. And the interesting thing that happens here is that you switch signs inside the absolute value. You're instead of taking an absolute value of a negative, you're now taking the absolute value as you cross this point of a positive. And that's why we see a switch in direction here of this line. And so you see the same thing happening right over here. So this point right over here, we know that our function, we know that our equation needs to evaluate out to zero."}, {"video_title": "Shifting absolute value graphs Mathematics II High School Math Khan Academy.mp3", "Sentence": "You're instead of taking an absolute value of a negative, you're now taking the absolute value as you cross this point of a positive. And that's why we see a switch in direction here of this line. And so you see the same thing happening right over here. So this point right over here, we know that our function, we know that our equation needs to evaluate out to zero. And this is where it's going to switch signs. And so we say, okay, well, this looks like an absolute value so it's going to have the form y is equal to the absolute value of something. And so you say, okay, how do I, if x is three, how do I make that equal to zero?"}, {"video_title": "Shifting absolute value graphs Mathematics II High School Math Khan Academy.mp3", "Sentence": "So this point right over here, we know that our function, we know that our equation needs to evaluate out to zero. And this is where it's going to switch signs. And so we say, okay, well, this looks like an absolute value so it's going to have the form y is equal to the absolute value of something. And so you say, okay, how do I, if x is three, how do I make that equal to zero? Well, I can subtract three from it. If I say y is equal to the absolute value of x minus three, well, let's try it out. Let's see if it makes sense."}, {"video_title": "Shifting absolute value graphs Mathematics II High School Math Khan Academy.mp3", "Sentence": "And so you say, okay, how do I, if x is three, how do I make that equal to zero? Well, I can subtract three from it. If I say y is equal to the absolute value of x minus three, well, let's try it out. Let's see if it makes sense. So when x is equal to three, three minus three is zero, absolute value of that is zero. That works out. When x is equal to four, four minus three is one, absolute value of one is indeed, is indeed one."}, {"video_title": "Shifting absolute value graphs Mathematics II High School Math Khan Academy.mp3", "Sentence": "Let's see if it makes sense. So when x is equal to three, three minus three is zero, absolute value of that is zero. That works out. When x is equal to four, four minus three is one, absolute value of one is indeed, is indeed one. And if x is equal to two, well, two minus three is negative one, but the absolute value of that is one. So once again, I'm showing you this by really trying out numbers, trying to give you a little bit of an intuition because that wasn't obvious to me when I first learned this, that if I'm shifting to the right, which it looks like I'm increasing an x value, that what I would really do is replace my x with an x minus the amount that I'm shifting to the right. But I encourage you to try the numbers and think about what's happening here."}, {"video_title": "Shifting absolute value graphs Mathematics II High School Math Khan Academy.mp3", "Sentence": "When x is equal to four, four minus three is one, absolute value of one is indeed, is indeed one. And if x is equal to two, well, two minus three is negative one, but the absolute value of that is one. So once again, I'm showing you this by really trying out numbers, trying to give you a little bit of an intuition because that wasn't obvious to me when I first learned this, that if I'm shifting to the right, which it looks like I'm increasing an x value, that what I would really do is replace my x with an x minus the amount that I'm shifting to the right. But I encourage you to try the numbers and think about what's happening here. At this vertex right over here, whatever was in the absolute value sign was equaling zero. It's when whatever was in the absolute value side is switching from negative signs to positive signs. So once again, if you shift three to the right, that has to happen at x equals three."}, {"video_title": "Shifting absolute value graphs Mathematics II High School Math Khan Academy.mp3", "Sentence": "But I encourage you to try the numbers and think about what's happening here. At this vertex right over here, whatever was in the absolute value sign was equaling zero. It's when whatever was in the absolute value side is switching from negative signs to positive signs. So once again, if you shift three to the right, that has to happen at x equals three. So whatever is inside the absolute value sign has to be equal to zero at x equals three. And this, you know, pause this video and really think about this if it isn't making sense. And even as you get more and more familiar with this, I encourage you to try out the numbers."}, {"video_title": "Shifting absolute value graphs Mathematics II High School Math Khan Academy.mp3", "Sentence": "So once again, if you shift three to the right, that has to happen at x equals three. So whatever is inside the absolute value sign has to be equal to zero at x equals three. And this, you know, pause this video and really think about this if it isn't making sense. And even as you get more and more familiar with this, I encourage you to try out the numbers. That will give you more, instead of just memorizing, hey, if I shift to the right, I replace x with x minus the amount that I shift. Always try out the numbers and try to get an intuition for that why that works. But now, let's start from here and shift four up."}, {"video_title": "Shifting absolute value graphs Mathematics II High School Math Khan Academy.mp3", "Sentence": "And even as you get more and more familiar with this, I encourage you to try out the numbers. That will give you more, instead of just memorizing, hey, if I shift to the right, I replace x with x minus the amount that I shift. Always try out the numbers and try to get an intuition for that why that works. But now, let's start from here and shift four up. And shifting four up is, in some ways, a lot more intuitive. So let me do that. Let me shift four up."}, {"video_title": "Shifting absolute value graphs Mathematics II High School Math Khan Academy.mp3", "Sentence": "But now, let's start from here and shift four up. And shifting four up is, in some ways, a lot more intuitive. So let me do that. Let me shift four up. Alright, so let me move. I'm gonna go up one, two, three, and four. I think I got that right."}, {"video_title": "Shifting absolute value graphs Mathematics II High School Math Khan Academy.mp3", "Sentence": "Let me shift four up. Alright, so let me move. I'm gonna go up one, two, three, and four. I think I got that right. So now I've shifted four up. And just as a reminder of what we've even done, the first part, we shifted three to the right, and now we are shifting four up. So now, now we are shifting four up."}, {"video_title": "Shifting absolute value graphs Mathematics II High School Math Khan Academy.mp3", "Sentence": "I think I got that right. So now I've shifted four up. And just as a reminder of what we've even done, the first part, we shifted three to the right, and now we are shifting four up. So now, now we are shifting four up. So before, y equaled zero here, but now y needs to be equal to four. So whatever this was evaluating to, we now have to add four to it. So when we just shifted three to the right, our equation was y is equal to the absolute value of x minus three."}, {"video_title": "Shifting absolute value graphs Mathematics II High School Math Khan Academy.mp3", "Sentence": "So now, now we are shifting four up. So before, y equaled zero here, but now y needs to be equal to four. So whatever this was evaluating to, we now have to add four to it. So when we just shifted three to the right, our equation was y is equal to the absolute value of x minus three. But now, whatever we were getting before, we now have to add four to it. We're going up in the vertical direction. So we just have to add four."}, {"video_title": "Shifting absolute value graphs Mathematics II High School Math Khan Academy.mp3", "Sentence": "So when we just shifted three to the right, our equation was y is equal to the absolute value of x minus three. But now, whatever we were getting before, we now have to add four to it. We're going up in the vertical direction. So we just have to add four. Now this makes a little bit more sense. If you're shifting in the vertical direction, if you shift up in the vertical direction, well, you just add a constant by the amount you're shifting. If you shift down in the vertical direction, well, you would subtract."}, {"video_title": "Shifting absolute value graphs Mathematics II High School Math Khan Academy.mp3", "Sentence": "So we just have to add four. Now this makes a little bit more sense. If you're shifting in the vertical direction, if you shift up in the vertical direction, well, you just add a constant by the amount you're shifting. If you shift down in the vertical direction, well, you would subtract. So if we said shift down four, you would subtract four right over here. The less intuitive thing is what we did with x, is when you shift to the right, you actually replace your x with x minus the amount that you shifted. But once again, try out numbers until it really makes some intuitive sense for you."}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "Let's see if we can stumble our way to another logarithm property. So let's say that the log base x of a is equal to b. Right? That's the same thing as saying, that is the exact same thing as saying that x to the b is equal to a. Fair enough. So what I want to do is experiment. What happens if I multiply this expression by another variable, let's call it c. Right?"}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "That's the same thing as saying, that is the exact same thing as saying that x to the b is equal to a. Fair enough. So what I want to do is experiment. What happens if I multiply this expression by another variable, let's call it c. Right? So I'm going to multiply both sides of this equation times c. I'm going to switch colors just to keep things interesting. That's not an x, that's a c. I should probably just do a dot instead. Times c. Right?"}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "What happens if I multiply this expression by another variable, let's call it c. Right? So I'm going to multiply both sides of this equation times c. I'm going to switch colors just to keep things interesting. That's not an x, that's a c. I should probably just do a dot instead. Times c. Right? So I'm going to multiply both sides of this equation times c. So I get c times log base x of a is equal to, multiply both sides of the equation, right? Is equal to b times c. Fair enough. I think you realize I have not done anything profound just yet."}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "Times c. Right? So I'm going to multiply both sides of this equation times c. So I get c times log base x of a is equal to, multiply both sides of the equation, right? Is equal to b times c. Fair enough. I think you realize I have not done anything profound just yet. Let's go back. We said that this is the same thing as this. So let's experiment with something."}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "I think you realize I have not done anything profound just yet. Let's go back. We said that this is the same thing as this. So let's experiment with something. Let's raise this side to the power c. So I'm going to raise this side to the power c. That's a kind of a caret. And when you type exponents, that's what you'd use, a caret, right? So I'm going to raise it to the power c. So then this side is x to the b to the c power is equal to a to the c, right?"}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "So let's experiment with something. Let's raise this side to the power c. So I'm going to raise this side to the power c. That's a kind of a caret. And when you type exponents, that's what you'd use, a caret, right? So I'm going to raise it to the power c. So then this side is x to the b to the c power is equal to a to the c, right? All I did is I raised both sides of this equation to the c-th power. And what do we know about when you raise something to an exponent and you raise that whole thing to another exponent? What happens to the exponents?"}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "So I'm going to raise it to the power c. So then this side is x to the b to the c power is equal to a to the c, right? All I did is I raised both sides of this equation to the c-th power. And what do we know about when you raise something to an exponent and you raise that whole thing to another exponent? What happens to the exponents? Well, that's just an exponent rule and you just multiply those two exponents. So this just becomes, this just implies that x to the bc is equal to a to the c. What can we do now? Well, I don't know."}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "What happens to the exponents? Well, that's just an exponent rule and you just multiply those two exponents. So this just becomes, this just implies that x to the bc is equal to a to the c. What can we do now? Well, I don't know. Let's take the logarithm of both sides. Or let's just write this. Let's not take the logarithm of both sides."}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, I don't know. Let's take the logarithm of both sides. Or let's just write this. Let's not take the logarithm of both sides. Let's write this as a logarithm expression. We know that x to the bc is equal to a to the c. Well, that's the exact same thing as saying that the logarithm base x of a to the c is equal to bc, correct? Because all I did is I rewrote this as a logarithm expression."}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "Let's not take the logarithm of both sides. Let's write this as a logarithm expression. We know that x to the bc is equal to a to the c. Well, that's the exact same thing as saying that the logarithm base x of a to the c is equal to bc, correct? Because all I did is I rewrote this as a logarithm expression. And I think now you realize that something interesting has happened. That bc, this bc, well, of course it's the same thing as this bc, right? So this expression must be equal to this expression."}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "Because all I did is I rewrote this as a logarithm expression. And I think now you realize that something interesting has happened. That bc, this bc, well, of course it's the same thing as this bc, right? So this expression must be equal to this expression. And I think we have another logarithm property. That if I have some kind of coefficient in front of the logarithm, I'm multiplying the logarithm. So if I have c clog base x of a."}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "So this expression must be equal to this expression. And I think we have another logarithm property. That if I have some kind of coefficient in front of the logarithm, I'm multiplying the logarithm. So if I have c clog base x of a. But that's c times the logarithm base x of a. That equals the log base x of a to the c. So you could take this coefficient and instead make it an exponent on the term inside the logarithm. That is another logarithm property."}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "So if I have c clog base x of a. But that's c times the logarithm base x of a. That equals the log base x of a to the c. So you could take this coefficient and instead make it an exponent on the term inside the logarithm. That is another logarithm property. So let's review what we know so far about logarithms. We know that if I write, let me use a different, let me say, if I write, well, let me just, the letters I've been using, c times logarithm base x of a is equal to logarithm base x of a to the c. We know that. And we know, we just learned, that logarithm base x of a plus logarithm base x of b is equal to the logarithm base x of a times b."}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "That is another logarithm property. So let's review what we know so far about logarithms. We know that if I write, let me use a different, let me say, if I write, well, let me just, the letters I've been using, c times logarithm base x of a is equal to logarithm base x of a to the c. We know that. And we know, we just learned, that logarithm base x of a plus logarithm base x of b is equal to the logarithm base x of a times b. Now let me ask you a question. What happens if instead of a positive sign here, we put a negative sign? Well, you could probably figure it out yourself, but we could do that same exact proof that we did in the beginning, but in this time we'll set it up with a negative."}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "And we know, we just learned, that logarithm base x of a plus logarithm base x of b is equal to the logarithm base x of a times b. Now let me ask you a question. What happens if instead of a positive sign here, we put a negative sign? Well, you could probably figure it out yourself, but we could do that same exact proof that we did in the beginning, but in this time we'll set it up with a negative. So if I said that, let's just say that log base x of a is equal to l. Let's say that log base x of b is equal to m. And let's say that log base x of a divided by b is equal to n, right? How can we write all of these expressions as exponents? Well, this just says that x to the l is equal to a."}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, you could probably figure it out yourself, but we could do that same exact proof that we did in the beginning, but in this time we'll set it up with a negative. So if I said that, let's just say that log base x of a is equal to l. Let's say that log base x of b is equal to m. And let's say that log base x of a divided by b is equal to n, right? How can we write all of these expressions as exponents? Well, this just says that x to the l is equal to a. Let me switch colors. That keeps it interesting. This is just saying that x to the m is equal to b."}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, this just says that x to the l is equal to a. Let me switch colors. That keeps it interesting. This is just saying that x to the m is equal to b. And this is just saying that x to the n is equal to a over b. So what can we do here? Well, what's another way of writing a over b?"}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "This is just saying that x to the m is equal to b. And this is just saying that x to the n is equal to a over b. So what can we do here? Well, what's another way of writing a over b? Well, that's just the same thing as writing x to the l, because that's a, x to the l over x to the m. That's b, right? And this we know from our exponent rules, right? This could also be written as x to the l, x to the negative m, right?"}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, what's another way of writing a over b? Well, that's just the same thing as writing x to the l, because that's a, x to the l over x to the m. That's b, right? And this we know from our exponent rules, right? This could also be written as x to the l, x to the negative m, right? Or that also equals x to the l minus m. So what do we know? We know that x to the n is equal to x to the l minus m. Right? x to the n is equal to x to the l minus m. Those equal each other."}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "This could also be written as x to the l, x to the negative m, right? Or that also equals x to the l minus m. So what do we know? We know that x to the n is equal to x to the l minus m. Right? x to the n is equal to x to the l minus m. Those equal each other. I just made a big equal chain here. So we know that n is equal to l minus m. Well, what does that do for us? Well, what's another way of writing n?"}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "x to the n is equal to x to the l minus m. Those equal each other. I just made a big equal chain here. So we know that n is equal to l minus m. Well, what does that do for us? Well, what's another way of writing n? I'm going to do it up here, because I think we have stumbled upon another logarithm rule. What's another way of writing n? I did it right here."}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, what's another way of writing n? I'm going to do it up here, because I think we have stumbled upon another logarithm rule. What's another way of writing n? I did it right here. This is another way of writing n, right? So logarithm base x of a over b, this is an x over here, is equal to l. l is this right here. Log base x of a is equal to l. The log base x of a minus m. I wrote m right here."}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "I did it right here. This is another way of writing n, right? So logarithm base x of a over b, this is an x over here, is equal to l. l is this right here. Log base x of a is equal to l. The log base x of a minus m. I wrote m right here. That's log base x of b. There you go. I probably didn't have to prove it."}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "Log base x of a is equal to l. The log base x of a minus m. I wrote m right here. That's log base x of b. There you go. I probably didn't have to prove it. You could have probably tried it out with dividing it, putting it in your number, whatever. But you now are hopefully satisfied that we have this new logarithm property right there. I have one more logarithm property to show you, but I don't think I have time to show it in this video."}, {"video_title": "Graphing a line given point and slope Linear equations & graphs Algebra I Khan Academy.mp3", "Sentence": "We are told graph a line with a slope of negative two that contains the point four comma negative three. And we have our little Khan Academy graphing widget right over here where we just have to find two points on that line and then that will graph the line for us. So pause this video and even if you don't have access to the widget right now, although it's all available on Khan Academy, see how you would, think at least, think about how you would approach this. And if you have paper and pencil handy, I encourage you to try to graph this line on your own before I work through it with this little widget. All right, now let's do it together. So we do know that it contains the point four comma negative three, so that's, I guess you could say, the easy part, we just have to find the point four, x is four, y is negative three, so it's from the origin, four to the right, three down. But then we have to figure out, well, where could another point be?"}, {"video_title": "Graphing a line given point and slope Linear equations & graphs Algebra I Khan Academy.mp3", "Sentence": "And if you have paper and pencil handy, I encourage you to try to graph this line on your own before I work through it with this little widget. All right, now let's do it together. So we do know that it contains the point four comma negative three, so that's, I guess you could say, the easy part, we just have to find the point four, x is four, y is negative three, so it's from the origin, four to the right, three down. But then we have to figure out, well, where could another point be? Because if we can figure out another point, then we would have graphed a line. And the clue here is that they say a slope of negative two. So one way to think about it is we can start at the point that we know is on the line, and a slope of negative two tells us that as x increases by one, y goes down by two."}, {"video_title": "Graphing a line given point and slope Linear equations & graphs Algebra I Khan Academy.mp3", "Sentence": "But then we have to figure out, well, where could another point be? Because if we can figure out another point, then we would have graphed a line. And the clue here is that they say a slope of negative two. So one way to think about it is we can start at the point that we know is on the line, and a slope of negative two tells us that as x increases by one, y goes down by two. The change in y would be negative two. And so this could be another point on that line. So I could graph it like this."}, {"video_title": "Graphing a line given point and slope Linear equations & graphs Algebra I Khan Academy.mp3", "Sentence": "So one way to think about it is we can start at the point that we know is on the line, and a slope of negative two tells us that as x increases by one, y goes down by two. The change in y would be negative two. And so this could be another point on that line. So I could graph it like this. As x goes up by one, as x goes from four to five, y will go, or y will change by negative two. So y will go from negative three to negative five. So this, we would be done."}, {"video_title": "Graphing a line given point and slope Linear equations & graphs Algebra I Khan Academy.mp3", "Sentence": "So I could graph it like this. As x goes up by one, as x goes from four to five, y will go, or y will change by negative two. So y will go from negative three to negative five. So this, we would be done. We have just graphed that line. Now another way that you could do it, because sometimes you might not have space on the paper or on the widget to be able to go to the right for x to increase, is to go the other way. If you have a slope of negative two, another way to think about it is if x goes down by one, if x goes down by one, then y goes up by two."}, {"video_title": "Graphing a line given point and slope Linear equations & graphs Algebra I Khan Academy.mp3", "Sentence": "So this, we would be done. We have just graphed that line. Now another way that you could do it, because sometimes you might not have space on the paper or on the widget to be able to go to the right for x to increase, is to go the other way. If you have a slope of negative two, another way to think about it is if x goes down by one, if x goes down by one, then y goes up by two. Because remember, slope is change of y over change in x. So you could either say you have a positive change in y of two when x has a negative one change, or you could think of it when x has a positive one change, y has a negative two change. But either way, notice you got the same line."}, {"video_title": "Rewriting a quadratic function to find roots and vertex Algebra I Khan Academy.mp3", "Sentence": "I have a function here defined as x squared minus 5x plus 6. What I want us to think about is what other forms we can write this function in if we, say, wanted to find the zeros of this function. If we wanted to figure out where does this function intersect the x-axis, what form would we put this in? And then another form for maybe finding out what's the minimum value of this. We see that we have a positive coefficient on the x squared term. This is going to be an upward opening parabola. But what's the minimum point of this?"}, {"video_title": "Rewriting a quadratic function to find roots and vertex Algebra I Khan Academy.mp3", "Sentence": "And then another form for maybe finding out what's the minimum value of this. We see that we have a positive coefficient on the x squared term. This is going to be an upward opening parabola. But what's the minimum point of this? Or even better, what's the vertex of this parabola right over here? So if the function looks something like this, we could use one form of the function to figure out where does it intersect the x-axis. So where does it intersect the x-axis?"}, {"video_title": "Rewriting a quadratic function to find roots and vertex Algebra I Khan Academy.mp3", "Sentence": "But what's the minimum point of this? Or even better, what's the vertex of this parabola right over here? So if the function looks something like this, we could use one form of the function to figure out where does it intersect the x-axis. So where does it intersect the x-axis? And maybe we can manipulate it to get another form to figure out what's the minimum point. What's this point right over here of this function? And I don't even know if the function looks like this."}, {"video_title": "Rewriting a quadratic function to find roots and vertex Algebra I Khan Academy.mp3", "Sentence": "So where does it intersect the x-axis? And maybe we can manipulate it to get another form to figure out what's the minimum point. What's this point right over here of this function? And I don't even know if the function looks like this. So I encourage you to pause this video and try to manipulate this into those two different forms. So let's work on it. So in order to find the roots, the easiest thing I could think of doing is trying to factor this quadratic expression, which is being used to define this function."}, {"video_title": "Rewriting a quadratic function to find roots and vertex Algebra I Khan Academy.mp3", "Sentence": "And I don't even know if the function looks like this. So I encourage you to pause this video and try to manipulate this into those two different forms. So let's work on it. So in order to find the roots, the easiest thing I could think of doing is trying to factor this quadratic expression, which is being used to define this function. So we could think about, well, let's think of two numbers whose product is positive 6 and whose sum is negative 5. So since their product is positive, we know that they have the same sign. And if they have the same sign but we get to a negative value, that means they both must be negative."}, {"video_title": "Rewriting a quadratic function to find roots and vertex Algebra I Khan Academy.mp3", "Sentence": "So in order to find the roots, the easiest thing I could think of doing is trying to factor this quadratic expression, which is being used to define this function. So we could think about, well, let's think of two numbers whose product is positive 6 and whose sum is negative 5. So since their product is positive, we know that they have the same sign. And if they have the same sign but we get to a negative value, that means they both must be negative. So let's see, negative 2 times negative 3 is positive 6. Negative 2 plus negative 3 is negative 5. So we could rewrite f of x."}, {"video_title": "Rewriting a quadratic function to find roots and vertex Algebra I Khan Academy.mp3", "Sentence": "And if they have the same sign but we get to a negative value, that means they both must be negative. So let's see, negative 2 times negative 3 is positive 6. Negative 2 plus negative 3 is negative 5. So we could rewrite f of x. And so let me write it this way. We could write f of x as being equal to x minus 2 times x minus 3. Now, how does this help us find the zeros?"}, {"video_title": "Rewriting a quadratic function to find roots and vertex Algebra I Khan Academy.mp3", "Sentence": "So we could rewrite f of x. And so let me write it this way. We could write f of x as being equal to x minus 2 times x minus 3. Now, how does this help us find the zeros? Well, in what situations is this right-hand expression on the right-hand going to be equal to 0? Well, it's the product of these two expressions. If either one of these is equal to 0, 0 times anything is 0."}, {"video_title": "Rewriting a quadratic function to find roots and vertex Algebra I Khan Academy.mp3", "Sentence": "Now, how does this help us find the zeros? Well, in what situations is this right-hand expression on the right-hand going to be equal to 0? Well, it's the product of these two expressions. If either one of these is equal to 0, 0 times anything is 0. 0 times anything else is 0. So this whole thing is going to be 0 if x minus 2 is equal to 0 or x minus 3 is equal to 0. Add 2 to both sides of this equation, you get x is equal to 2 or x is equal to 3."}, {"video_title": "Rewriting a quadratic function to find roots and vertex Algebra I Khan Academy.mp3", "Sentence": "If either one of these is equal to 0, 0 times anything is 0. 0 times anything else is 0. So this whole thing is going to be 0 if x minus 2 is equal to 0 or x minus 3 is equal to 0. Add 2 to both sides of this equation, you get x is equal to 2 or x is equal to 3. So those are the two zeros for this function, I guess you could say. And we could already think about it a little bit in terms of graphing it. So let's try to graph this thing."}, {"video_title": "Rewriting a quadratic function to find roots and vertex Algebra I Khan Academy.mp3", "Sentence": "Add 2 to both sides of this equation, you get x is equal to 2 or x is equal to 3. So those are the two zeros for this function, I guess you could say. And we could already think about it a little bit in terms of graphing it. So let's try to graph this thing. So this is x equals 1. This is x equals 2. This is x equals 3 right over there."}, {"video_title": "Rewriting a quadratic function to find roots and vertex Algebra I Khan Academy.mp3", "Sentence": "So let's try to graph this thing. So this is x equals 1. This is x equals 2. This is x equals 3 right over there. So that's our x-axis. That, you could say, is our y is equal to f of x-axis. And we're seeing that we intersect both here and here."}, {"video_title": "Rewriting a quadratic function to find roots and vertex Algebra I Khan Academy.mp3", "Sentence": "This is x equals 3 right over there. So that's our x-axis. That, you could say, is our y is equal to f of x-axis. And we're seeing that we intersect both here and here. When x is equal to 2, this f of x is equal to 0. When x is equal to 3, f of x is equal to 0. And you could substitute either of these values into the original expression, and you'll see it's going to get you to 0, because that is the same thing as that."}, {"video_title": "Rewriting a quadratic function to find roots and vertex Algebra I Khan Academy.mp3", "Sentence": "And we're seeing that we intersect both here and here. When x is equal to 2, this f of x is equal to 0. When x is equal to 3, f of x is equal to 0. And you could substitute either of these values into the original expression, and you'll see it's going to get you to 0, because that is the same thing as that. Now, what about the vertex? What form could we write this original thing in order to pick out the vertex? Well, we're already a little familiar with completing the square."}, {"video_title": "Rewriting a quadratic function to find roots and vertex Algebra I Khan Academy.mp3", "Sentence": "And you could substitute either of these values into the original expression, and you'll see it's going to get you to 0, because that is the same thing as that. Now, what about the vertex? What form could we write this original thing in order to pick out the vertex? Well, we're already a little familiar with completing the square. And when you put it in that form, or when you complete the square with this expression, that seems to be a pretty good way of thinking about what the minimum value of this function is. So let's just do that right over here. So I'm just going to rewrite it."}, {"video_title": "Rewriting a quadratic function to find roots and vertex Algebra I Khan Academy.mp3", "Sentence": "Well, we're already a little familiar with completing the square. And when you put it in that form, or when you complete the square with this expression, that seems to be a pretty good way of thinking about what the minimum value of this function is. So let's just do that right over here. So I'm just going to rewrite it. So we get f of x is equal to x squared minus 5x. And I'm just going to throw the plus 6 right over here. And I'm giving myself some real estate, because what I need to do, what I want to think about doing, is adding and subtracting the same value."}, {"video_title": "Rewriting a quadratic function to find roots and vertex Algebra I Khan Academy.mp3", "Sentence": "So I'm just going to rewrite it. So we get f of x is equal to x squared minus 5x. And I'm just going to throw the plus 6 right over here. And I'm giving myself some real estate, because what I need to do, what I want to think about doing, is adding and subtracting the same value. So I'm going to add it here, and I'm going to subtract it there. And I can do that, because then I've just added 0. I haven't changed the value of this right-hand side."}, {"video_title": "Rewriting a quadratic function to find roots and vertex Algebra I Khan Academy.mp3", "Sentence": "And I'm giving myself some real estate, because what I need to do, what I want to think about doing, is adding and subtracting the same value. So I'm going to add it here, and I'm going to subtract it there. And I can do that, because then I've just added 0. I haven't changed the value of this right-hand side. But I want to do that so that this part that I've underlined, or in this magenta color, so that this part right over here is a perfect square. And we've done this multiple times when we've completed the square. I encourage you to watch those videos if you need a little bit of a review on it."}, {"video_title": "Rewriting a quadratic function to find roots and vertex Algebra I Khan Academy.mp3", "Sentence": "I haven't changed the value of this right-hand side. But I want to do that so that this part that I've underlined, or in this magenta color, so that this part right over here is a perfect square. And we've done this multiple times when we've completed the square. I encourage you to watch those videos if you need a little bit of a review on it. But the general idea is this is going to be a perfect square. If we take this coefficient right over here, we take negative 5, we take half of that, which is negative 5 halves, and we square it. So we could write this as plus negative 5 halves squared."}, {"video_title": "Rewriting a quadratic function to find roots and vertex Algebra I Khan Academy.mp3", "Sentence": "I encourage you to watch those videos if you need a little bit of a review on it. But the general idea is this is going to be a perfect square. If we take this coefficient right over here, we take negative 5, we take half of that, which is negative 5 halves, and we square it. So we could write this as plus negative 5 halves squared. So I could write this negative 5 halves squared. Well, if we square a negative number, it's just going to be a positive. So it's going to be the same thing as 5 halves squared."}, {"video_title": "Rewriting a quadratic function to find roots and vertex Algebra I Khan Academy.mp3", "Sentence": "So we could write this as plus negative 5 halves squared. So I could write this negative 5 halves squared. Well, if we square a negative number, it's just going to be a positive. So it's going to be the same thing as 5 halves squared. 5 squared is 25, 2 squared is 4. So this is going to be plus 25 over 4. Now once again, if we want this equality to be true, we either have to add the same thing to both sides, or if we're just operating on one side, if we added it to that side, we could just subtract it from that side."}, {"video_title": "Rewriting a quadratic function to find roots and vertex Algebra I Khan Academy.mp3", "Sentence": "So it's going to be the same thing as 5 halves squared. 5 squared is 25, 2 squared is 4. So this is going to be plus 25 over 4. Now once again, if we want this equality to be true, we either have to add the same thing to both sides, or if we're just operating on one side, if we added it to that side, we could just subtract it from that side. And we haven't changed the total value on that side. So we added 25 over 4, and we subtracted 25 over 4. So what does this part right over here become, the part that I've underlined in magenta?"}, {"video_title": "Rewriting a quadratic function to find roots and vertex Algebra I Khan Academy.mp3", "Sentence": "Now once again, if we want this equality to be true, we either have to add the same thing to both sides, or if we're just operating on one side, if we added it to that side, we could just subtract it from that side. And we haven't changed the total value on that side. So we added 25 over 4, and we subtracted 25 over 4. So what does this part right over here become, the part that I've underlined in magenta? The whole reason why we engineered it in this way is so that this could be x minus 5 over 2 squared. And I encourage you to verify this. And we go into more detail about why taking half the coefficient here and then squaring it, adding it there, and then subtracting there, why that works."}, {"video_title": "Rewriting a quadratic function to find roots and vertex Algebra I Khan Academy.mp3", "Sentence": "So what does this part right over here become, the part that I've underlined in magenta? The whole reason why we engineered it in this way is so that this could be x minus 5 over 2 squared. And I encourage you to verify this. And we go into more detail about why taking half the coefficient here and then squaring it, adding it there, and then subtracting there, why that works. We do that in the completing the square videos. But these two things, you can verify that they are equivalent. So that's that part."}, {"video_title": "Rewriting a quadratic function to find roots and vertex Algebra I Khan Academy.mp3", "Sentence": "And we go into more detail about why taking half the coefficient here and then squaring it, adding it there, and then subtracting there, why that works. We do that in the completing the square videos. But these two things, you can verify that they are equivalent. So that's that part. And now we just have to simplify 6 minus 25 over 4. So 6 could be rewritten as 24 over 4. 24 over 4 minus 25 over 4 is negative 1 4th."}, {"video_title": "Rewriting a quadratic function to find roots and vertex Algebra I Khan Academy.mp3", "Sentence": "So that's that part. And now we just have to simplify 6 minus 25 over 4. So 6 could be rewritten as 24 over 4. 24 over 4 minus 25 over 4 is negative 1 4th. So minus 1 4th, just like that. So we've rewritten our original function as f of x is equal to x minus 5 halves squared minus 1 4th. Now why is this form interesting?"}, {"video_title": "Rewriting a quadratic function to find roots and vertex Algebra I Khan Academy.mp3", "Sentence": "24 over 4 minus 25 over 4 is negative 1 4th. So minus 1 4th, just like that. So we've rewritten our original function as f of x is equal to x minus 5 halves squared minus 1 4th. Now why is this form interesting? Well, one way to think about it is this part is always going to be non-negative. The minimum value of this part in magenta is going to be 0. Why?"}, {"video_title": "Rewriting a quadratic function to find roots and vertex Algebra I Khan Academy.mp3", "Sentence": "Now why is this form interesting? Well, one way to think about it is this part is always going to be non-negative. The minimum value of this part in magenta is going to be 0. Why? Because we're squaring this thing. If you're taking something like this, and we're just dealing with real numbers, and you're squaring it, you're not going to be able to get a negative value. At the minimum value, this is going to be 0."}, {"video_title": "Rewriting a quadratic function to find roots and vertex Algebra I Khan Academy.mp3", "Sentence": "Why? Because we're squaring this thing. If you're taking something like this, and we're just dealing with real numbers, and you're squaring it, you're not going to be able to get a negative value. At the minimum value, this is going to be 0. And then it obviously could be positive values as well. So if we want to think about when does this thing hit its minimum value? Well, it hits its minimum value when you're squaring 0."}, {"video_title": "Rewriting a quadratic function to find roots and vertex Algebra I Khan Academy.mp3", "Sentence": "At the minimum value, this is going to be 0. And then it obviously could be positive values as well. So if we want to think about when does this thing hit its minimum value? Well, it hits its minimum value when you're squaring 0. And when are you squaring 0? Well, you're squaring 0 when x minus 5 halves is equal to 0 or when x is equal to 5 halves, if you just want to add 5 halves to both sides of that equation. So this thing hits its minimum value when x is equal to 5 halves."}, {"video_title": "Rewriting a quadratic function to find roots and vertex Algebra I Khan Academy.mp3", "Sentence": "Well, it hits its minimum value when you're squaring 0. And when are you squaring 0? Well, you're squaring 0 when x minus 5 halves is equal to 0 or when x is equal to 5 halves, if you just want to add 5 halves to both sides of that equation. So this thing hits its minimum value when x is equal to 5 halves. And then what is y, or what is f of x, when x is equal to 5 halves? f of 5 halves. And once again, you could use any of those forms to evaluate 5 halves, but it's really easy in this form."}, {"video_title": "Rewriting a quadratic function to find roots and vertex Algebra I Khan Academy.mp3", "Sentence": "So this thing hits its minimum value when x is equal to 5 halves. And then what is y, or what is f of x, when x is equal to 5 halves? f of 5 halves. And once again, you could use any of those forms to evaluate 5 halves, but it's really easy in this form. When x is equal to 5 halves, this term right over here becomes 0, 0 squared, 0, you're just left with negative 1 4th. You're just left with negative 1 4th. So another way to think about it is our vertex is at the point."}, {"video_title": "Rewriting a quadratic function to find roots and vertex Algebra I Khan Academy.mp3", "Sentence": "And once again, you could use any of those forms to evaluate 5 halves, but it's really easy in this form. When x is equal to 5 halves, this term right over here becomes 0, 0 squared, 0, you're just left with negative 1 4th. You're just left with negative 1 4th. So another way to think about it is our vertex is at the point. So we could say our vertex is at the point x equals 5 halves, y equals negative 1 4th. So x equals 5 halves, that's the same thing as 2 and 1 half. So x equals 5 halves, and y is equal to negative 1 4th."}, {"video_title": "Rewriting a quadratic function to find roots and vertex Algebra I Khan Academy.mp3", "Sentence": "So another way to think about it is our vertex is at the point. So we could say our vertex is at the point x equals 5 halves, y equals negative 1 4th. So x equals 5 halves, that's the same thing as 2 and 1 half. So x equals 5 halves, and y is equal to negative 1 4th. So if that is negative 1, 1 4th would be something like that. So that right over there is a vertex, that is the point. Let me make it clear, that's the point 5 halves, comma, negative 1 4th."}, {"video_title": "Rewriting a quadratic function to find roots and vertex Algebra I Khan Academy.mp3", "Sentence": "So x equals 5 halves, and y is equal to negative 1 4th. So if that is negative 1, 1 4th would be something like that. So that right over there is a vertex, that is the point. Let me make it clear, that's the point 5 halves, comma, negative 1 4th. And what's cool is we've just used this form to figure out the minimum point, to figure out the vertex in this case. And then we can use the roots as two other points to get a rough sketch of what this parabola will actually look like. So the interesting, or I guess the takeaway from this video is just to realize that we can rewrite this in different forms, depending on what we're trying to understand about this function."}, {"video_title": "Parabola vertex and axis of symmetry Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "And just as a bit of a refresher, if a parabola looks like this, the vertex is the lowest point here, it's this minimum point here for an upward opening parabola. If the parabola opens downward like this, the vertex is the topmost point, right like that, it's the maximum point. And the axis of symmetry is the line that you could reflect the parabola around, and it's symmetric. So that's the axis of symmetry. That is a reflection of the left-hand side along that axis of symmetry. Same thing if it's a downward opening parabola. And the general way of telling the difference between an upward opening and a downward opening parabola is that this will have a positive coefficient on the x squared term, and this will have a negative coefficient, and we'll see that in a little bit more detail."}, {"video_title": "Parabola vertex and axis of symmetry Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So that's the axis of symmetry. That is a reflection of the left-hand side along that axis of symmetry. Same thing if it's a downward opening parabola. And the general way of telling the difference between an upward opening and a downward opening parabola is that this will have a positive coefficient on the x squared term, and this will have a negative coefficient, and we'll see that in a little bit more detail. So let's just work on this. Now, in order to figure out the vertex, there's kind of a quick and dirty formula, but I'm not going to do the formula here because the formula really tells you nothing about how you got it. But I'll show you how to apply the formula at the end of this video if you see this on a math test and just want to do it really quickly."}, {"video_title": "Parabola vertex and axis of symmetry Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "And the general way of telling the difference between an upward opening and a downward opening parabola is that this will have a positive coefficient on the x squared term, and this will have a negative coefficient, and we'll see that in a little bit more detail. So let's just work on this. Now, in order to figure out the vertex, there's kind of a quick and dirty formula, but I'm not going to do the formula here because the formula really tells you nothing about how you got it. But I'll show you how to apply the formula at the end of this video if you see this on a math test and just want to do it really quickly. But we're going to do it the slow and intuitive way first. So let's think about how we can find either the maximum or the minimum point of this parabola. So the best way I can think of doing it is to complete the square, and it might seem like a very foreign concept right now, but let's just do it one step at a time."}, {"video_title": "Parabola vertex and axis of symmetry Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "But I'll show you how to apply the formula at the end of this video if you see this on a math test and just want to do it really quickly. But we're going to do it the slow and intuitive way first. So let's think about how we can find either the maximum or the minimum point of this parabola. So the best way I can think of doing it is to complete the square, and it might seem like a very foreign concept right now, but let's just do it one step at a time. So I can rewrite this as y is equal to, well, I could factor out a negative 2. It's equal to negative 2 times x squared minus 4x minus 4, and I'm going to put the minus 4 out here, and this is where I'm going to complete the square. Now, what I want to do is express this stuff in the parentheses as a sum of a perfect square and then some number over here."}, {"video_title": "Parabola vertex and axis of symmetry Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So the best way I can think of doing it is to complete the square, and it might seem like a very foreign concept right now, but let's just do it one step at a time. So I can rewrite this as y is equal to, well, I could factor out a negative 2. It's equal to negative 2 times x squared minus 4x minus 4, and I'm going to put the minus 4 out here, and this is where I'm going to complete the square. Now, what I want to do is express this stuff in the parentheses as a sum of a perfect square and then some number over here. So I have x squared minus 4x. If I wanted this to be a perfect square, if I had a positive 4 over here, if I had a positive 4 over there, then this would be a perfect square. It would be x minus 2 squared, and I got the 4 because I said, well, I want whatever half of this number is."}, {"video_title": "Parabola vertex and axis of symmetry Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Now, what I want to do is express this stuff in the parentheses as a sum of a perfect square and then some number over here. So I have x squared minus 4x. If I wanted this to be a perfect square, if I had a positive 4 over here, if I had a positive 4 over there, then this would be a perfect square. It would be x minus 2 squared, and I got the 4 because I said, well, I want whatever half of this number is. So half of negative 4 is negative 2. Let me square it. That will give me a positive 4 right there, but I can't just add a 4 willy-nilly to one side of an equation."}, {"video_title": "Parabola vertex and axis of symmetry Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "It would be x minus 2 squared, and I got the 4 because I said, well, I want whatever half of this number is. So half of negative 4 is negative 2. Let me square it. That will give me a positive 4 right there, but I can't just add a 4 willy-nilly to one side of an equation. I would either have to add it to the other side, or I would have to then just subtract it. So here I haven't changed the equation. I added 4, and then I subtracted 4."}, {"video_title": "Parabola vertex and axis of symmetry Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "That will give me a positive 4 right there, but I can't just add a 4 willy-nilly to one side of an equation. I would either have to add it to the other side, or I would have to then just subtract it. So here I haven't changed the equation. I added 4, and then I subtracted 4. I just added 0 to this little expression here, so it didn't change it. But what it does allow me to do is express this part right here as a perfect square. x squared minus 4x plus 4 is x minus 2 squared."}, {"video_title": "Parabola vertex and axis of symmetry Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "I added 4, and then I subtracted 4. I just added 0 to this little expression here, so it didn't change it. But what it does allow me to do is express this part right here as a perfect square. x squared minus 4x plus 4 is x minus 2 squared. And then you have this negative 2 out front, multiplying everything, and then you have a negative 4 minus negative 4 minus 8, just like that. So you have y is equal to negative 2 times this entire thing, and now we can multiply out the negative 2 again. So we can distribute it."}, {"video_title": "Parabola vertex and axis of symmetry Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "x squared minus 4x plus 4 is x minus 2 squared. And then you have this negative 2 out front, multiplying everything, and then you have a negative 4 minus negative 4 minus 8, just like that. So you have y is equal to negative 2 times this entire thing, and now we can multiply out the negative 2 again. So we can distribute it. y is equal to negative 2 times x minus 2 squared, and the negative 2 times negative 8 is plus 16. Now, all I did is algebraically rearrange this equation, but what this allows us to do is think about what the maximum or minimum point of this equation is. So let's just look at, let's explore this a little bit."}, {"video_title": "Parabola vertex and axis of symmetry Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So we can distribute it. y is equal to negative 2 times x minus 2 squared, and the negative 2 times negative 8 is plus 16. Now, all I did is algebraically rearrange this equation, but what this allows us to do is think about what the maximum or minimum point of this equation is. So let's just look at, let's explore this a little bit. This quantity right here, x minus 2 squared, if you're squaring anything, this is always going to be a positive quantity. That right there is always positive, but it's being multiplied by a negative number. So if you look at the larger context, if you look at the always positive multiplied by the negative 2, that's going to be always negative."}, {"video_title": "Parabola vertex and axis of symmetry Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So let's just look at, let's explore this a little bit. This quantity right here, x minus 2 squared, if you're squaring anything, this is always going to be a positive quantity. That right there is always positive, but it's being multiplied by a negative number. So if you look at the larger context, if you look at the always positive multiplied by the negative 2, that's going to be always negative. And the more positive that this number becomes, the more positive this number becomes, when you multiply it by a negative, the more negative this entire expression becomes. So if you think about it, this is going to be a downward opening parabola. We have a negative coefficient out here."}, {"video_title": "Parabola vertex and axis of symmetry Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So if you look at the larger context, if you look at the always positive multiplied by the negative 2, that's going to be always negative. And the more positive that this number becomes, the more positive this number becomes, when you multiply it by a negative, the more negative this entire expression becomes. So if you think about it, this is going to be a downward opening parabola. We have a negative coefficient out here. And the maximum point on this downward opening parabola is when this expression right here is as small as possible. If this gets any larger, it gets multiplied by a negative number, and then you subtract it from 16. So if this expression right here is 0, then we have our maximum y value, which is 16."}, {"video_title": "Parabola vertex and axis of symmetry Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "We have a negative coefficient out here. And the maximum point on this downward opening parabola is when this expression right here is as small as possible. If this gets any larger, it gets multiplied by a negative number, and then you subtract it from 16. So if this expression right here is 0, then we have our maximum y value, which is 16. So how do we get x is equal to 0 here? Well, the way to get x minus 2 equal to 0, so let's just do it. x minus 2 is equal to 0, so that happens when x is equal to 2."}, {"video_title": "Parabola vertex and axis of symmetry Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So if this expression right here is 0, then we have our maximum y value, which is 16. So how do we get x is equal to 0 here? Well, the way to get x minus 2 equal to 0, so let's just do it. x minus 2 is equal to 0, so that happens when x is equal to 2. So when x is equal to 2, this expression is 0, 0 times a negative number. It's all 0, and then y is equal to 16. This is our vertex."}, {"video_title": "Parabola vertex and axis of symmetry Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "x minus 2 is equal to 0, so that happens when x is equal to 2. So when x is equal to 2, this expression is 0, 0 times a negative number. It's all 0, and then y is equal to 16. This is our vertex. This is our maximum point. We just reasoned through it, just looking at the algebra, that the highest value this can take on is 16. As x moves away from 2 in the positive or negative direction, this quantity right here, it might be negative or positive, but when you square it, it's going to be positive."}, {"video_title": "Parabola vertex and axis of symmetry Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "This is our vertex. This is our maximum point. We just reasoned through it, just looking at the algebra, that the highest value this can take on is 16. As x moves away from 2 in the positive or negative direction, this quantity right here, it might be negative or positive, but when you square it, it's going to be positive. When you multiply it by a negative 2, it's going to become negative, and it's going to subtract from 16. So our vertex right here is x is equal to 2. Actually, let's say each of these units are 2."}, {"video_title": "Parabola vertex and axis of symmetry Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "As x moves away from 2 in the positive or negative direction, this quantity right here, it might be negative or positive, but when you square it, it's going to be positive. When you multiply it by a negative 2, it's going to become negative, and it's going to subtract from 16. So our vertex right here is x is equal to 2. Actually, let's say each of these units are 2. So this is 2, and this is 2, 4, 6, 8, 10, 12, 14, 16. So my vertex is here. That is the absolute maximum point for this parabola."}, {"video_title": "Parabola vertex and axis of symmetry Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Actually, let's say each of these units are 2. So this is 2, and this is 2, 4, 6, 8, 10, 12, 14, 16. So my vertex is here. That is the absolute maximum point for this parabola. Its axis of symmetry is going to be along the line x is equal to 2, along the vertical line x is equal to 2. That is going to be its axis of symmetry. Now, if we're just curious for a couple of other points, just because we want to plot this thing, we could say, well, what happens when, I don't know, what happens when x is equal to 0?"}, {"video_title": "Parabola vertex and axis of symmetry Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "That is the absolute maximum point for this parabola. Its axis of symmetry is going to be along the line x is equal to 2, along the vertical line x is equal to 2. That is going to be its axis of symmetry. Now, if we're just curious for a couple of other points, just because we want to plot this thing, we could say, well, what happens when, I don't know, what happens when x is equal to 0? That's an easy one. When x is equal to 0, y is equal to 8. So when x is equal to 0, we have 1, 2, 3, 4, oh, well, these are 2, 2, 4, 6, 8."}, {"video_title": "Parabola vertex and axis of symmetry Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Now, if we're just curious for a couple of other points, just because we want to plot this thing, we could say, well, what happens when, I don't know, what happens when x is equal to 0? That's an easy one. When x is equal to 0, y is equal to 8. So when x is equal to 0, we have 1, 2, 3, 4, oh, well, these are 2, 2, 4, 6, 8. It's right there. This is an axis of symmetry. So when x is equal to 3, y is also going to be equal to 8."}, {"video_title": "Parabola vertex and axis of symmetry Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So when x is equal to 0, we have 1, 2, 3, 4, oh, well, these are 2, 2, 4, 6, 8. It's right there. This is an axis of symmetry. So when x is equal to 3, y is also going to be equal to 8. So this parabola is going to be, it's a really steep and narrow one that looks something like this, where this right here is the maximum point. Now, I told you this is kind of the slow and intuitive way to do the problem. If you wanted the quick and dirty way to figure out a vertex, there is a formula."}, {"video_title": "Parabola vertex and axis of symmetry Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So when x is equal to 3, y is also going to be equal to 8. So this parabola is going to be, it's a really steep and narrow one that looks something like this, where this right here is the maximum point. Now, I told you this is kind of the slow and intuitive way to do the problem. If you wanted the quick and dirty way to figure out a vertex, there is a formula. You can derive it, actually, doing this exact same process we just did. But the formula for the vertex is x is, or the x value of the vertex, or the axis of symmetry is x is equal to negative b over 2a. So if we just apply this, but, you know, this is just kind of mindless application of a formula."}, {"video_title": "Parabola vertex and axis of symmetry Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "If you wanted the quick and dirty way to figure out a vertex, there is a formula. You can derive it, actually, doing this exact same process we just did. But the formula for the vertex is x is, or the x value of the vertex, or the axis of symmetry is x is equal to negative b over 2a. So if we just apply this, but, you know, this is just kind of mindless application of a formula. I wanted to show you the intuition why this formula even exists. But if you just mindlessly apply this, you'll get, what's b here? So x is equal to negative, b here is 8, 8 over 2 times a. a right here is a negative 2, 2 times negative 2."}, {"video_title": "Parabola vertex and axis of symmetry Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So if we just apply this, but, you know, this is just kind of mindless application of a formula. I wanted to show you the intuition why this formula even exists. But if you just mindlessly apply this, you'll get, what's b here? So x is equal to negative, b here is 8, 8 over 2 times a. a right here is a negative 2, 2 times negative 2. So what is that going to be equal to? It is negative 8 over negative 4, which is equal to 2, which is the exact same thing we got by reasoning it out. And when x is equal to 2, y is equal to 16."}, {"video_title": "Subtracting rational expressions unlike denominators High School Math Khan Academy.mp3", "Sentence": "I encourage you to pause the video and see what this would result in. So actually do the subtraction. All right, now let's do this together. And if we're subtracting two rational expressions, we'd like to have them have the same denominator. And they clearly don't have the same denominator, and so we need to find a common denominator. And a common denominator is one that is going to be divisible by either of these, and then we can multiply them by an appropriate expression or number so that it becomes the common denominator. So the easiest common denominator I can think of, especially because these factors, these two expressions have no factors in common, would just be their product."}, {"video_title": "Subtracting rational expressions unlike denominators High School Math Khan Academy.mp3", "Sentence": "And if we're subtracting two rational expressions, we'd like to have them have the same denominator. And they clearly don't have the same denominator, and so we need to find a common denominator. And a common denominator is one that is going to be divisible by either of these, and then we can multiply them by an appropriate expression or number so that it becomes the common denominator. So the easiest common denominator I can think of, especially because these factors, these two expressions have no factors in common, would just be their product. So this is going to be equal to, so we could just multiply these two. So this is going to be, actually let me do this one right over here in magenta. So this is going to be equal to the common denominator."}, {"video_title": "Subtracting rational expressions unlike denominators High School Math Khan Academy.mp3", "Sentence": "So the easiest common denominator I can think of, especially because these factors, these two expressions have no factors in common, would just be their product. So this is going to be equal to, so we could just multiply these two. So this is going to be, actually let me do this one right over here in magenta. So this is going to be equal to the common denominator. If I say, if I want to just multiply those two denominators, for this one I'll have my eight x plus seven, and now I'm going to multiply it by three x plus one. I'm multiplying it by the other denominator. And I had negative five x in the numerator, but if I'm going to multiply the denominator by three x plus one, and I don't want to change the value of the expression, I'll have to multiply the numerator by three x plus one as well."}, {"video_title": "Subtracting rational expressions unlike denominators High School Math Khan Academy.mp3", "Sentence": "So this is going to be equal to the common denominator. If I say, if I want to just multiply those two denominators, for this one I'll have my eight x plus seven, and now I'm going to multiply it by three x plus one. I'm multiplying it by the other denominator. And I had negative five x in the numerator, but if I'm going to multiply the denominator by three x plus one, and I don't want to change the value of the expression, I'll have to multiply the numerator by three x plus one as well. Notice three x plus one divided by three x plus one is just one, and you'd be left with what we started with. And from that, we are going to subtract all of this. Now there's a couple of ways you could think of the subtraction."}, {"video_title": "Subtracting rational expressions unlike denominators High School Math Khan Academy.mp3", "Sentence": "And I had negative five x in the numerator, but if I'm going to multiply the denominator by three x plus one, and I don't want to change the value of the expression, I'll have to multiply the numerator by three x plus one as well. Notice three x plus one divided by three x plus one is just one, and you'd be left with what we started with. And from that, we are going to subtract all of this. Now there's a couple of ways you could think of the subtraction. I could just write a minus sign right over here, and do the same thing that I just did for the first term. Or another way to think about it, and actually for this particular case, I like thinking about it better this way, is to just add the negative of this. And so if I just multiplied negative one times this expression, I get negative six x third over three x plus one."}, {"video_title": "Subtracting rational expressions unlike denominators High School Math Khan Academy.mp3", "Sentence": "Now there's a couple of ways you could think of the subtraction. I could just write a minus sign right over here, and do the same thing that I just did for the first term. Or another way to think about it, and actually for this particular case, I like thinking about it better this way, is to just add the negative of this. And so if I just multiplied negative one times this expression, I get negative six x third over three x plus one. If I had more terms up here in the numerator, I would have to be careful to distribute that negative sign. But here I only have one term, so I just made it negative. And so I could say this is going to be plus, and let me do this in a new color."}, {"video_title": "Subtracting rational expressions unlike denominators High School Math Khan Academy.mp3", "Sentence": "And so if I just multiplied negative one times this expression, I get negative six x third over three x plus one. If I had more terms up here in the numerator, I would have to be careful to distribute that negative sign. But here I only have one term, so I just made it negative. And so I could say this is going to be plus, and let me do this in a new color. Do this in green. Our common denominator we already established is just the product of our two denominators. So it is going to be eight x plus seven times three x plus one."}, {"video_title": "Subtracting rational expressions unlike denominators High School Math Khan Academy.mp3", "Sentence": "And so I could say this is going to be plus, and let me do this in a new color. Do this in green. Our common denominator we already established is just the product of our two denominators. So it is going to be eight x plus seven times three x plus one. Now if we multiply the denominator here was three x plus one, we're multiplying it by eight x plus seven. So that means we have to multiply the numerator by eight x plus seven as well. Eight x plus seven times negative six x to the third power."}, {"video_title": "Subtracting rational expressions unlike denominators High School Math Khan Academy.mp3", "Sentence": "So it is going to be eight x plus seven times three x plus one. Now if we multiply the denominator here was three x plus one, we're multiplying it by eight x plus seven. So that means we have to multiply the numerator by eight x plus seven as well. Eight x plus seven times negative six x to the third power. Notice eight x plus seven divided by eight x plus seven is one. If you were to do that, you would get back to your original expression right over here, the negative six x to the third over three x plus one. And now we're ready to add."}, {"video_title": "Subtracting rational expressions unlike denominators High School Math Khan Academy.mp3", "Sentence": "Eight x plus seven times negative six x to the third power. Notice eight x plus seven divided by eight x plus seven is one. If you were to do that, you would get back to your original expression right over here, the negative six x to the third over three x plus one. And now we're ready to add. This is all going to be equal to, I'll write the denominator in white. So we have our common denominator, eight x plus seven times three x plus one. Now in the magenta, I would want to distribute the negative five x."}, {"video_title": "Subtracting rational expressions unlike denominators High School Math Khan Academy.mp3", "Sentence": "And now we're ready to add. This is all going to be equal to, I'll write the denominator in white. So we have our common denominator, eight x plus seven times three x plus one. Now in the magenta, I would want to distribute the negative five x. So negative five x times positive three x is negative 15 x squared. And then negative five x times one is minus five x. And then in the green, I would have, let's see, I'll distribute the negative six x to the third power."}, {"video_title": "Subtracting rational expressions unlike denominators High School Math Khan Academy.mp3", "Sentence": "Now in the magenta, I would want to distribute the negative five x. So negative five x times positive three x is negative 15 x squared. And then negative five x times one is minus five x. And then in the green, I would have, let's see, I'll distribute the negative six x to the third power. So negative six x to the third times positive eight x is going to be negative 48 x to the fourth power. And then negative six x to the third times positive seven is going to be negative 42 x to the third. And I think I'm done because there's no more, I only have one fourth degree term, one third degree term, one second degree term, one first degree term, and that's it."}, {"video_title": "Subtracting rational expressions unlike denominators High School Math Khan Academy.mp3", "Sentence": "And then in the green, I would have, let's see, I'll distribute the negative six x to the third power. So negative six x to the third times positive eight x is going to be negative 48 x to the fourth power. And then negative six x to the third times positive seven is going to be negative 42 x to the third. And I think I'm done because there's no more, I only have one fourth degree term, one third degree term, one second degree term, one first degree term, and that's it. There's no more simplification here. Some of you might want to just write it in descending degree order. So you could write it as negative 48 x to the fourth minus 42 x to the third minus 15 x squared minus five x, all of that over, all of that over eight x plus seven times three x plus one."}, {"video_title": "Subtracting rational expressions unlike denominators High School Math Khan Academy.mp3", "Sentence": "And I think I'm done because there's no more, I only have one fourth degree term, one third degree term, one second degree term, one first degree term, and that's it. There's no more simplification here. Some of you might want to just write it in descending degree order. So you could write it as negative 48 x to the fourth minus 42 x to the third minus 15 x squared minus five x, all of that over, all of that over eight x plus seven times three x plus one. But either way, we are all done. And it looks like up here, yeah, there's nothing to factor out. These two are divisible by five, these are divisible by six, but even if I were to factor that out, nothing over here, down here, no five or six to factor out."}, {"video_title": "Multi-step inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "And we have 5x plus 7 is greater than 3 times x plus 1. So let's just try to isolate x on one side of this inequality. But before we do that, let's just simplify this right-hand side. So we get 5x plus 7 is greater than, let's distribute this 3. So 3 times x plus 1 is the same thing as 3 times x plus 3 times 1. So it's going to be 3x plus 3 times 1 is 3. Now if we want to put our x's on the left-hand side, we can subtract 3x from both sides."}, {"video_title": "Multi-step inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "So we get 5x plus 7 is greater than, let's distribute this 3. So 3 times x plus 1 is the same thing as 3 times x plus 3 times 1. So it's going to be 3x plus 3 times 1 is 3. Now if we want to put our x's on the left-hand side, we can subtract 3x from both sides. That'll get rid of this 3x on the right-hand side. So let's do that. Let's subtract 3x from both sides."}, {"video_title": "Multi-step inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "Now if we want to put our x's on the left-hand side, we can subtract 3x from both sides. That'll get rid of this 3x on the right-hand side. So let's do that. Let's subtract 3x from both sides. And we get, on the left-hand side, 5x minus 3x is 2x plus 7 is greater than 3x minus 3x. Those cancel out. That was the whole point behind subtracting 3x from both sides."}, {"video_title": "Multi-step inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "Let's subtract 3x from both sides. And we get, on the left-hand side, 5x minus 3x is 2x plus 7 is greater than 3x minus 3x. Those cancel out. That was the whole point behind subtracting 3x from both sides. Is greater than 3. Now we can subtract 7 from both sides to get rid of this positive 7 right over here. So let's subtract 7 from both sides."}, {"video_title": "Multi-step inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "That was the whole point behind subtracting 3x from both sides. Is greater than 3. Now we can subtract 7 from both sides to get rid of this positive 7 right over here. So let's subtract 7 from both sides. And we get, on the left-hand side, 2x plus 7 minus 7 is just 2x. Is greater than 3 minus 7, which is negative 4. And then let's see."}, {"video_title": "Multi-step inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "So let's subtract 7 from both sides. And we get, on the left-hand side, 2x plus 7 minus 7 is just 2x. Is greater than 3 minus 7, which is negative 4. And then let's see. We have 2x is greater than negative 4. If we just wanted x over here, we can divide both sides by 2. Since 2 is a positive number, we don't have to swap the inequality."}, {"video_title": "Multi-step inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "And then let's see. We have 2x is greater than negative 4. If we just wanted x over here, we can divide both sides by 2. Since 2 is a positive number, we don't have to swap the inequality. So let's just divide both sides by 2. And we get x is greater than negative 4 divided by 2 is negative 2. So the solution will look like this."}, {"video_title": "Multi-step inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "Since 2 is a positive number, we don't have to swap the inequality. So let's just divide both sides by 2. And we get x is greater than negative 4 divided by 2 is negative 2. So the solution will look like this. Draw the number line. I can draw a straighter number line than that. There we go."}, {"video_title": "Multi-step inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "So the solution will look like this. Draw the number line. I can draw a straighter number line than that. There we go. Still not that great, but it'll serve our purposes. Let's say that's negative 3, negative 2, negative 1, 0, 1, 2, 3. x is greater than negative 2. It does not include negative 2."}, {"video_title": "Multi-step inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "There we go. Still not that great, but it'll serve our purposes. Let's say that's negative 3, negative 2, negative 1, 0, 1, 2, 3. x is greater than negative 2. It does not include negative 2. It is not greater than or equal to negative 2. So we have to exclude negative 2. And we exclude negative 2 by drawing an open circle at negative 2."}, {"video_title": "Multi-step inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "It does not include negative 2. It is not greater than or equal to negative 2. So we have to exclude negative 2. And we exclude negative 2 by drawing an open circle at negative 2. But then all of the values greater than that are valid x's that would satisfy this inequality. So anything above it will work. And let's just try something that should work."}, {"video_title": "Multi-step inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "And we exclude negative 2 by drawing an open circle at negative 2. But then all of the values greater than that are valid x's that would satisfy this inequality. So anything above it will work. And let's just try something that should work. And let's try something else that shouldn't work. So 0 should work. It is greater than negative 2."}, {"video_title": "Multi-step inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "And let's just try something that should work. And let's try something else that shouldn't work. So 0 should work. It is greater than negative 2. It's right over here. So let's verify that. 5 times 0 plus 7 should be greater than 3 times 0 plus 1."}, {"video_title": "Multi-step inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "It is greater than negative 2. It's right over here. So let's verify that. 5 times 0 plus 7 should be greater than 3 times 0 plus 1. So this is 7, because this is just a 0. 7 should be greater than 3. 3 times 1."}, {"video_title": "Multi-step inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "5 times 0 plus 7 should be greater than 3 times 0 plus 1. So this is 7, because this is just a 0. 7 should be greater than 3. 3 times 1. So 7 should be greater than 3. And it definitely is. Now let's try something that should not work."}, {"video_title": "Multi-step inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "3 times 1. So 7 should be greater than 3. And it definitely is. Now let's try something that should not work. Let's try negative 3. So 5 times negative 3 plus 7. Let's see if it's greater than 3 times negative 3 plus 1."}, {"video_title": "Multi-step inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "Now let's try something that should not work. Let's try negative 3. So 5 times negative 3 plus 7. Let's see if it's greater than 3 times negative 3 plus 1. So this is negative 15 plus 7 is negative 8. That is negative 8. Let's see if that is greater than negative 3 plus 1 is negative 2 times 3 is negative 6."}, {"video_title": "Multi-step inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "Let's see if it's greater than 3 times negative 3 plus 1. So this is negative 15 plus 7 is negative 8. That is negative 8. Let's see if that is greater than negative 3 plus 1 is negative 2 times 3 is negative 6. Negative 8 is not greater than negative 6. Negative 8 is more negative than negative 6. It's less than."}, {"video_title": "Multi-step inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "Let's see if that is greater than negative 3 plus 1 is negative 2 times 3 is negative 6. Negative 8 is not greater than negative 6. Negative 8 is more negative than negative 6. It's less than. So it's good that negative 3 didn't work, because we didn't include that in our solution set. So we tried something that is in our solution set, and it did work, and something that's not. And it didn't work, so we're feeling pretty good."}, {"video_title": "Example 3 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "We could factor it and just figure out the values of x that satisfy it and just count them. That would be the number of solutions. We could just apply the quadratic formula. But what I want to do here is actually explore the quadratic formula and think about how we can determine the number of solutions without even maybe necessarily finding them explicitly. So the quadratic formula tells us that if we have an equation of the form ax squared plus bx plus c is equal to 0, that the solutions are going to be, or the solution, if it exists, is going to be negative b plus or minus the square root of b squared minus 4ac, all of that over 2a. Now, the reason why this is two solutions, or the reason why this can be two solutions, is that we have a plus or a minus here. If this b squared minus 4ac is a positive number, so let's think about this a little bit."}, {"video_title": "Example 3 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "But what I want to do here is actually explore the quadratic formula and think about how we can determine the number of solutions without even maybe necessarily finding them explicitly. So the quadratic formula tells us that if we have an equation of the form ax squared plus bx plus c is equal to 0, that the solutions are going to be, or the solution, if it exists, is going to be negative b plus or minus the square root of b squared minus 4ac, all of that over 2a. Now, the reason why this is two solutions, or the reason why this can be two solutions, is that we have a plus or a minus here. If this b squared minus 4ac is a positive number, so let's think about this a little bit. If b squared minus 4ac is greater than 0, what's going to happen? Well, then it's a positive number. It's going to have a square root."}, {"video_title": "Example 3 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "If this b squared minus 4ac is a positive number, so let's think about this a little bit. If b squared minus 4ac is greater than 0, what's going to happen? Well, then it's a positive number. It's going to have a square root. And then when you add it to negative b over 2a, or when you add it to negative b, you're going to get one value for the numerator. And when you subtract it from negative b, you're going to get another value in the numerator. So this is going to lead to two solutions."}, {"video_title": "Example 3 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "It's going to have a square root. And then when you add it to negative b over 2a, or when you add it to negative b, you're going to get one value for the numerator. And when you subtract it from negative b, you're going to get another value in the numerator. So this is going to lead to two solutions. Now, what happens if b squared minus 4ac is equal to 0? If this expression under the radical is equal to 0, you're just going to have the square root of 0. So it's going to be negative b plus or minus 0."}, {"video_title": "Example 3 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So this is going to lead to two solutions. Now, what happens if b squared minus 4ac is equal to 0? If this expression under the radical is equal to 0, you're just going to have the square root of 0. So it's going to be negative b plus or minus 0. And it doesn't matter whether you add or subtract 0, you're going to get the same value. So in that situation, the actual solution of the equation is going to be negative b over 2a. There's not going to be this plus or minus."}, {"video_title": "Example 3 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So it's going to be negative b plus or minus 0. And it doesn't matter whether you add or subtract 0, you're going to get the same value. So in that situation, the actual solution of the equation is going to be negative b over 2a. There's not going to be this plus or minus. It's not going to be relevant. So you're only going to have one solution. So if b squared minus 4ac is equal to 0, you only have one solution."}, {"video_title": "Example 3 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "There's not going to be this plus or minus. It's not going to be relevant. So you're only going to have one solution. So if b squared minus 4ac is equal to 0, you only have one solution. And then what happens if b squared minus 4ac is less than 0? Well, if b squared minus 4ac is less than 0, this is going to be a negative number right here. And you're going to have to take the square root of a negative number."}, {"video_title": "Example 3 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So if b squared minus 4ac is equal to 0, you only have one solution. And then what happens if b squared minus 4ac is less than 0? Well, if b squared minus 4ac is less than 0, this is going to be a negative number right here. And you're going to have to take the square root of a negative number. And we know, if we're dealing with real numbers, you can't take the square root. There is no real number squared that becomes a negative number. So in this situation, there is no solutions or no real solution."}, {"video_title": "Example 3 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "And you're going to have to take the square root of a negative number. And we know, if we're dealing with real numbers, you can't take the square root. There is no real number squared that becomes a negative number. So in this situation, there is no solutions or no real solution. So let's think about it in the context of this equation right here. And just in case you're curious if whether this expression right here, b squared minus 4ac, has a name, it does. It's called the discriminant."}, {"video_title": "Example 3 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So in this situation, there is no solutions or no real solution. So let's think about it in the context of this equation right here. And just in case you're curious if whether this expression right here, b squared minus 4ac, has a name, it does. It's called the discriminant. This is the discriminant. That's that part of the quadratic equation. It determines the number of solutions we have."}, {"video_title": "Example 3 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "It's called the discriminant. This is the discriminant. That's that part of the quadratic equation. It determines the number of solutions we have. So if we want to figure out the number of solutions in this equation, we don't have to go through the whole quadratic equation, although it's not that much work. We just have to evaluate b squared minus 4ac. So what is b squared minus 4ac?"}, {"video_title": "Example 3 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "It determines the number of solutions we have. So if we want to figure out the number of solutions in this equation, we don't have to go through the whole quadratic equation, although it's not that much work. We just have to evaluate b squared minus 4ac. So what is b squared minus 4ac? So b is right here. It's 14. So it's 14 squared minus 4 times a, which is 1, times c, which is 49."}, {"video_title": "Example 3 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So what is b squared minus 4ac? So b is right here. It's 14. So it's 14 squared minus 4 times a, which is 1, times c, which is 49. That's c right there, times 49. Well, what's 14 times 14? 14 times 14."}, {"video_title": "Example 3 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So it's 14 squared minus 4 times a, which is 1, times c, which is 49. That's c right there, times 49. Well, what's 14 times 14? 14 times 14. 4 times 4 is 16. 4 times 1 is 4, plus 1 is 56. Put a 0."}, {"video_title": "Example 3 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "14 times 14. 4 times 4 is 16. 4 times 1 is 4, plus 1 is 56. Put a 0. 1 times 14 is 14. It is 691. It's 196."}, {"video_title": "Example 3 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Put a 0. 1 times 14 is 14. It is 691. It's 196. So this right here is 196. And that's what's, we can ignore the 1. What's 4 times 49?"}, {"video_title": "Example 3 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "It's 196. So this right here is 196. And that's what's, we can ignore the 1. What's 4 times 49? So 49 times 4. 4 times 9 is 6. Sorry, it's 36, not 46."}, {"video_title": "Example 3 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "What's 4 times 49? So 49 times 4. 4 times 9 is 6. Sorry, it's 36, not 46. It's 36. 4 times 4 is 16, plus 3 is 19. So you get 196."}, {"video_title": "Example 3 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Sorry, it's 36, not 46. It's 36. 4 times 4 is 16, plus 3 is 19. So you get 196. So this right here is 196. So b squared minus 4ac is 196 minus 196. So 196 minus 196 is equal to 0."}, {"video_title": "Example 3 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So you get 196. So this right here is 196. So b squared minus 4ac is 196 minus 196. So 196 minus 196 is equal to 0. So we're dealing with a situation where the discriminant is equal to 0. We only have one solution. And if you want, you could try to find that one solution."}, {"video_title": "Example 3 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So 196 minus 196 is equal to 0. So we're dealing with a situation where the discriminant is equal to 0. We only have one solution. And if you want, you could try to find that one solution. This whole part is just going to be the square root of 0. It's just going to be 0. So the solution is going to be negative b over 2a."}, {"video_title": "Example 3 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "And if you want, you could try to find that one solution. This whole part is just going to be the square root of 0. It's just going to be 0. So the solution is going to be negative b over 2a. And negative b is, we could just solve it. Negative b is negative 14 over 2 times a. a is just 1, over 2. So it's equal to negative 7."}, {"video_title": "Example 3 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So the solution is going to be negative b over 2a. And negative b is, we could just solve it. Negative b is negative 14 over 2 times a. a is just 1, over 2. So it's equal to negative 7. That's the only solution to this equation. But if you just wanted to know how many solutions, you just have to find out that b squared minus 4ac is 0. So it's only going to have one solution."}, {"video_title": "Introduction to point-slope form Algebra I Khan Academy.mp3", "Sentence": "So what I've drawn here in yellow is a line. And let's say we know two things about this line. We know that it has a slope of m, and we know that the point a comma b is on this line. And so the question that we're going to try to answer is can we easily come up with an equation for this line using this information? Well, let's try it out. So any point on this line, or any x comma y on this line, would have to satisfy the condition that the slope between that point, so let's say that this is some point x comma y, it's an arbitrary point on the line. The fact that it's on the line tells us that the slope between a comma b and x comma y must be equal to m. So let's use that knowledge to actually construct an equation."}, {"video_title": "Introduction to point-slope form Algebra I Khan Academy.mp3", "Sentence": "And so the question that we're going to try to answer is can we easily come up with an equation for this line using this information? Well, let's try it out. So any point on this line, or any x comma y on this line, would have to satisfy the condition that the slope between that point, so let's say that this is some point x comma y, it's an arbitrary point on the line. The fact that it's on the line tells us that the slope between a comma b and x comma y must be equal to m. So let's use that knowledge to actually construct an equation. So what is the slope between a comma b and x comma y? Well, our change in y, remember, slope is just change in y over change in x. Let me write that."}, {"video_title": "Introduction to point-slope form Algebra I Khan Academy.mp3", "Sentence": "The fact that it's on the line tells us that the slope between a comma b and x comma y must be equal to m. So let's use that knowledge to actually construct an equation. So what is the slope between a comma b and x comma y? Well, our change in y, remember, slope is just change in y over change in x. Let me write that. Slope is equal to change in y over change in x. This little triangle character, that's the Greek letter delta, shorthand for change in. Our change in y, well, let's see."}, {"video_title": "Introduction to point-slope form Algebra I Khan Academy.mp3", "Sentence": "Let me write that. Slope is equal to change in y over change in x. This little triangle character, that's the Greek letter delta, shorthand for change in. Our change in y, well, let's see. If we start at y equals, we're starting at y is equal to b. And if we end up at y equals this arbitrary y right over here, this change in y right over here is going to be y minus b. Let me write it in those same colors."}, {"video_title": "Introduction to point-slope form Algebra I Khan Academy.mp3", "Sentence": "Our change in y, well, let's see. If we start at y equals, we're starting at y is equal to b. And if we end up at y equals this arbitrary y right over here, this change in y right over here is going to be y minus b. Let me write it in those same colors. So this is going to be y minus my little orange b. And that's going to be over our change in x. And the exact same logic, we start at x equals a."}, {"video_title": "Introduction to point-slope form Algebra I Khan Academy.mp3", "Sentence": "Let me write it in those same colors. So this is going to be y minus my little orange b. And that's going to be over our change in x. And the exact same logic, we start at x equals a. We finish at x equals this arbitrary x, whatever x we happen to be at. So that change in x is going to be that ending point minus our starting point, minus a. And we know this is the slope between these two points."}, {"video_title": "Introduction to point-slope form Algebra I Khan Academy.mp3", "Sentence": "And the exact same logic, we start at x equals a. We finish at x equals this arbitrary x, whatever x we happen to be at. So that change in x is going to be that ending point minus our starting point, minus a. And we know this is the slope between these two points. That's the slope between any two points on this line. And that's going to be equal to m. So this is going to be equal to m. And so what we've already done here, what we've already done right here, is actually create an equation that describes this line. It might not be in any form that you're used to seeing, but this is an equation that describes any xy that satisfies this equation right over here will be on the line."}, {"video_title": "Introduction to point-slope form Algebra I Khan Academy.mp3", "Sentence": "And we know this is the slope between these two points. That's the slope between any two points on this line. And that's going to be equal to m. So this is going to be equal to m. And so what we've already done here, what we've already done right here, is actually create an equation that describes this line. It might not be in any form that you're used to seeing, but this is an equation that describes any xy that satisfies this equation right over here will be on the line. Because any xy that satisfies this, the slope between that xy and this point right over here, between the point a, b, is going to be equal to m. So let's actually now convert this into forms that we might recognize more easily. So let me paste that. So to simplify this expression a little bit, or at least to get rid of the x minus a in the denominator, let's multiply both sides by x minus a."}, {"video_title": "Introduction to point-slope form Algebra I Khan Academy.mp3", "Sentence": "It might not be in any form that you're used to seeing, but this is an equation that describes any xy that satisfies this equation right over here will be on the line. Because any xy that satisfies this, the slope between that xy and this point right over here, between the point a, b, is going to be equal to m. So let's actually now convert this into forms that we might recognize more easily. So let me paste that. So to simplify this expression a little bit, or at least to get rid of the x minus a in the denominator, let's multiply both sides by x minus a. Let me do that same. So if we multiply both sides by x minus a, so x minus a on the left-hand side, x minus a on the left, and x minus a on the right, let me put some parentheses around it. So we're going to multiply both sides by x minus a."}, {"video_title": "Introduction to point-slope form Algebra I Khan Academy.mp3", "Sentence": "So to simplify this expression a little bit, or at least to get rid of the x minus a in the denominator, let's multiply both sides by x minus a. Let me do that same. So if we multiply both sides by x minus a, so x minus a on the left-hand side, x minus a on the left, and x minus a on the right, let me put some parentheses around it. So we're going to multiply both sides by x minus a. The whole point of that is you have x minus a divided by x minus a, which is just going to be equal to 1. And then on the right-hand side, you just have m times x minus a. So this whole thing has simplified to y minus b is equal to m times x minus a."}, {"video_title": "Introduction to point-slope form Algebra I Khan Academy.mp3", "Sentence": "So we're going to multiply both sides by x minus a. The whole point of that is you have x minus a divided by x minus a, which is just going to be equal to 1. And then on the right-hand side, you just have m times x minus a. So this whole thing has simplified to y minus b is equal to m times x minus a. And right here, this is a form that mathematicians have categorized as point-slope form. So this right over here is the point-slope form of the equation that describes this line. Now, why is it called point-slope form?"}, {"video_title": "Introduction to point-slope form Algebra I Khan Academy.mp3", "Sentence": "So this whole thing has simplified to y minus b is equal to m times x minus a. And right here, this is a form that mathematicians have categorized as point-slope form. So this right over here is the point-slope form of the equation that describes this line. Now, why is it called point-slope form? Well, it's very easy to inspect this and say, OK, well, look. This is the slope of the line in green. That's the slope of the line."}, {"video_title": "Introduction to point-slope form Algebra I Khan Academy.mp3", "Sentence": "Now, why is it called point-slope form? Well, it's very easy to inspect this and say, OK, well, look. This is the slope of the line in green. That's the slope of the line. And I can put the two points in. If the point a comma b is on this line, I'll have the slope times x minus a is equal to y minus b. Now, let's see why this is useful, why people like to use this type of thing."}, {"video_title": "Introduction to point-slope form Algebra I Khan Academy.mp3", "Sentence": "That's the slope of the line. And I can put the two points in. If the point a comma b is on this line, I'll have the slope times x minus a is equal to y minus b. Now, let's see why this is useful, why people like to use this type of thing. Let's not use just a, b, and a slope of m anymore. Let's make this a little bit more concrete. Let's say that someone tells you that I'm dealing with some line where the slope is equal to 2, and it goes through the point."}, {"video_title": "Introduction to point-slope form Algebra I Khan Academy.mp3", "Sentence": "Now, let's see why this is useful, why people like to use this type of thing. Let's not use just a, b, and a slope of m anymore. Let's make this a little bit more concrete. Let's say that someone tells you that I'm dealing with some line where the slope is equal to 2, and it goes through the point. It goes through the point, I don't know, let's say it goes through the point negative 7 comma 5. So very quickly, you could use this information and your knowledge of point-slope form to write this in this form. You would just say, well, an equation that contains this point and has this slope would be y minus b, which is 5, y minus the y-coordinate of the point that this line contains, is equal to my slope times x minus the x-coordinate that this line contains."}, {"video_title": "Introduction to point-slope form Algebra I Khan Academy.mp3", "Sentence": "Let's say that someone tells you that I'm dealing with some line where the slope is equal to 2, and it goes through the point. It goes through the point, I don't know, let's say it goes through the point negative 7 comma 5. So very quickly, you could use this information and your knowledge of point-slope form to write this in this form. You would just say, well, an equation that contains this point and has this slope would be y minus b, which is 5, y minus the y-coordinate of the point that this line contains, is equal to my slope times x minus the x-coordinate that this line contains. So x minus negative 7. And just like that, we have written an equation that has a slope of 2 and that contains this point right over here. And if we don't like the x minus negative 7 right over here, we can obviously rewrite that as x plus 7."}, {"video_title": "Introduction to point-slope form Algebra I Khan Academy.mp3", "Sentence": "You would just say, well, an equation that contains this point and has this slope would be y minus b, which is 5, y minus the y-coordinate of the point that this line contains, is equal to my slope times x minus the x-coordinate that this line contains. So x minus negative 7. And just like that, we have written an equation that has a slope of 2 and that contains this point right over here. And if we don't like the x minus negative 7 right over here, we can obviously rewrite that as x plus 7. But this is kind of the purest point-slope form. If you want to simplify it a little bit, you could write it as y minus 5 is equal to 2 times x plus 7. And if you want to see that this is just one way of expressing the equation of this line, there are many others."}, {"video_title": "Introduction to point-slope form Algebra I Khan Academy.mp3", "Sentence": "And if we don't like the x minus negative 7 right over here, we can obviously rewrite that as x plus 7. But this is kind of the purest point-slope form. If you want to simplify it a little bit, you could write it as y minus 5 is equal to 2 times x plus 7. And if you want to see that this is just one way of expressing the equation of this line, there are many others. And the one that we're most familiar with is y-intercept form. This can easily be converted to y-intercept form. To do that, we just have to distribute this 2."}, {"video_title": "Introduction to point-slope form Algebra I Khan Academy.mp3", "Sentence": "And if you want to see that this is just one way of expressing the equation of this line, there are many others. And the one that we're most familiar with is y-intercept form. This can easily be converted to y-intercept form. To do that, we just have to distribute this 2. So we get y minus 5 is equal to 2 times x plus 2 times 7. So it's equal to 14. And then we can get rid of this negative 5 on the left by adding 5 to both sides of this equation."}, {"video_title": "Introduction to point-slope form Algebra I Khan Academy.mp3", "Sentence": "To do that, we just have to distribute this 2. So we get y minus 5 is equal to 2 times x plus 2 times 7. So it's equal to 14. And then we can get rid of this negative 5 on the left by adding 5 to both sides of this equation. And then we are left with, on the left-hand side, y. And on the right-hand side, 2x plus 19. So this right over here is slope-intercept form."}, {"video_title": "Introduction to point-slope form Algebra I Khan Academy.mp3", "Sentence": "And then we can get rid of this negative 5 on the left by adding 5 to both sides of this equation. And then we are left with, on the left-hand side, y. And on the right-hand side, 2x plus 19. So this right over here is slope-intercept form. You have your slope and your y-intercept. So this is slope-intercept form. And this right up here is point-slope form."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "Let me show you what I'm talking about. Let's have the equation the square root of x is equal to 2 times x minus 6. Now one of the things you're going to see whenever we do these radical equations is we want to isolate at least one of the radicals. There's only one of them in this equation. And when you isolate one of the radicals on one side of the equation, this one starts off like that. I have the square root of x isolated on the left-hand side, then we square both sides of the equation. So let us square both sides of the equation."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "There's only one of them in this equation. And when you isolate one of the radicals on one side of the equation, this one starts off like that. I have the square root of x isolated on the left-hand side, then we square both sides of the equation. So let us square both sides of the equation. So I'll just rewrite it again. We'll do this one slowly. I'm going to square that and that's going to be equal to 2x minus 6 squared."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "So let us square both sides of the equation. So I'll just rewrite it again. We'll do this one slowly. I'm going to square that and that's going to be equal to 2x minus 6 squared. And squaring seems like a valid operation. If that is equal to that, then that squared should also be equal to that squared. So we just keep on going."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "I'm going to square that and that's going to be equal to 2x minus 6 squared. And squaring seems like a valid operation. If that is equal to that, then that squared should also be equal to that squared. So we just keep on going. So when you take the square root of x and you square it, that'll just be x. And we get x is equal to this squared is going to be 2x squared, which is 4x squared. It's 2x squared, the whole thing."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "So we just keep on going. So when you take the square root of x and you square it, that'll just be x. And we get x is equal to this squared is going to be 2x squared, which is 4x squared. It's 2x squared, the whole thing. 4x squared, and then you multiply these two, which is negative 12x, and it's going to be twice that. So minus 24x. And then negative 6 squared is plus 36."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "It's 2x squared, the whole thing. 4x squared, and then you multiply these two, which is negative 12x, and it's going to be twice that. So minus 24x. And then negative 6 squared is plus 36. If you found going from this to this difficult, you might want to review multiplying polynomial expressions or multiplying binomials, or actually the special case where we square binomials. But the general view is this squared, which is that. And then you have minus 2 times the product of these two."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "And then negative 6 squared is plus 36. If you found going from this to this difficult, you might want to review multiplying polynomial expressions or multiplying binomials, or actually the special case where we square binomials. But the general view is this squared, which is that. And then you have minus 2 times the product of these two. The product of those two is minus 12x, or negative 12x. 2 times that is negative 24x. And then that squared."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "And then you have minus 2 times the product of these two. The product of those two is minus 12x, or negative 12x. 2 times that is negative 24x. And then that squared. So this is what our equation has, I guess we could say, simplified to. And let's see what happens if we subtract x from both sides of this equation. So if you subtract x from both sides of this equation, the left-hand side becomes 0."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "And then that squared. So this is what our equation has, I guess we could say, simplified to. And let's see what happens if we subtract x from both sides of this equation. So if you subtract x from both sides of this equation, the left-hand side becomes 0. And the right-hand side becomes 4x squared minus 25x plus 36. So this radical equation has simplified to just a standard quadratic equation. And just for simplicity, not having to worry how to factor it and group it and all of that, let's just use the quadratic formula."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "So if you subtract x from both sides of this equation, the left-hand side becomes 0. And the right-hand side becomes 4x squared minus 25x plus 36. So this radical equation has simplified to just a standard quadratic equation. And just for simplicity, not having to worry how to factor it and group it and all of that, let's just use the quadratic formula. So the quadratic formula tells us that our solutions to this x can be negative b, negative 25. The negative of negative 25 is positive 25 plus or minus the square root of 25 squared. 25 squared is 625 minus 4 times a, which is 4, times c, which is 36."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "And just for simplicity, not having to worry how to factor it and group it and all of that, let's just use the quadratic formula. So the quadratic formula tells us that our solutions to this x can be negative b, negative 25. The negative of negative 25 is positive 25 plus or minus the square root of 25 squared. 25 squared is 625 minus 4 times a, which is 4, times c, which is 36. All of that over 2 times 4. All of that over 8. So let's get our calculator out to figure out what this is over here."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "25 squared is 625 minus 4 times a, which is 4, times c, which is 36. All of that over 2 times 4. All of that over 8. So let's get our calculator out to figure out what this is over here. So let's say we have 625 minus, let's see, this is going to be 16 times 36. 16 times 36 is equal to 49. That's nice."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "So let's get our calculator out to figure out what this is over here. So let's say we have 625 minus, let's see, this is going to be 16 times 36. 16 times 36 is equal to 49. That's nice. It's a nice, perfect square. We know what the square root of 49 is. It's 7."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "That's nice. It's a nice, perfect square. We know what the square root of 49 is. It's 7. So let me go back to the problem. So this in here simplified to 49. So x is equal to 25 plus or minus the square root of 49, which is 7."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "It's 7. So let me go back to the problem. So this in here simplified to 49. So x is equal to 25 plus or minus the square root of 49, which is 7. All of that over 8. So our two solutions here, if we add 7, we get x is equal to 25 plus 7 is 32. 32 over 8, which is equal to 4."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "So x is equal to 25 plus or minus the square root of 49, which is 7. All of that over 8. So our two solutions here, if we add 7, we get x is equal to 25 plus 7 is 32. 32 over 8, which is equal to 4. And then our other solution, let me do that in a different color, x is equal to 25 minus 7, which is 18 over 8. 8 goes into 18 two times, remainder 2. So this is equal to 2 and 2 eighths, or 2 and 1 fourth, or 2.25, just like that."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "32 over 8, which is equal to 4. And then our other solution, let me do that in a different color, x is equal to 25 minus 7, which is 18 over 8. 8 goes into 18 two times, remainder 2. So this is equal to 2 and 2 eighths, or 2 and 1 fourth, or 2.25, just like that. Now, I'm going to show you an interesting phenomena that occurs. And maybe you might want to pause it after I show you this conundrum, although I'm going to tell you why this conundrum pops up. Let's try out to see if our solutions actually work."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "So this is equal to 2 and 2 eighths, or 2 and 1 fourth, or 2.25, just like that. Now, I'm going to show you an interesting phenomena that occurs. And maybe you might want to pause it after I show you this conundrum, although I'm going to tell you why this conundrum pops up. Let's try out to see if our solutions actually work. So let's try x is equal to 4. So if x is equal to 4 works, we get the principal root of 4 should be equal to 2 times 4 minus 6. The principal root of 4 is positive 2."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "Let's try out to see if our solutions actually work. So let's try x is equal to 4. So if x is equal to 4 works, we get the principal root of 4 should be equal to 2 times 4 minus 6. The principal root of 4 is positive 2. Positive 2 should be equal to 2 times 4, which is 8 minus 6, which it is. This is true. So 4 works."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "The principal root of 4 is positive 2. Positive 2 should be equal to 2 times 4, which is 8 minus 6, which it is. This is true. So 4 works. Now, let's try to do the same with 2.25. According to this, we should be able to take the square root, the principal root of 2.25 should be equal to 2 times 2.25 minus 6. Now, you may or may not be able to do this in your head."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "So 4 works. Now, let's try to do the same with 2.25. According to this, we should be able to take the square root, the principal root of 2.25 should be equal to 2 times 2.25 minus 6. Now, you may or may not be able to do this in your head. You might know that the square root of 225 is 15. And then from that, you might be able to figure out that the square root of 2.25 is 1.5. But let me just use a calculator to verify that for you."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "Now, you may or may not be able to do this in your head. You might know that the square root of 225 is 15. And then from that, you might be able to figure out that the square root of 2.25 is 1.5. But let me just use a calculator to verify that for you. So 2.25, take the square root, it's 1.5. The principal root is 1.5. Another square root is negative 1.5."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "But let me just use a calculator to verify that for you. So 2.25, take the square root, it's 1.5. The principal root is 1.5. Another square root is negative 1.5. So it's 1.5. And then according to this, this should be equal to 2 times 2.25 is 4.5 minus 6. Now, is this true?"}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "Another square root is negative 1.5. So it's 1.5. And then according to this, this should be equal to 2 times 2.25 is 4.5 minus 6. Now, is this true? This is telling us that 1.5 is equal to negative 1.5. This is not true. 2.5 did not work for this radical equation."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "Now, is this true? This is telling us that 1.5 is equal to negative 1.5. This is not true. 2.5 did not work for this radical equation. We call this an extraneous solution. So 2.25 is extraneous. Now, here's the conundrum."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "2.5 did not work for this radical equation. We call this an extraneous solution. So 2.25 is extraneous. Now, here's the conundrum. Why did we get 2.25 as an answer? It looks like we did very valid things the whole way down, and we got a quadratic, and we got 2.25. And there's a hint here."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "Now, here's the conundrum. Why did we get 2.25 as an answer? It looks like we did very valid things the whole way down, and we got a quadratic, and we got 2.25. And there's a hint here. When we substitute 2.25, we get 1.5 is equal to negative 1.5. So there's something here that something we did gave us this solution that doesn't quite apply over here. And I'll give you another hint."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "And there's a hint here. When we substitute 2.25, we get 1.5 is equal to negative 1.5. So there's something here that something we did gave us this solution that doesn't quite apply over here. And I'll give you another hint. Let's try it at this step. If you look at this step, you're going to see that both solutions actually work. So you could try it out if you like."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "And I'll give you another hint. Let's try it at this step. If you look at this step, you're going to see that both solutions actually work. So you could try it out if you like. Actually, try it out on your own time. Put in 2.25 for x here, you're going to see that it works. Put in 4 for x here, and you see that they both work here."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "So you could try it out if you like. Actually, try it out on your own time. Put in 2.25 for x here, you're going to see that it works. Put in 4 for x here, and you see that they both work here. So they're both valid solutions to that. So something happened when we squared that made the equation a little bit different. There's something slightly different about this equation than that equation."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "Put in 4 for x here, and you see that they both work here. So they're both valid solutions to that. So something happened when we squared that made the equation a little bit different. There's something slightly different about this equation than that equation. And the answer is, there's two ways you could think about it. To go back from this equation to that equation, we take the square root. But to be more particular about it, we are taking the principal root of both sides."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "There's something slightly different about this equation than that equation. And the answer is, there's two ways you could think about it. To go back from this equation to that equation, we take the square root. But to be more particular about it, we are taking the principal root of both sides. Now, you could take the negative square root as well. Notice, this is only taking the principal square root. Going from this right here, let me be very clear."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "But to be more particular about it, we are taking the principal root of both sides. Now, you could take the negative square root as well. Notice, this is only taking the principal square root. Going from this right here, let me be very clear. This statement, we already established that both of these solutions, both the valid solution and the extraneous solution to this radical equation, satisfy this right here, only the valid one satisfies the original problem. So let me write the equation that both of them satisfy. Because this is really an interesting conundrum."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "Going from this right here, let me be very clear. This statement, we already established that both of these solutions, both the valid solution and the extraneous solution to this radical equation, satisfy this right here, only the valid one satisfies the original problem. So let me write the equation that both of them satisfy. Because this is really an interesting conundrum. And I think it gives you a little bit of a nuance and kind of tells you what's happening when we take principal roots of things. And why, when you square both sides, you are, to some degree, you can either think of it as losing or gaining some information. Now, this could be written as x is equal to 2x minus 6 squared."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "Because this is really an interesting conundrum. And I think it gives you a little bit of a nuance and kind of tells you what's happening when we take principal roots of things. And why, when you square both sides, you are, to some degree, you can either think of it as losing or gaining some information. Now, this could be written as x is equal to 2x minus 6 squared. This is one valid interpretation of this equation right here. But there's a completely other legitimate interpretation of this equation. This could also be x is equal to negative 1 times 2x minus 6 squared."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "Now, this could be written as x is equal to 2x minus 6 squared. This is one valid interpretation of this equation right here. But there's a completely other legitimate interpretation of this equation. This could also be x is equal to negative 1 times 2x minus 6 squared. And why are these equal interpretations? Because when you square the negative 1, the negative 1 will disappear. These are equivalent statements."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "This could also be x is equal to negative 1 times 2x minus 6 squared. And why are these equal interpretations? Because when you square the negative 1, the negative 1 will disappear. These are equivalent statements. And another way of writing this one, another way of writing this right here, is that x is equal to, you multiply negative 1 times that, you get negative 2x plus 6, or 6 minus 2x squared. This and this are two ways of writing that. Now, when we took our square root, or when we, I guess there's two ways you can think about it, when we squared it, we're assuming that this was the only interpretation."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "These are equivalent statements. And another way of writing this one, another way of writing this right here, is that x is equal to, you multiply negative 1 times that, you get negative 2x plus 6, or 6 minus 2x squared. This and this are two ways of writing that. Now, when we took our square root, or when we, I guess there's two ways you can think about it, when we squared it, we're assuming that this was the only interpretation. But this was the other one. So we found two solutions to this, but only 4 satisfies this interpretation right here. And I hope you get what I'm saying, because we're kind of only taking, you can kind of imagine it, the positive square root."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "Now, when we took our square root, or when we, I guess there's two ways you can think about it, when we squared it, we're assuming that this was the only interpretation. But this was the other one. So we found two solutions to this, but only 4 satisfies this interpretation right here. And I hope you get what I'm saying, because we're kind of only taking, you can kind of imagine it, the positive square root. We're not considering the negative square root of this. Because when you take the square root of both sides to get here, we're only taking the principal root. Another way to view it, let me rewrite the original equation."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "And I hope you get what I'm saying, because we're kind of only taking, you can kind of imagine it, the positive square root. We're not considering the negative square root of this. Because when you take the square root of both sides to get here, we're only taking the principal root. Another way to view it, let me rewrite the original equation. Let me rewrite the original equation. We had the square root of x is equal to 2x minus 6. Now, we said 4 is a solution."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "Another way to view it, let me rewrite the original equation. Let me rewrite the original equation. We had the square root of x is equal to 2x minus 6. Now, we said 4 is a solution. 2.25 isn't a solution. 2.25 would have been a solution if we said both of the square roots of x is equal to 2x minus 6. Now you try it out, and 2.25 will have a valid solution here."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "Now, we said 4 is a solution. 2.25 isn't a solution. 2.25 would have been a solution if we said both of the square roots of x is equal to 2x minus 6. Now you try it out, and 2.25 will have a valid solution here. If you take the negative square root of 2.25, that is equal to 2 times 2.25. So that is equal to 4.5 minus 6, which is negative 1.5. That is true."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "Now you try it out, and 2.25 will have a valid solution here. If you take the negative square root of 2.25, that is equal to 2 times 2.25. So that is equal to 4.5 minus 6, which is negative 1.5. That is true. The positive version is where you get x is equal to 4. So that's why we got two solutions. And if you square this, maybe this is an easier way to remember it."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "That is true. The positive version is where you get x is equal to 4. So that's why we got two solutions. And if you square this, maybe this is an easier way to remember it. If you square this, you actually get this equation that both solutions are valid. Now, you might have found that a little bit confusing and all that. My intention is not to confuse you."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "And if you square this, maybe this is an easier way to remember it. If you square this, you actually get this equation that both solutions are valid. Now, you might have found that a little bit confusing and all that. My intention is not to confuse you. The simple thing to think about when you're solving radical equations is, look, isolate radicals, square, keep on solving, you might get more than one answer. Plug your answers back in. Answers that don't work, they're extraneous solutions."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "My intention is not to confuse you. The simple thing to think about when you're solving radical equations is, look, isolate radicals, square, keep on solving, you might get more than one answer. Plug your answers back in. Answers that don't work, they're extraneous solutions. But most of my explanation in this video is really why does that extraneous solution pop up? And hopefully I gave you some intuition. That our equation is the square root of x."}, {"video_title": "Applying rational equations 3 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Used together, the two hoses take 12 minutes to fill the pond. If used alone, one hose is able to fill the pond 10 minutes faster than the other. So 10 minutes faster than the other hose. How long does each hose take to fill the pond by itself? So let's think about each of the hoses. We have a faster hose, we have a faster hose, we have a slower hose. And let's say that the faster hose fills the pond, let's say it takes him F minutes per pond."}, {"video_title": "Applying rational equations 3 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "How long does each hose take to fill the pond by itself? So let's think about each of the hoses. We have a faster hose, we have a faster hose, we have a slower hose. And let's say that the faster hose fills the pond, let's say it takes him F minutes per pond. Now, how long is it going to take the slower hose? Well, the faster hose does it in 10 minutes less. It's 10 minutes faster."}, {"video_title": "Applying rational equations 3 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And let's say that the faster hose fills the pond, let's say it takes him F minutes per pond. Now, how long is it going to take the slower hose? Well, the faster hose does it in 10 minutes less. It's 10 minutes faster. So the slower hose is going to take 10 minutes more. The slower hose is going to take F plus 10 minutes per fish pond. Now, this is in minutes per pond, but if we want to be able to add rates together, we should really think about it in terms of ponds per minute."}, {"video_title": "Applying rational equations 3 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "It's 10 minutes faster. So the slower hose is going to take 10 minutes more. The slower hose is going to take F plus 10 minutes per fish pond. Now, this is in minutes per pond, but if we want to be able to add rates together, we should really think about it in terms of ponds per minute. So let's rewrite each of these statements as ponds per minute. You could write this as F minutes per one pond, or F plus 10 minutes per one pond. And if you just take the inverse of each of these statements, these ratios are equivalent to saying one pond per F minutes."}, {"video_title": "Applying rational equations 3 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Now, this is in minutes per pond, but if we want to be able to add rates together, we should really think about it in terms of ponds per minute. So let's rewrite each of these statements as ponds per minute. You could write this as F minutes per one pond, or F plus 10 minutes per one pond. And if you just take the inverse of each of these statements, these ratios are equivalent to saying one pond per F minutes. So it's really not saying anything else, I'm just inverting the ratio. Or you could think of it as 1 over F ponds per minute. Same logic right here."}, {"video_title": "Applying rational equations 3 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And if you just take the inverse of each of these statements, these ratios are equivalent to saying one pond per F minutes. So it's really not saying anything else, I'm just inverting the ratio. Or you could think of it as 1 over F ponds per minute. Same logic right here. We could essentially rewrite this ratio as 1 over F plus 10 ponds per minute. So now we have the rate of the faster hose, we have the rate of the slower hose, how many ponds per minute for the faster hose, how many ponds per minute for the slower hose. If we add these two rates, we'll know the ponds per minute when they're acting together."}, {"video_title": "Applying rational equations 3 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Same logic right here. We could essentially rewrite this ratio as 1 over F plus 10 ponds per minute. So now we have the rate of the faster hose, we have the rate of the slower hose, how many ponds per minute for the faster hose, how many ponds per minute for the slower hose. If we add these two rates, we'll know the ponds per minute when they're acting together. So if we have 1 over F ponds per minute plus 1 over F plus 10 ponds per minute, this is the faster hose, this is the slower hose, this will tell us how many ponds per minute they can do together. Now, we know that information. They say the two hoses take 12 minutes."}, {"video_title": "Applying rational equations 3 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "If we add these two rates, we'll know the ponds per minute when they're acting together. So if we have 1 over F ponds per minute plus 1 over F plus 10 ponds per minute, this is the faster hose, this is the slower hose, this will tell us how many ponds per minute they can do together. Now, we know that information. They say the two hoses take 12 minutes. So let me write that over here. The combined take 12 minutes per pond. So what is their combined rate in terms of ponds per minute?"}, {"video_title": "Applying rational equations 3 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "They say the two hoses take 12 minutes. So let me write that over here. The combined take 12 minutes per pond. So what is their combined rate in terms of ponds per minute? So you could use this 12 minutes per 1 pond. You could take the inverse of this, or take the ratio in terms of ponds per minute instead, and you get 1 over 12 ponds per minute. So if you take it combined, they'll fill 1 twelfth of a pond, which makes complete sense, because it takes them 12 minutes to fill the whole thing, so in 1 minute they'll only do 1 twelfth of it."}, {"video_title": "Applying rational equations 3 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So what is their combined rate in terms of ponds per minute? So you could use this 12 minutes per 1 pond. You could take the inverse of this, or take the ratio in terms of ponds per minute instead, and you get 1 over 12 ponds per minute. So if you take it combined, they'll fill 1 twelfth of a pond, which makes complete sense, because it takes them 12 minutes to fill the whole thing, so in 1 minute they'll only do 1 twelfth of it. So this is their combined rate in ponds per minute. This is also their combined rate in ponds per minute, so this is going to be equal to 1 over 12. And now we just have to solve for F, and then F plus 10 is going to be how long it takes the slower hose."}, {"video_title": "Applying rational equations 3 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So if you take it combined, they'll fill 1 twelfth of a pond, which makes complete sense, because it takes them 12 minutes to fill the whole thing, so in 1 minute they'll only do 1 twelfth of it. So this is their combined rate in ponds per minute. This is also their combined rate in ponds per minute, so this is going to be equal to 1 over 12. And now we just have to solve for F, and then F plus 10 is going to be how long it takes the slower hose. So let's multiply both sides of this equation times F, and times F plus 10. So let's do that. So I'm going to multiply both sides of this equation times F, and F plus 10, times both sides of this equation."}, {"video_title": "Applying rational equations 3 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And now we just have to solve for F, and then F plus 10 is going to be how long it takes the slower hose. So let's multiply both sides of this equation times F, and times F plus 10. So let's do that. So I'm going to multiply both sides of this equation times F, and F plus 10, times both sides of this equation. So F and F plus 10. Scroll down a little bit. Scroll to the left so we have some real estate."}, {"video_title": "Applying rational equations 3 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So I'm going to multiply both sides of this equation times F, and F plus 10, times both sides of this equation. So F and F plus 10. Scroll down a little bit. Scroll to the left so we have some real estate. So let's distribute this F times F plus 10. So if we multiply F times F plus 10 times 1 over F, that F and that F will cancel out, and we're just going to be left with an F plus 10. That's when you multiply that term times the F times F plus 10."}, {"video_title": "Applying rational equations 3 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Scroll to the left so we have some real estate. So let's distribute this F times F plus 10. So if we multiply F times F plus 10 times 1 over F, that F and that F will cancel out, and we're just going to be left with an F plus 10. That's when you multiply that term times the F times F plus 10. Now when you multiply this term, when you multiply 1 over F plus 10 times F times F plus 10, this and this will cancel out, and you're just left with an F. So you have plus F is equal to, and you have over 12. Actually, well, in a second, let me multiply all sides of this equation by 12. I'll do that next in a second."}, {"video_title": "Applying rational equations 3 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "That's when you multiply that term times the F times F plus 10. Now when you multiply this term, when you multiply 1 over F plus 10 times F times F plus 10, this and this will cancel out, and you're just left with an F. So you have plus F is equal to, and you have over 12. Actually, well, in a second, let me multiply all sides of this equation by 12. I'll do that next in a second. So let's just say this is going to be equal to 1 over 12 times F squared, times F squared. F times F is F squared, plus 10F. And now let's just multiply both sides of this equation by 12."}, {"video_title": "Applying rational equations 3 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "I'll do that next in a second. So let's just say this is going to be equal to 1 over 12 times F squared, times F squared. F times F is F squared, plus 10F. And now let's just multiply both sides of this equation by 12. I could have done it in the last step so that we don't have any fractions here. And so the left-hand side, we get 12 times F. We get 12F plus 120 plus 12F. The right-hand side, that and that cancels out, and you are left with F squared plus 10F."}, {"video_title": "Applying rational equations 3 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And now let's just multiply both sides of this equation by 12. I could have done it in the last step so that we don't have any fractions here. And so the left-hand side, we get 12 times F. We get 12F plus 120 plus 12F. The right-hand side, that and that cancels out, and you are left with F squared plus 10F. And now we have a quadratic. We just have to get into a form that we know how to manipulate or deal with. And before that, we can simplify it."}, {"video_title": "Applying rational equations 3 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "The right-hand side, that and that cancels out, and you are left with F squared plus 10F. And now we have a quadratic. We just have to get into a form that we know how to manipulate or deal with. And before that, we can simplify it. We have a 12F and a 12F. So this becomes 24F plus 120 is equal to F squared plus 10F. And then let's get all of this stuff out of the left-hand side."}, {"video_title": "Applying rational equations 3 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And before that, we can simplify it. We have a 12F and a 12F. So this becomes 24F plus 120 is equal to F squared plus 10F. And then let's get all of this stuff out of the left-hand side. Let's get it all on the right-hand side. So from both sides of this equation, let's subtract 24F and a negative 120, or minus 120. So you have minus 24F minus 120."}, {"video_title": "Applying rational equations 3 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And then let's get all of this stuff out of the left-hand side. Let's get it all on the right-hand side. So from both sides of this equation, let's subtract 24F and a negative 120, or minus 120. So you have minus 24F minus 120. The left-hand side just becomes 0. That was the whole point. The right-hand side is F squared."}, {"video_title": "Applying rational equations 3 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So you have minus 24F minus 120. The left-hand side just becomes 0. That was the whole point. The right-hand side is F squared. 10 minus 24F is negative 14F minus 120. Now we could factor this. Let's see."}, {"video_title": "Applying rational equations 3 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "The right-hand side is F squared. 10 minus 24F is negative 14F minus 120. Now we could factor this. Let's see. If you do 20 times 6, yeah, that looks like it would work. So negative 20 and 6, when you take their product, give you negative 120. And negative 20 plus 6 is negative 14."}, {"video_title": "Applying rational equations 3 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Let's see. If you do 20 times 6, yeah, that looks like it would work. So negative 20 and 6, when you take their product, give you negative 120. And negative 20 plus 6 is negative 14. So we could factor this right-hand side. 0 is equal to F minus 20 times F plus 6. When you multiply negative 20 times 6, you get negative 120."}, {"video_title": "Applying rational equations 3 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And negative 20 plus 6 is negative 14. So we could factor this right-hand side. 0 is equal to F minus 20 times F plus 6. When you multiply negative 20 times 6, you get negative 120. Negative 20 plus 6 is negative 14. And the only way that that's going to be equal to 0 is if F minus 20 is equal to 0, or F plus 6 is equal to 0. Add 20 to both sides of this equation."}, {"video_title": "Applying rational equations 3 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "When you multiply negative 20 times 6, you get negative 120. Negative 20 plus 6 is negative 14. And the only way that that's going to be equal to 0 is if F minus 20 is equal to 0, or F plus 6 is equal to 0. Add 20 to both sides of this equation. You get F is equal to 20. Remember, F is how many minutes does it take for the fast hose to fill the pond. And then if you take this one, you subtract 6 from both sides, you get F is equal to negative 6."}, {"video_title": "Applying rational equations 3 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Add 20 to both sides of this equation. You get F is equal to 20. Remember, F is how many minutes does it take for the fast hose to fill the pond. And then if you take this one, you subtract 6 from both sides, you get F is equal to negative 6. Now, when we're talking about how many minutes does it take for the fast hose to fill the pond, it doesn't make any sense to say that it takes it negative 6 minutes to fill the pond. So we can't use this answer. We need a positive answer."}, {"video_title": "Applying rational equations 3 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And then if you take this one, you subtract 6 from both sides, you get F is equal to negative 6. Now, when we're talking about how many minutes does it take for the fast hose to fill the pond, it doesn't make any sense to say that it takes it negative 6 minutes to fill the pond. So we can't use this answer. We need a positive answer. So this is how many minutes it takes the fast hose to fill the pond. F is equal to 20. So this right here, the faster hose, takes 20 minutes per pond."}, {"video_title": "Applying rational equations 3 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "We need a positive answer. So this is how many minutes it takes the fast hose to fill the pond. F is equal to 20. So this right here, the faster hose, takes 20 minutes per pond. So 20, I'll write it here, 20 minutes per pond is the fast hose. And then the slower hose takes 10 minutes more. It's F plus 10."}, {"video_title": "Applying rational equations 3 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So this right here, the faster hose, takes 20 minutes per pond. So 20, I'll write it here, 20 minutes per pond is the fast hose. And then the slower hose takes 10 minutes more. It's F plus 10. So it takes 30 minutes per pond. And we're done. I don't want to confuse you with this stuff."}, {"video_title": "Applying rational equations 3 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "It's F plus 10. So it takes 30 minutes per pond. And we're done. I don't want to confuse you with this stuff. The faster hose takes 20 minutes. The slower hose takes 30 minutes per pond. If they were to do it together, it would take 12 minutes, which is a little bit more than half."}, {"video_title": "Applying rational equations 3 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "I don't want to confuse you with this stuff. The faster hose takes 20 minutes. The slower hose takes 30 minutes per pond. If they were to do it together, it would take 12 minutes, which is a little bit more than half. If you had two faster hoses, it would take 10 minutes. But this guy is a little bit slower, so it's taking you a little bit more than 10 minutes. So it makes sense."}, {"video_title": "Extraneous solutions of radical equations Mathematics III High School Math Khan Academy.mp3", "Sentence": "So we already have the radical isolated on one side of the equation, so we might say, well, let's just get rid of the radical. Let's square both sides of this equation. So we might say that this is the same thing as two x minus one squared is equal to the square root of eight minus x, eight minus x squared. And then we would get, let's see, two x minus one squared is four x squared minus four x plus one is equal to eight minus x. Now we have to be very, very, very careful here. We might feel like, hey, we did legitimate operations. We did the same thing to both sides, that these are equivalent equations, but they aren't quite equivalent because when you're squaring something, one way to think about it is when you're squaring it, you're losing information."}, {"video_title": "Extraneous solutions of radical equations Mathematics III High School Math Khan Academy.mp3", "Sentence": "And then we would get, let's see, two x minus one squared is four x squared minus four x plus one is equal to eight minus x. Now we have to be very, very, very careful here. We might feel like, hey, we did legitimate operations. We did the same thing to both sides, that these are equivalent equations, but they aren't quite equivalent because when you're squaring something, one way to think about it is when you're squaring it, you're losing information. So for example, this would be true even if the original equation were two x, let me do this in a different color, even if the original equation were two x minus one is equal to the negative of the square root of eight minus x. Because if you squared both sides of this, you would also get, you would also get that right over there because the negative squared would be equal to a positive. So when we're finding a solution to this, we need to test our solution to make sure it's truly the solution to this first yellow equation here and not the solution to this up here."}, {"video_title": "Extraneous solutions of radical equations Mathematics III High School Math Khan Academy.mp3", "Sentence": "We did the same thing to both sides, that these are equivalent equations, but they aren't quite equivalent because when you're squaring something, one way to think about it is when you're squaring it, you're losing information. So for example, this would be true even if the original equation were two x, let me do this in a different color, even if the original equation were two x minus one is equal to the negative of the square root of eight minus x. Because if you squared both sides of this, you would also get, you would also get that right over there because the negative squared would be equal to a positive. So when we're finding a solution to this, we need to test our solution to make sure it's truly the solution to this first yellow equation here and not the solution to this up here. If it's a solution to this right-hand side one and not the yellow one, then we would call that an extraneous solution. So let's see if we can solve this. So let's write this as kind of a standard quadratic."}, {"video_title": "Extraneous solutions of radical equations Mathematics III High School Math Khan Academy.mp3", "Sentence": "So when we're finding a solution to this, we need to test our solution to make sure it's truly the solution to this first yellow equation here and not the solution to this up here. If it's a solution to this right-hand side one and not the yellow one, then we would call that an extraneous solution. So let's see if we can solve this. So let's write this as kind of a standard quadratic. Let's subtract eight from both sides. So let's subtract eight from both sides to get rid of this eight over here, and let's add x to both sides. So plus x, plus x, and we are going to get, we are going to get four x squared minus three x minus seven, minus seven is equal to, is equal to zero."}, {"video_title": "Extraneous solutions of radical equations Mathematics III High School Math Khan Academy.mp3", "Sentence": "So let's write this as kind of a standard quadratic. Let's subtract eight from both sides. So let's subtract eight from both sides to get rid of this eight over here, and let's add x to both sides. So plus x, plus x, and we are going to get, we are going to get four x squared minus three x minus seven, minus seven is equal to, is equal to zero. And let's see, we would want to factor this right over here, and let's see, maybe I could do this by, if I do it by, well, I'll just use a quadratic formula here. So the solutions are going to be x is going to be equal to negative b, so three, plus or minus the square root of b squared, so negative three squared is nine, minus four times a, which is four, times c, which is negative seven. So I could just say times, well, I'll just write, I'll write a seven here, then that negative is going to make this a positive."}, {"video_title": "Extraneous solutions of radical equations Mathematics III High School Math Khan Academy.mp3", "Sentence": "So plus x, plus x, and we are going to get, we are going to get four x squared minus three x minus seven, minus seven is equal to, is equal to zero. And let's see, we would want to factor this right over here, and let's see, maybe I could do this by, if I do it by, well, I'll just use a quadratic formula here. So the solutions are going to be x is going to be equal to negative b, so three, plus or minus the square root of b squared, so negative three squared is nine, minus four times a, which is four, times c, which is negative seven. So I could just say times, well, I'll just write, I'll write a seven here, then that negative is going to make this a positive. All of that over two a, so two times four is eight. So this is going to be three plus or minus the square root of, let's see, four times four is 16 times seven. 16 times seven is going to be 70 plus 42."}, {"video_title": "Extraneous solutions of radical equations Mathematics III High School Math Khan Academy.mp3", "Sentence": "So I could just say times, well, I'll just write, I'll write a seven here, then that negative is going to make this a positive. All of that over two a, so two times four is eight. So this is going to be three plus or minus the square root of, let's see, four times four is 16 times seven. 16 times seven is going to be 70 plus 42. Let me make sure I'm doing this right. So 16 times seven, two, four, so it's 112, plus nine, so 121, that worked out nicely. So plus or minus the square root of 121, all of that over eight, well, that is equal to three plus or minus 11, all of that over eight."}, {"video_title": "Extraneous solutions of radical equations Mathematics III High School Math Khan Academy.mp3", "Sentence": "16 times seven is going to be 70 plus 42. Let me make sure I'm doing this right. So 16 times seven, two, four, so it's 112, plus nine, so 121, that worked out nicely. So plus or minus the square root of 121, all of that over eight, well, that is equal to three plus or minus 11, all of that over eight. So that is equal to, if we add 11, that is 14 eighths, or if we subtract 11, three minus 11 is negative eight. Negative eight divided by eight is negative one. So we have to think about, you might say, okay, I found two solutions to the radical equation, but remember, one of these might be solutions to this alternate radical equation that got lost when we squared both sides."}, {"video_title": "Extraneous solutions of radical equations Mathematics III High School Math Khan Academy.mp3", "Sentence": "So plus or minus the square root of 121, all of that over eight, well, that is equal to three plus or minus 11, all of that over eight. So that is equal to, if we add 11, that is 14 eighths, or if we subtract 11, three minus 11 is negative eight. Negative eight divided by eight is negative one. So we have to think about, you might say, okay, I found two solutions to the radical equation, but remember, one of these might be solutions to this alternate radical equation that got lost when we squared both sides. We have to make sure that they're legitimate or maybe one of these is an extraneous solution. In fact, one is very likely a solution to this radical equation, which wasn't our original goal. So let's see."}, {"video_title": "Extraneous solutions of radical equations Mathematics III High School Math Khan Academy.mp3", "Sentence": "So we have to think about, you might say, okay, I found two solutions to the radical equation, but remember, one of these might be solutions to this alternate radical equation that got lost when we squared both sides. We have to make sure that they're legitimate or maybe one of these is an extraneous solution. In fact, one is very likely a solution to this radical equation, which wasn't our original goal. So let's see. Let's try out x equals negative one. If x equals negative one, we would have two times negative one minus one is equal to the square root of eight minus negative one. So that would be negative two minus one is equal to the square root of, is equal to the square root of nine."}, {"video_title": "Extraneous solutions of radical equations Mathematics III High School Math Khan Academy.mp3", "Sentence": "So let's see. Let's try out x equals negative one. If x equals negative one, we would have two times negative one minus one is equal to the square root of eight minus negative one. So that would be negative two minus one is equal to the square root of, is equal to the square root of nine. And so we'd have negative three is equal to the square root of nine. The principle root of nine, this is the positive square root. This is not true."}, {"video_title": "Extraneous solutions of radical equations Mathematics III High School Math Khan Academy.mp3", "Sentence": "So that would be negative two minus one is equal to the square root of, is equal to the square root of nine. And so we'd have negative three is equal to the square root of nine. The principle root of nine, this is the positive square root. This is not true. So this right over here, that is an extraneous solution. Extraneous. Extraneous solution."}, {"video_title": "Extraneous solutions of radical equations Mathematics III High School Math Khan Academy.mp3", "Sentence": "This is not true. So this right over here, that is an extraneous solution. Extraneous. Extraneous solution. It is a solution to this one right over here. Because notice, for that one, if you substitute two times negative one minus one is equal to the negative of eight minus negative one. So this is negative three is equal to the negative of three."}, {"video_title": "Extraneous solutions of radical equations Mathematics III High School Math Khan Academy.mp3", "Sentence": "Extraneous solution. It is a solution to this one right over here. Because notice, for that one, if you substitute two times negative one minus one is equal to the negative of eight minus negative one. So this is negative three is equal to the negative of three. So it checks out for this one. So this one right over here is the extraneous solution. This one right over here is going to be the actual solution for our original equation."}, {"video_title": "Exponential equation with rational answer Exponent expressions Algebra I Khan Academy.mp3", "Sentence": "So right here we've got 3 to the a power, or eighth power I guess, is, I don't want to confuse it with the number 8, 3 to the a power, is equal to the fifth root of 3 squared. And what we need to figure out is what would a be equal to? Let's solve for a, and I encourage you right now to pause this video and try it on your own. Well if you have a fifth root right over here, one thing that you might be tempted to do to undo the fifth root is to raise it to the fifth power. And of course we can't just raise one side of an equation to the fifth power. Whatever we do to one side, we have to do to the other side if we want this to still be equal. So let's raise both sides of this equation to the fifth power."}, {"video_title": "Exponential equation with rational answer Exponent expressions Algebra I Khan Academy.mp3", "Sentence": "Well if you have a fifth root right over here, one thing that you might be tempted to do to undo the fifth root is to raise it to the fifth power. And of course we can't just raise one side of an equation to the fifth power. Whatever we do to one side, we have to do to the other side if we want this to still be equal. So let's raise both sides of this equation to the fifth power. Now this left-hand side, we just have to remember a little bit of our exponent properties. 3a to the fifth power, and if we want to just remind ourselves where that comes from, it's the same thing as 3 to the a times 3 to the a times 3 to the a times 3 to the a times 3 to the a. Well what's that going to be equal to?"}, {"video_title": "Exponential equation with rational answer Exponent expressions Algebra I Khan Academy.mp3", "Sentence": "So let's raise both sides of this equation to the fifth power. Now this left-hand side, we just have to remember a little bit of our exponent properties. 3a to the fifth power, and if we want to just remind ourselves where that comes from, it's the same thing as 3 to the a times 3 to the a times 3 to the a times 3 to the a times 3 to the a. Well what's that going to be equal to? That's going to be 3 to the a plus a plus a plus a plus a power, which is the same thing as 3 to the 5a power. So the exponent property here is if you raise a base to some exponent and then raise that whole thing to another exponent, that's the equivalent of raising the base to an exponent that is the product of these two exponents. So we could rewrite this left-hand side as 3 to the 5a power is going to be equal to, well if you take something that's a fifth root and you raise it to the fifth power, then you're just left with what you had under the radical."}, {"video_title": "Exponential equation with rational answer Exponent expressions Algebra I Khan Academy.mp3", "Sentence": "Well what's that going to be equal to? That's going to be 3 to the a plus a plus a plus a plus a power, which is the same thing as 3 to the 5a power. So the exponent property here is if you raise a base to some exponent and then raise that whole thing to another exponent, that's the equivalent of raising the base to an exponent that is the product of these two exponents. So we could rewrite this left-hand side as 3 to the 5a power is going to be equal to, well if you take something that's a fifth root and you raise it to the fifth power, then you're just left with what you had under the radical. That's going to be equal to 3 squared. So now things become a lot clearer. 3 to the 5a needs to be equal to 3 squared."}, {"video_title": "Exponential equation with rational answer Exponent expressions Algebra I Khan Academy.mp3", "Sentence": "So we could rewrite this left-hand side as 3 to the 5a power is going to be equal to, well if you take something that's a fifth root and you raise it to the fifth power, then you're just left with what you had under the radical. That's going to be equal to 3 squared. So now things become a lot clearer. 3 to the 5a needs to be equal to 3 squared. Or another way of thinking about it, we have the same base on both sides, so this exponent needs to be equal to this exponent right over there. Or we could write that 5 times a needs to be equal to 2. And of course now we can just divide both sides by 5, and we get a is equal to 2 fifths."}, {"video_title": "Exponential equation with rational answer Exponent expressions Algebra I Khan Academy.mp3", "Sentence": "3 to the 5a needs to be equal to 3 squared. Or another way of thinking about it, we have the same base on both sides, so this exponent needs to be equal to this exponent right over there. Or we could write that 5 times a needs to be equal to 2. And of course now we can just divide both sides by 5, and we get a is equal to 2 fifths. And this is an interesting result. What's neat about this example, it kind of shows you the motivation for how we define rational exponents. So let's just put this back into the original expression."}, {"video_title": "Exponential equation with rational answer Exponent expressions Algebra I Khan Academy.mp3", "Sentence": "And of course now we can just divide both sides by 5, and we get a is equal to 2 fifths. And this is an interesting result. What's neat about this example, it kind of shows you the motivation for how we define rational exponents. So let's just put this back into the original expression. We've just said, or we've just solved for a, and we've gotten that 3 to the 2 fifths power to the 2, and actually let me color code it a little bit, because I think that will be interesting. 3 to the 2 over 5 power is equal to the fifth root. Notice the fifth root."}, {"video_title": "Exponential equation with rational answer Exponent expressions Algebra I Khan Academy.mp3", "Sentence": "So let's just put this back into the original expression. We've just said, or we've just solved for a, and we've gotten that 3 to the 2 fifths power to the 2, and actually let me color code it a little bit, because I think that will be interesting. 3 to the 2 over 5 power is equal to the fifth root. Notice the fifth root. So the denominator here, that's the root. So the fifth root of 3 squared. So if you take this base 3, you square it, but then you take the fifth root of that, that's the same thing as raising it to the 2 fifths power."}, {"video_title": "Exponential equation with rational answer Exponent expressions Algebra I Khan Academy.mp3", "Sentence": "Notice the fifth root. So the denominator here, that's the root. So the fifth root of 3 squared. So if you take this base 3, you square it, but then you take the fifth root of that, that's the same thing as raising it to the 2 fifths power. Notice, take this 3, take it to the second power, and then you find the fifth root of it. Or we could, if you use this property that we just saw right over here, you could rewrite this. This is the same thing as 3 squared, and then you raise that to the 1 fifth power."}, {"video_title": "Exponential equation with rational answer Exponent expressions Algebra I Khan Academy.mp3", "Sentence": "So if you take this base 3, you square it, but then you take the fifth root of that, that's the same thing as raising it to the 2 fifths power. Notice, take this 3, take it to the second power, and then you find the fifth root of it. Or we could, if you use this property that we just saw right over here, you could rewrite this. This is the same thing as 3 squared, and then you raise that to the 1 fifth power. We saw that property at play over here. You could just multiply these two exponents. You'd get 3 to the 2 fifths power, and that's the same thing as 3 squared, and then find the fifth root of it."}, {"video_title": "Factoring completely with a common factor Algebra 1 Khan Academy.mp3", "Sentence": "So factor this completely. Pause the video and have a go at that. All right, now let's work through this together. So the way that I like to think about it, I first try to see is there any common factor to all the terms? And I try to find the greatest of the common factor, possible common factors to all of the terms. So let's see, they're all divisible by two, so two would be a common factor. But let's see, they're also all divisible by four."}, {"video_title": "Factoring completely with a common factor Algebra 1 Khan Academy.mp3", "Sentence": "So the way that I like to think about it, I first try to see is there any common factor to all the terms? And I try to find the greatest of the common factor, possible common factors to all of the terms. So let's see, they're all divisible by two, so two would be a common factor. But let's see, they're also all divisible by four. Four is divisible by four, eight is divisible by four, 12 is divisible by four. And that looks like the greatest common factor. They're not all divisible by x, so I can't throw an x in there."}, {"video_title": "Factoring completely with a common factor Algebra 1 Khan Academy.mp3", "Sentence": "But let's see, they're also all divisible by four. Four is divisible by four, eight is divisible by four, 12 is divisible by four. And that looks like the greatest common factor. They're not all divisible by x, so I can't throw an x in there. So what I wanna do is factor out a four. So I could rewrite this as four times, now what would it be, four times what? Well, if I factor four out of four x squared, I'm just going to be left with an x squared."}, {"video_title": "Factoring completely with a common factor Algebra 1 Khan Academy.mp3", "Sentence": "They're not all divisible by x, so I can't throw an x in there. So what I wanna do is factor out a four. So I could rewrite this as four times, now what would it be, four times what? Well, if I factor four out of four x squared, I'm just going to be left with an x squared. If I factor a four out of negative eight x, negative eight x divided by four is negative two, so I'm going to have negative two x. And if I factor a four out of negative 12, negative 12 divided by four is negative three. Now, am I done factoring?"}, {"video_title": "Factoring completely with a common factor Algebra 1 Khan Academy.mp3", "Sentence": "Well, if I factor four out of four x squared, I'm just going to be left with an x squared. If I factor a four out of negative eight x, negative eight x divided by four is negative two, so I'm going to have negative two x. And if I factor a four out of negative 12, negative 12 divided by four is negative three. Now, am I done factoring? Well, it looks like I could factor this thing a little bit more. Can I think of two numbers that add up to negative two, and when I multiply it, I get negative three. And since when I multiply, I get a negative value, one of them's going to be positive, and one of them's going to be negative."}, {"video_title": "Factoring completely with a common factor Algebra 1 Khan Academy.mp3", "Sentence": "Now, am I done factoring? Well, it looks like I could factor this thing a little bit more. Can I think of two numbers that add up to negative two, and when I multiply it, I get negative three. And since when I multiply, I get a negative value, one of them's going to be positive, and one of them's going to be negative. I could think about it this way. A plus B is equal to negative two. A times B needs to be equal to negative three."}, {"video_title": "Factoring completely with a common factor Algebra 1 Khan Academy.mp3", "Sentence": "And since when I multiply, I get a negative value, one of them's going to be positive, and one of them's going to be negative. I could think about it this way. A plus B is equal to negative two. A times B needs to be equal to negative three. So let's see, A could be equal to negative three, and B could be equal to one, because negative three plus one is negative two, and negative three times one is negative three. So I could rewrite all of this as four times x plus negative three, or I could just write that as x minus three, times x plus one, x plus one. And now I have actually factored this completely."}, {"video_title": "Factoring completely with a common factor Algebra 1 Khan Academy.mp3", "Sentence": "A times B needs to be equal to negative three. So let's see, A could be equal to negative three, and B could be equal to one, because negative three plus one is negative two, and negative three times one is negative three. So I could rewrite all of this as four times x plus negative three, or I could just write that as x minus three, times x plus one, x plus one. And now I have actually factored this completely. Let's do another example. So let's say that we had the expression negative three x squared plus 21x minus 30. Pause the video and see if you can factor this completely."}, {"video_title": "Factoring completely with a common factor Algebra 1 Khan Academy.mp3", "Sentence": "And now I have actually factored this completely. Let's do another example. So let's say that we had the expression negative three x squared plus 21x minus 30. Pause the video and see if you can factor this completely. All right, now let's do this together. So what would be the greatest common factor? So let's see, they're all divisible by three, so you could factor out a three, but let's see what happens if you factor out a three."}, {"video_title": "Factoring completely with a common factor Algebra 1 Khan Academy.mp3", "Sentence": "Pause the video and see if you can factor this completely. All right, now let's do this together. So what would be the greatest common factor? So let's see, they're all divisible by three, so you could factor out a three, but let's see what happens if you factor out a three. This is the same thing as three times, well, negative three x squared divided by three is negative x squared. 21x divided by three is seven x, so plus seven x. And then negative 30 divided by three is negative 10."}, {"video_title": "Factoring completely with a common factor Algebra 1 Khan Academy.mp3", "Sentence": "So let's see, they're all divisible by three, so you could factor out a three, but let's see what happens if you factor out a three. This is the same thing as three times, well, negative three x squared divided by three is negative x squared. 21x divided by three is seven x, so plus seven x. And then negative 30 divided by three is negative 10. You could do it this way, but having this negative out on the x squared term still makes it a little bit confusing on how you would factor this further. You can do it, but it still takes a little bit more of a mental load. So instead of just factoring out a three, let's factor out a negative three."}, {"video_title": "Factoring completely with a common factor Algebra 1 Khan Academy.mp3", "Sentence": "And then negative 30 divided by three is negative 10. You could do it this way, but having this negative out on the x squared term still makes it a little bit confusing on how you would factor this further. You can do it, but it still takes a little bit more of a mental load. So instead of just factoring out a three, let's factor out a negative three. So we could write it this way. If we factor out a negative three, what does that become? Well then, if you factor out a negative three out of this term, you're just left with an x squared."}, {"video_title": "Factoring completely with a common factor Algebra 1 Khan Academy.mp3", "Sentence": "So instead of just factoring out a three, let's factor out a negative three. So we could write it this way. If we factor out a negative three, what does that become? Well then, if you factor out a negative three out of this term, you're just left with an x squared. If you factor out a negative three from this term, 21 divided by negative three is negative seven x. And if you factor out a negative three out of negative 30, you're left with a positive 10, positive 10. And now, let's see if we can factor this thing a little bit more."}, {"video_title": "Factoring completely with a common factor Algebra 1 Khan Academy.mp3", "Sentence": "Well then, if you factor out a negative three out of this term, you're just left with an x squared. If you factor out a negative three from this term, 21 divided by negative three is negative seven x. And if you factor out a negative three out of negative 30, you're left with a positive 10, positive 10. And now, let's see if we can factor this thing a little bit more. Can I think of two numbers where if I were to add them, I get to negative seven, and if I were to multiply them, I get to 10. And let's see, they'd have to have the same sign because their product is positive. So see, a could be equal to negative five, and then b is equal to negative two."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "Let's do some equations that deal with absolute values. And just as a bit of a review, when you take the absolute value of a number, let's say I take the absolute value of negative 1. What you're really doing is you're saying how far is that number from 0? And in the case of negative 1, if we draw a number line right there, that's a very badly drawn number line. So if we draw a number line right there that's 0, you have a negative 1 right there. Well, it's 1 away from 0. So the absolute value of negative 1 is 1."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "And in the case of negative 1, if we draw a number line right there, that's a very badly drawn number line. So if we draw a number line right there that's 0, you have a negative 1 right there. Well, it's 1 away from 0. So the absolute value of negative 1 is 1. And the absolute value of 1 is also 1 away from 0. It's also equal to 1. So on some level, an absolute value is a distance from 0."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "So the absolute value of negative 1 is 1. And the absolute value of 1 is also 1 away from 0. It's also equal to 1. So on some level, an absolute value is a distance from 0. But another, I guess, simpler way to think of it, it always results in a positive version of the number. The absolute value of negative 7,346 is equal to 7,346. So with that in mind, let's try to solve some equations with absolute values in them."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "So on some level, an absolute value is a distance from 0. But another, I guess, simpler way to think of it, it always results in a positive version of the number. The absolute value of negative 7,346 is equal to 7,346. So with that in mind, let's try to solve some equations with absolute values in them. So let's add the equation. The absolute value of x minus 5 is equal to 10. And one way you can interpret this, and I want you to think about this, this is actually saying that the distance between x and 5 is equal to 10."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "So with that in mind, let's try to solve some equations with absolute values in them. So let's add the equation. The absolute value of x minus 5 is equal to 10. And one way you can interpret this, and I want you to think about this, this is actually saying that the distance between x and 5 is equal to 10. So how many numbers that are exactly 10 away from 5? And you can already think of the solution to this equation. But I'll show you how to solve it systematically."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "And one way you can interpret this, and I want you to think about this, this is actually saying that the distance between x and 5 is equal to 10. So how many numbers that are exactly 10 away from 5? And you can already think of the solution to this equation. But I'll show you how to solve it systematically. Now, this is going to be true in two situations. Either x minus 5 is equal to positive 10. If this evaluates out to positive 10, then when you take the absolute value of it, you're going to get positive 10."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "But I'll show you how to solve it systematically. Now, this is going to be true in two situations. Either x minus 5 is equal to positive 10. If this evaluates out to positive 10, then when you take the absolute value of it, you're going to get positive 10. Or x minus 5 might evaluate to negative 10. If x minus 5 evaluated to negative 10, when you take the absolute value of it, you would get 10 again. So x minus 5 could also be equal to negative 10."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "If this evaluates out to positive 10, then when you take the absolute value of it, you're going to get positive 10. Or x minus 5 might evaluate to negative 10. If x minus 5 evaluated to negative 10, when you take the absolute value of it, you would get 10 again. So x minus 5 could also be equal to negative 10. Both of these would satisfy this equation. Now, to solve this one, add 5 to both sides of this equation. You get x is equal to 15."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "So x minus 5 could also be equal to negative 10. Both of these would satisfy this equation. Now, to solve this one, add 5 to both sides of this equation. You get x is equal to 15. To solve this one, add 5 to both sides of this equation. x is equal to negative 5. So our solution, there's two x's that satisfy this equation."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "You get x is equal to 15. To solve this one, add 5 to both sides of this equation. x is equal to negative 5. So our solution, there's two x's that satisfy this equation. x could be 15. 15 minus 5 is 10. Take the absolute value, you're going to get 10."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "So our solution, there's two x's that satisfy this equation. x could be 15. 15 minus 5 is 10. Take the absolute value, you're going to get 10. Or x could be negative 5. Negative 5 minus 5 is negative 10. Take the absolute value, you get 10."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "Take the absolute value, you're going to get 10. Or x could be negative 5. Negative 5 minus 5 is negative 10. Take the absolute value, you get 10. And notice, both of these numbers are exactly 10 away from the number 5. Let's do another one of these. Let's do another one."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "Take the absolute value, you get 10. And notice, both of these numbers are exactly 10 away from the number 5. Let's do another one of these. Let's do another one. Let's say we have the absolute value of x plus 2 is equal to 6. So what does that tell us? That tells us that either x plus 2, that the thing inside the absolute value sign, is equal to 6, or the thing inside of the absolute value sign, the x plus 2, could also be negative 6."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "Let's do another one. Let's say we have the absolute value of x plus 2 is equal to 6. So what does that tell us? That tells us that either x plus 2, that the thing inside the absolute value sign, is equal to 6, or the thing inside of the absolute value sign, the x plus 2, could also be negative 6. If this whole thing evaluated to negative 6, you take the absolute value, you'd get 6. So or x plus 2 could equal negative 6. And then if you subtract 2 from both sides of this equation, you get x could be equal to 4."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "That tells us that either x plus 2, that the thing inside the absolute value sign, is equal to 6, or the thing inside of the absolute value sign, the x plus 2, could also be negative 6. If this whole thing evaluated to negative 6, you take the absolute value, you'd get 6. So or x plus 2 could equal negative 6. And then if you subtract 2 from both sides of this equation, you get x could be equal to 4. If you subtract 2 from both sides of this equation, you get x could be equal to negative 8. So these are the two solutions to the equation. And just to kind of have it gel in your mind that absolute value, you can kind of view it as a distance, you could rewrite this problem as the absolute value of x minus negative 2 is equal to 6."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "And then if you subtract 2 from both sides of this equation, you get x could be equal to 4. If you subtract 2 from both sides of this equation, you get x could be equal to negative 8. So these are the two solutions to the equation. And just to kind of have it gel in your mind that absolute value, you can kind of view it as a distance, you could rewrite this problem as the absolute value of x minus negative 2 is equal to 6. And so this is asking me, what are the x's that are exactly 6 away from negative 2? Remember, up here we said, what are the x's that are exactly 10 away from positive 5? Whatever number you're subtracting from positive 5."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "And just to kind of have it gel in your mind that absolute value, you can kind of view it as a distance, you could rewrite this problem as the absolute value of x minus negative 2 is equal to 6. And so this is asking me, what are the x's that are exactly 6 away from negative 2? Remember, up here we said, what are the x's that are exactly 10 away from positive 5? Whatever number you're subtracting from positive 5. These are both 10 away from positive 5. This is asking, what is exactly 6 away from negative 2? And it's going to be 4 or negative 8."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "Whatever number you're subtracting from positive 5. These are both 10 away from positive 5. This is asking, what is exactly 6 away from negative 2? And it's going to be 4 or negative 8. You could try those numbers out for yourself. Let's do another one of these. Let's do another one."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "And it's going to be 4 or negative 8. You could try those numbers out for yourself. Let's do another one of these. Let's do another one. We'll do it in purple. Let's say we have the absolute value of 4x. I'm going to change this problem up a little bit."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "Let's do another one. We'll do it in purple. Let's say we have the absolute value of 4x. I'm going to change this problem up a little bit. 4x minus 1 is equal to 19. So just like the last few problems, 4x minus 1 could be equal to 19. 4x minus 1 could be equal to 19."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "I'm going to change this problem up a little bit. 4x minus 1 is equal to 19. So just like the last few problems, 4x minus 1 could be equal to 19. 4x minus 1 could be equal to 19. Or 4x minus 1 might evaluate to negative 19, because then when we take the absolute value, you're going to get 19 again. Or 4x minus 1 could be equal to negative 19. Then you just solve these two equations."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "4x minus 1 could be equal to 19. Or 4x minus 1 might evaluate to negative 19, because then when we take the absolute value, you're going to get 19 again. Or 4x minus 1 could be equal to negative 19. Then you just solve these two equations. Add 1 to both sides of this equation. We could do them simultaneously even. Add 1 to both sides of this."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "Then you just solve these two equations. Add 1 to both sides of this equation. We could do them simultaneously even. Add 1 to both sides of this. You get 4x is equal to 20. Add 1 to both sides of this equation. You get 4x is equal to negative 18."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "Add 1 to both sides of this. You get 4x is equal to 20. Add 1 to both sides of this equation. You get 4x is equal to negative 18. Divide both sides of this by 4. You get x is equal to 5. Divide both sides of this by 4."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "You get 4x is equal to negative 18. Divide both sides of this by 4. You get x is equal to 5. Divide both sides of this by 4. You get x is equal to negative 18 over 4, which is equal to negative 9 halves. So both of these x values satisfy the equation. Try it out."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "Divide both sides of this by 4. You get x is equal to negative 18 over 4, which is equal to negative 9 halves. So both of these x values satisfy the equation. Try it out. Negative 9 halves times 4. This will become a negative 18. Negative 18 minus 1 is negative 19."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "Try it out. Negative 9 halves times 4. This will become a negative 18. Negative 18 minus 1 is negative 19. Take the absolute value. You get 19. You put a 5 here."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "Negative 18 minus 1 is negative 19. Take the absolute value. You get 19. You put a 5 here. 4 times 5 is 20 minus 1 is positive 19. So you take the absolute value. Once again, you'll get a 19."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "You put a 5 here. 4 times 5 is 20 minus 1 is positive 19. So you take the absolute value. Once again, you'll get a 19. Let's try to graph one of these just for fun. So let's say I have y is equal to the absolute value of x plus 3. So this is a function or a graph with an absolute value in it."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "Once again, you'll get a 19. Let's try to graph one of these just for fun. So let's say I have y is equal to the absolute value of x plus 3. So this is a function or a graph with an absolute value in it. So let's think about two scenarios. There's one scenario where the thing inside of the absolute value is positive. So you have the scenario where x plus 3 is greater than 0."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "So this is a function or a graph with an absolute value in it. So let's think about two scenarios. There's one scenario where the thing inside of the absolute value is positive. So you have the scenario where x plus 3 is greater than 0. And then you have the scenario where x plus 3 is less than 0. When x plus 3 is greater than 0, this graph or this line, or I guess we can't call it a line, this function, is the same thing as y is equal to x plus 3. If this thing over here is greater than 0, then the absolute value sign is irrelevant."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "So you have the scenario where x plus 3 is greater than 0. And then you have the scenario where x plus 3 is less than 0. When x plus 3 is greater than 0, this graph or this line, or I guess we can't call it a line, this function, is the same thing as y is equal to x plus 3. If this thing over here is greater than 0, then the absolute value sign is irrelevant. So then this thing is the same thing as y is equal to x plus 3. But when is x plus 3 greater than 0? Well, if you subtract 3 from both sides, you get x is greater than negative 3."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "If this thing over here is greater than 0, then the absolute value sign is irrelevant. So then this thing is the same thing as y is equal to x plus 3. But when is x plus 3 greater than 0? Well, if you subtract 3 from both sides, you get x is greater than negative 3. So when x is greater than negative 3, this graph is going to look just like y is equal to x plus 3. Now, when x plus 3 is less than 0, in the situation where the inside of our absolute value sign is negative, in that situation, this equation is going to be y is equal to the negative of x plus 3. How can I say that?"}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "Well, if you subtract 3 from both sides, you get x is greater than negative 3. So when x is greater than negative 3, this graph is going to look just like y is equal to x plus 3. Now, when x plus 3 is less than 0, in the situation where the inside of our absolute value sign is negative, in that situation, this equation is going to be y is equal to the negative of x plus 3. How can I say that? Well, look, if this is going to be a negative number, if x plus 3 is going to be a negative number, that's what we're assuming here. If it's going to be a negative number, then when you take the absolute value of a negative number, you're going to make it positive. That's just like multiplying it by negative 1."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "How can I say that? Well, look, if this is going to be a negative number, if x plus 3 is going to be a negative number, that's what we're assuming here. If it's going to be a negative number, then when you take the absolute value of a negative number, you're going to make it positive. That's just like multiplying it by negative 1. If you know you're taking the absolute value of a negative number, it's just like multiplying it by negative 1, because you're going to make it positive. And this is going to be the situation. x plus 3 is less than 0 if we subtract 3 from both sides when x is less than negative 3."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "That's just like multiplying it by negative 1. If you know you're taking the absolute value of a negative number, it's just like multiplying it by negative 1, because you're going to make it positive. And this is going to be the situation. x plus 3 is less than 0 if we subtract 3 from both sides when x is less than negative 3. So when x is less than negative 3, the graph will look like this. When x is greater than negative 3, the graph will look like that. So let's see what that would make the entire graph look like."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "x plus 3 is less than 0 if we subtract 3 from both sides when x is less than negative 3. So when x is less than negative 3, the graph will look like this. When x is greater than negative 3, the graph will look like that. So let's see what that would make the entire graph look like. Let me draw my axes. That's my x-axis. That's my y-axis."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "So let's see what that would make the entire graph look like. Let me draw my axes. That's my x-axis. That's my y-axis. So let me multiply this out, just so that we have it in mx plus b form. So this is equal to negative x minus 3. So let's just figure out what this graph would look like in general."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "That's my y-axis. So let me multiply this out, just so that we have it in mx plus b form. So this is equal to negative x minus 3. So let's just figure out what this graph would look like in general. Negative x minus 3. The y-intercept is negative 3. So 1, 2, 3."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "So let's just figure out what this graph would look like in general. Negative x minus 3. The y-intercept is negative 3. So 1, 2, 3. And negative x means it slopes downward. It has a downward slope of 1. So it would look like this."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "So 1, 2, 3. And negative x means it slopes downward. It has a downward slope of 1. So it would look like this. The x-intercept would be at x is equal to. So if you say y is equal to 0, that would happen when x is equal to negative 3. So it's going to go through that point right there."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "So it would look like this. The x-intercept would be at x is equal to. So if you say y is equal to 0, that would happen when x is equal to negative 3. So it's going to go through that point right there. And the graph, if we didn't have this constraint right here, would look something like this. That's if we didn't constrain it to a certain interval on the x-axis. Now this graph, what does it look like?"}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "So it's going to go through that point right there. And the graph, if we didn't have this constraint right here, would look something like this. That's if we didn't constrain it to a certain interval on the x-axis. Now this graph, what does it look like? Let's see. It has this y-intercept at positive 3. Just like that."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "Now this graph, what does it look like? Let's see. It has this y-intercept at positive 3. Just like that. And where's its x-intercept? When y is equal to 0, x is negative 3. So it also goes through that point right there."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "Just like that. And where's its x-intercept? When y is equal to 0, x is negative 3. So it also goes through that point right there. And it has a slope of 1. So it would look something like this. That's what this graph looks like."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "So it also goes through that point right there. And it has a slope of 1. So it would look something like this. That's what this graph looks like. Now what we figured out is that this absolute value function, it looks like this purple graph when x is less than negative 3. So when x is less than negative 3, that's x is equal to negative 3 right there. When x is less than negative 3, it looks like this purple graph right there."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "That's what this graph looks like. Now what we figured out is that this absolute value function, it looks like this purple graph when x is less than negative 3. So when x is less than negative 3, that's x is equal to negative 3 right there. When x is less than negative 3, it looks like this purple graph right there. So that's when x is less than negative 3. But when x is greater than negative 3, it looks like the green graph. It looks like that."}, {"video_title": "Absolute value equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "When x is less than negative 3, it looks like this purple graph right there. So that's when x is less than negative 3. But when x is greater than negative 3, it looks like the green graph. It looks like that. So this graph looks like this strange v. When x is greater than negative 3, this is positive. So we have the graph of a positive slope. But then when x is less than negative 3, we're essentially taking the negative of the function, if you want to view it that way."}, {"video_title": "i as the principal root of -1 (a little technical) Precalculus Khan Academy.mp3", "Sentence": "In your mathematical careers, you might encounter people who say it is wrong to say that i is equal to the principal square root of negative 1. And if you ask them why is this wrong, they'll show up with this kind of line of logic that actually seems pretty reasonable. They will tell you that, okay, well let's just start with negative 1. We know from definition that negative 1 is equal to i times i. Everything seems pretty straightforward right now. And then they'll say, well look, if you take this, if you assume this part right here, then we can replace each of these i's with the square root of negative 1. And they'd be right."}, {"video_title": "i as the principal root of -1 (a little technical) Precalculus Khan Academy.mp3", "Sentence": "We know from definition that negative 1 is equal to i times i. Everything seems pretty straightforward right now. And then they'll say, well look, if you take this, if you assume this part right here, then we can replace each of these i's with the square root of negative 1. And they'd be right. So then this would be the same thing as the square root of negative 1 times the square root of negative 1. And then they would tell you that, hey, look, just from straight up properties of the principal square root function, they will tell you that the square root of a times b is the same thing as the principal square root of a times the principal square root of b. And so if you have the principal square root of a times the principal square root of b, that's the same thing as the square root of a times b."}, {"video_title": "i as the principal root of -1 (a little technical) Precalculus Khan Academy.mp3", "Sentence": "And they'd be right. So then this would be the same thing as the square root of negative 1 times the square root of negative 1. And then they would tell you that, hey, look, just from straight up properties of the principal square root function, they will tell you that the square root of a times b is the same thing as the principal square root of a times the principal square root of b. And so if you have the principal square root of a times the principal square root of b, that's the same thing as the square root of a times b. So based on this property of the radical of the principal root, they will say that this over here is the same thing as the square root of negative 1 times negative 1. times negative 1. If I have the principal root of the product of two things, that's the same thing as the product of each of their principal roots. I'm doing this in the other order here."}, {"video_title": "i as the principal root of -1 (a little technical) Precalculus Khan Academy.mp3", "Sentence": "And so if you have the principal square root of a times the principal square root of b, that's the same thing as the square root of a times b. So based on this property of the radical of the principal root, they will say that this over here is the same thing as the square root of negative 1 times negative 1. times negative 1. If I have the principal root of the product of two things, that's the same thing as the product of each of their principal roots. I'm doing this in the other order here. Here I had the principal root of the products. Over here I have this on the right. And then from that, we all know that negative 1 times negative 1 is 1."}, {"video_title": "i as the principal root of -1 (a little technical) Precalculus Khan Academy.mp3", "Sentence": "I'm doing this in the other order here. Here I had the principal root of the products. Over here I have this on the right. And then from that, we all know that negative 1 times negative 1 is 1. So this should be equal to the principal square root of 1. And then the principal square root of 1, remember, this radical means principal square root, positive square root. That is just going to be positive 1."}, {"video_title": "i as the principal root of -1 (a little technical) Precalculus Khan Academy.mp3", "Sentence": "And then from that, we all know that negative 1 times negative 1 is 1. So this should be equal to the principal square root of 1. And then the principal square root of 1, remember, this radical means principal square root, positive square root. That is just going to be positive 1. And they'll say, this is wrong. Clearly, negative 1 and positive 1 are not the same thing. And they'll argue, therefore, you can't make this substitution that we did in this step."}, {"video_title": "i as the principal root of -1 (a little technical) Precalculus Khan Academy.mp3", "Sentence": "That is just going to be positive 1. And they'll say, this is wrong. Clearly, negative 1 and positive 1 are not the same thing. And they'll argue, therefore, you can't make this substitution that we did in this step. And what you should then point out is that this was not the incorrect step. That it is true, negative 1 is not equal to 1. But the faulty line of reasoning here was in using this property when both A and B are negative."}, {"video_title": "i as the principal root of -1 (a little technical) Precalculus Khan Academy.mp3", "Sentence": "And they'll argue, therefore, you can't make this substitution that we did in this step. And what you should then point out is that this was not the incorrect step. That it is true, negative 1 is not equal to 1. But the faulty line of reasoning here was in using this property when both A and B are negative. If both A and B are negative, this will never be true. So A and B both cannot be negative. In fact, normally when this property is given, sometimes it's given a little bit in the footnotes."}, {"video_title": "i as the principal root of -1 (a little technical) Precalculus Khan Academy.mp3", "Sentence": "But the faulty line of reasoning here was in using this property when both A and B are negative. If both A and B are negative, this will never be true. So A and B both cannot be negative. In fact, normally when this property is given, sometimes it's given a little bit in the footnotes. Or you might not even notice it, because it's not relevant when you're learning it the first time. But they'll usually give a little bit of a constraint there. They'll usually say for A, A and B greater than or equal to 0."}, {"video_title": "i as the principal root of -1 (a little technical) Precalculus Khan Academy.mp3", "Sentence": "In fact, normally when this property is given, sometimes it's given a little bit in the footnotes. Or you might not even notice it, because it's not relevant when you're learning it the first time. But they'll usually give a little bit of a constraint there. They'll usually say for A, A and B greater than or equal to 0. So that's where they list this property. This is true for A and B greater than or equal to 0. And in particular, it's false if both A and B are both negative."}, {"video_title": "i as the principal root of -1 (a little technical) Precalculus Khan Academy.mp3", "Sentence": "They'll usually say for A, A and B greater than or equal to 0. So that's where they list this property. This is true for A and B greater than or equal to 0. And in particular, it's false if both A and B are both negative. Now, I've said that I've just spent the last three minutes saying that people who tell you that this is wrong are wrong. But with that said, I will say that you have to be a little bit careful about it. You have to be a little bit careful about it."}, {"video_title": "i as the principal root of -1 (a little technical) Precalculus Khan Academy.mp3", "Sentence": "And in particular, it's false if both A and B are both negative. Now, I've said that I've just spent the last three minutes saying that people who tell you that this is wrong are wrong. But with that said, I will say that you have to be a little bit careful about it. You have to be a little bit careful about it. When we take traditional principal square roots, so when you take the principal square root of 4, we know that this is positive 2. That 4 actually has two square roots. There's negative 2 is also a square root of 4."}, {"video_title": "i as the principal root of -1 (a little technical) Precalculus Khan Academy.mp3", "Sentence": "You have to be a little bit careful about it. When we take traditional principal square roots, so when you take the principal square root of 4, we know that this is positive 2. That 4 actually has two square roots. There's negative 2 is also a square root of 4. If you have negative 2 times negative 2, it's also equal to 4. This radical symbol here means principal square root. Or when we're just dealing with real numbers, non-imaginary, non-complex numbers, you can really view it as the positive square root."}, {"video_title": "i as the principal root of -1 (a little technical) Precalculus Khan Academy.mp3", "Sentence": "There's negative 2 is also a square root of 4. If you have negative 2 times negative 2, it's also equal to 4. This radical symbol here means principal square root. Or when we're just dealing with real numbers, non-imaginary, non-complex numbers, you can really view it as the positive square root. This has two square roots, positive and negative 2. If you have this radical symbol right here, principal square roots, it means the positive square root of 2. So when you start thinking about taking square roots of negative numbers, or even in the future we'll do imaginary numbers and complex numbers and all the rest, you have to expand the definition of what this radical means."}, {"video_title": "i as the principal root of -1 (a little technical) Precalculus Khan Academy.mp3", "Sentence": "Or when we're just dealing with real numbers, non-imaginary, non-complex numbers, you can really view it as the positive square root. This has two square roots, positive and negative 2. If you have this radical symbol right here, principal square roots, it means the positive square root of 2. So when you start thinking about taking square roots of negative numbers, or even in the future we'll do imaginary numbers and complex numbers and all the rest, you have to expand the definition of what this radical means. So when you are taking the square root really of any negative number, you're really saying that this is no longer the traditional principal square root function. You're now talking that this is the principal complex square root function. Or this is now defined for complex inputs or the domain and can also generate imaginary or complex outputs."}, {"video_title": "i as the principal root of -1 (a little technical) Precalculus Khan Academy.mp3", "Sentence": "So when you start thinking about taking square roots of negative numbers, or even in the future we'll do imaginary numbers and complex numbers and all the rest, you have to expand the definition of what this radical means. So when you are taking the square root really of any negative number, you're really saying that this is no longer the traditional principal square root function. You're now talking that this is the principal complex square root function. Or this is now defined for complex inputs or the domain and can also generate imaginary or complex outputs. Or I guess you could call that the range. And if you assume that, then really straight from this you get that the square root of negative x is going to be equal to i times the square root of x. And this is only, and I'm going to make this clear, because I just told you that this will be false of both a and b are negative."}, {"video_title": "i as the principal root of -1 (a little technical) Precalculus Khan Academy.mp3", "Sentence": "Or this is now defined for complex inputs or the domain and can also generate imaginary or complex outputs. Or I guess you could call that the range. And if you assume that, then really straight from this you get that the square root of negative x is going to be equal to i times the square root of x. And this is only, and I'm going to make this clear, because I just told you that this will be false of both a and b are negative. So this is only true. So we can apply this. We can apply this when x is greater than or equal to 0."}, {"video_title": "i as the principal root of -1 (a little technical) Precalculus Khan Academy.mp3", "Sentence": "And this is only, and I'm going to make this clear, because I just told you that this will be false of both a and b are negative. So this is only true. So we can apply this. We can apply this when x is greater than or equal to 0. So if x is greater than or equal to 0, then negative x is clearly a negative number. Or I guess it could also be 0. It's a negative number."}, {"video_title": "i as the principal root of -1 (a little technical) Precalculus Khan Academy.mp3", "Sentence": "We can apply this when x is greater than or equal to 0. So if x is greater than or equal to 0, then negative x is clearly a negative number. Or I guess it could also be 0. It's a negative number. And then we can apply this right over here. If x was less than 0, then we'd be doing all of this nonsense up here and we would start to get nonsensical answers. And if you look at it this way and you say, hey, look, i can be the square root of negative 1 if we're taking the principal branch of the complex square root function, then you could rewrite this right over here as the square root of negative 1 times the square root of x."}, {"video_title": "i as the principal root of -1 (a little technical) Precalculus Khan Academy.mp3", "Sentence": "It's a negative number. And then we can apply this right over here. If x was less than 0, then we'd be doing all of this nonsense up here and we would start to get nonsensical answers. And if you look at it this way and you say, hey, look, i can be the square root of negative 1 if we're taking the principal branch of the complex square root function, then you could rewrite this right over here as the square root of negative 1 times the square root of x. And so really, the real fault in this logic, why when people say, hey, negative 1 can't be equal to 1, the real fault is using this property when both a and b, where both of these are negative numbers. That will come up with something that is unambiguously false. If you expand your definition of the principal root to include negative numbers in the domain and to include imaginary numbers, then you can do this."}, {"video_title": "i as the principal root of -1 (a little technical) Precalculus Khan Academy.mp3", "Sentence": "And if you look at it this way and you say, hey, look, i can be the square root of negative 1 if we're taking the principal branch of the complex square root function, then you could rewrite this right over here as the square root of negative 1 times the square root of x. And so really, the real fault in this logic, why when people say, hey, negative 1 can't be equal to 1, the real fault is using this property when both a and b, where both of these are negative numbers. That will come up with something that is unambiguously false. If you expand your definition of the principal root to include negative numbers in the domain and to include imaginary numbers, then you can do this. You can say the square root of negative x is the square root of negative 1 times the principal square root of negative x. I should be particular with my words. Is the same thing as the principal square root of negative 1 times the principal square root of x when x is greater than or equal to 0. And I don't want to confuse you."}, {"video_title": "Solving exponential equations using exponent properties High School Math Khan Academy.mp3", "Sentence": "We have 26 to the nine x plus five power equals one. So pause the video and see if you can tell me what x is going to be. Well, the key here is to realize that 26 to the zeroth power, to the zeroth power, is equal to one. Anything to the zeroth power is going to be equal to one. Zero to zeroth power, we can discuss at some other time, but anything other than zero to the zeroth power is going to be one. So we just have to say, well, nine x plus five needs to be equal to zero. Nine x plus five needs to be equal to zero."}, {"video_title": "Solving exponential equations using exponent properties High School Math Khan Academy.mp3", "Sentence": "Anything to the zeroth power is going to be equal to one. Zero to zeroth power, we can discuss at some other time, but anything other than zero to the zeroth power is going to be one. So we just have to say, well, nine x plus five needs to be equal to zero. Nine x plus five needs to be equal to zero. And this is pretty straightforward to solve. Subtract five from both sides and we get nine x is equal to negative five. Divide both sides by nine and we are left with x is equal to negative five."}, {"video_title": "Solving exponential equations using exponent properties High School Math Khan Academy.mp3", "Sentence": "Nine x plus five needs to be equal to zero. And this is pretty straightforward to solve. Subtract five from both sides and we get nine x is equal to negative five. Divide both sides by nine and we are left with x is equal to negative five. Let's do another one of these and let's make it a little bit more interesting. Let's say we have the exponential equation two to the three x plus five power is equal to 64 to the x minus seventh power. Once again, pause the video and see if you can tell me what x is going to be or what x needs to be to satisfy this exponential equation."}, {"video_title": "Solving exponential equations using exponent properties High School Math Khan Academy.mp3", "Sentence": "Divide both sides by nine and we are left with x is equal to negative five. Let's do another one of these and let's make it a little bit more interesting. Let's say we have the exponential equation two to the three x plus five power is equal to 64 to the x minus seventh power. Once again, pause the video and see if you can tell me what x is going to be or what x needs to be to satisfy this exponential equation. All right, so you might at first say, oh, maybe three x plus five needs to be equal to x minus seven but that wouldn't work because these are two different bases. You have two to the three x plus five power then you have 64 to the x minus seven. So the key here is to express both of these with the same base."}, {"video_title": "Solving exponential equations using exponent properties High School Math Khan Academy.mp3", "Sentence": "Once again, pause the video and see if you can tell me what x is going to be or what x needs to be to satisfy this exponential equation. All right, so you might at first say, oh, maybe three x plus five needs to be equal to x minus seven but that wouldn't work because these are two different bases. You have two to the three x plus five power then you have 64 to the x minus seven. So the key here is to express both of these with the same base. And lucky for us, 64 is a power of two. Two to the, let's see, two to the third is eight. So it's gonna be two to the third times two to the third."}, {"video_title": "Solving exponential equations using exponent properties High School Math Khan Academy.mp3", "Sentence": "So the key here is to express both of these with the same base. And lucky for us, 64 is a power of two. Two to the, let's see, two to the third is eight. So it's gonna be two to the third times two to the third. Eight times eight is 64. So it's two to the sixth is equal to 64. And you can verify that."}, {"video_title": "Solving exponential equations using exponent properties High School Math Khan Academy.mp3", "Sentence": "So it's gonna be two to the third times two to the third. Eight times eight is 64. So it's two to the sixth is equal to 64. And you can verify that. Take six twos and multiply them together. You're going to get 64. This is just a little bit easier for me."}, {"video_title": "Solving exponential equations using exponent properties High School Math Khan Academy.mp3", "Sentence": "And you can verify that. Take six twos and multiply them together. You're going to get 64. This is just a little bit easier for me. Eight times eight, and this is the same thing as two to the sixth power is 64. And I knew it was two to the sixth power because I just added the exponents because I had the same base. All right, so I can rewrite 64."}, {"video_title": "Solving exponential equations using exponent properties High School Math Khan Academy.mp3", "Sentence": "This is just a little bit easier for me. Eight times eight, and this is the same thing as two to the sixth power is 64. And I knew it was two to the sixth power because I just added the exponents because I had the same base. All right, so I can rewrite 64. Let me rewrite the whole thing. So this is two to the three x plus five power is equal to, instead of writing 64, I am going to write two to the sixth power. Two to the sixth power."}, {"video_title": "Solving exponential equations using exponent properties High School Math Khan Academy.mp3", "Sentence": "All right, so I can rewrite 64. Let me rewrite the whole thing. So this is two to the three x plus five power is equal to, instead of writing 64, I am going to write two to the sixth power. Two to the sixth power. And then that to the x minus seventh power. X minus seventh power. And to simplify this a little bit, we just have to remind ourselves that if I raise something to one power and then I raise that to another power, this is the same thing as raising my base to the product of these powers."}, {"video_title": "Solving exponential equations using exponent properties High School Math Khan Academy.mp3", "Sentence": "Two to the sixth power. And then that to the x minus seventh power. X minus seventh power. And to simplify this a little bit, we just have to remind ourselves that if I raise something to one power and then I raise that to another power, this is the same thing as raising my base to the product of these powers. A to the bc power. So this equation I can rewrite as two to the three x plus five is equal to two to the, and I just multiply six times x minus seven. So it's going to be six x, six x minus, six times seven is 42."}, {"video_title": "Solving exponential equations using exponent properties High School Math Khan Academy.mp3", "Sentence": "And to simplify this a little bit, we just have to remind ourselves that if I raise something to one power and then I raise that to another power, this is the same thing as raising my base to the product of these powers. A to the bc power. So this equation I can rewrite as two to the three x plus five is equal to two to the, and I just multiply six times x minus seven. So it's going to be six x, six x minus, six times seven is 42. I'll just write the whole thing in yellow. So six x minus 42. I just multiplied the six times the entire expression, x minus seven."}, {"video_title": "Solving exponential equations using exponent properties High School Math Khan Academy.mp3", "Sentence": "So it's going to be six x, six x minus, six times seven is 42. I'll just write the whole thing in yellow. So six x minus 42. I just multiplied the six times the entire expression, x minus seven. And so now it's interesting. I have two to the three x plus five power has to be equal to two to the six x minus 42 power. So these need to be the same exponent."}, {"video_title": "Solving exponential equations using exponent properties High School Math Khan Academy.mp3", "Sentence": "I just multiplied the six times the entire expression, x minus seven. And so now it's interesting. I have two to the three x plus five power has to be equal to two to the six x minus 42 power. So these need to be the same exponent. So three x plus five needs to be equal to six x minus 42. So there we go. It sets up a nice little linear equation for us."}, {"video_title": "Solving exponential equations using exponent properties High School Math Khan Academy.mp3", "Sentence": "So these need to be the same exponent. So three x plus five needs to be equal to six x minus 42. So there we go. It sets up a nice little linear equation for us. Three x plus five is equal to six x minus 42. Let's see. We could get all of our, since, I'll put all my x's on the right-hand side since I have more x's on the right already."}, {"video_title": "Solving exponential equations using exponent properties High School Math Khan Academy.mp3", "Sentence": "It sets up a nice little linear equation for us. Three x plus five is equal to six x minus 42. Let's see. We could get all of our, since, I'll put all my x's on the right-hand side since I have more x's on the right already. So let me subtract three x from both sides. And let me, I want to get rid of this 42 here. So let's add 42 to both sides."}, {"video_title": "Solving exponential equations using exponent properties High School Math Khan Academy.mp3", "Sentence": "We could get all of our, since, I'll put all my x's on the right-hand side since I have more x's on the right already. So let me subtract three x from both sides. And let me, I want to get rid of this 42 here. So let's add 42 to both sides. And we are going to be left with five plus 42 is 47, is equal to, 47 is equal to three x. Now we just divide both sides by three, and we are left with x is equal to 47 over three. X is equal to 47 over three."}, {"video_title": "Even and odd functions Tables Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "According to the table, is f even, odd, or neither? So pause this video and see if you can figure that out on your own. All right, now let's work on this together. So let's just remind ourselves the definition of even and odd. One definition that we can think of is that f of x, if f of x is equal to f of negative x, then we are dealing with an even function. And if f of x is equal to the negative of f of negative x, or another way of saying that, if f of negative x, if f of negative x, instead of it being equal to f of x, it's equal to negative f of x, these last two are equivalent, then in these situations, we are dealing with an odd function. And if neither of these are true, then we're dealing with neither."}, {"video_title": "Even and odd functions Tables Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So let's just remind ourselves the definition of even and odd. One definition that we can think of is that f of x, if f of x is equal to f of negative x, then we are dealing with an even function. And if f of x is equal to the negative of f of negative x, or another way of saying that, if f of negative x, if f of negative x, instead of it being equal to f of x, it's equal to negative f of x, these last two are equivalent, then in these situations, we are dealing with an odd function. And if neither of these are true, then we're dealing with neither. So what about what's going on over here? So let's see. F of negative seven is equal to negative one."}, {"video_title": "Even and odd functions Tables Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And if neither of these are true, then we're dealing with neither. So what about what's going on over here? So let's see. F of negative seven is equal to negative one. What about f of the negative of negative seven? Well, that would be f of seven. And we see f of seven here is also equal to negative one."}, {"video_title": "Even and odd functions Tables Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "F of negative seven is equal to negative one. What about f of the negative of negative seven? Well, that would be f of seven. And we see f of seven here is also equal to negative one. So at least in that case and that case, if we think of x as seven, f of x is equal to f of negative x. So it works for that. It also works for negative three and three."}, {"video_title": "Even and odd functions Tables Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And we see f of seven here is also equal to negative one. So at least in that case and that case, if we think of x as seven, f of x is equal to f of negative x. So it works for that. It also works for negative three and three. F of three is equal to f of negative three. They're both equal to two. And you can see, and you can kind of visualize in your head that we have this symmetry around the y-axis."}, {"video_title": "Even and odd functions Tables Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "It also works for negative three and three. F of three is equal to f of negative three. They're both equal to two. And you can see, and you can kind of visualize in your head that we have this symmetry around the y-axis. And so this looks like an even function. So I will circle that in. Let's do another example."}, {"video_title": "Even and odd functions Tables Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And you can see, and you can kind of visualize in your head that we have this symmetry around the y-axis. And so this looks like an even function. So I will circle that in. Let's do another example. So here, once again, the table defines function f. It's a different function f. Is this function even, odd, or neither? So pause this video and try to think about it. All right, so let's just try a few examples."}, {"video_title": "Even and odd functions Tables Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Let's do another example. So here, once again, the table defines function f. It's a different function f. Is this function even, odd, or neither? So pause this video and try to think about it. All right, so let's just try a few examples. So here we have f of five is equal to two. F of five is equal to two. What is f of negative five?"}, {"video_title": "Even and odd functions Tables Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "All right, so let's just try a few examples. So here we have f of five is equal to two. F of five is equal to two. What is f of negative five? F of negative five, not only is it not equal to two, it would have to be equal to two if this was an even function. And it would be equal to negative two if this was an odd function, but it's neither. So we very clearly see, just looking at that data point, that this can neither be even nor odd."}, {"video_title": "Even and odd functions Tables Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "What is f of negative five? F of negative five, not only is it not equal to two, it would have to be equal to two if this was an even function. And it would be equal to negative two if this was an odd function, but it's neither. So we very clearly see, just looking at that data point, that this can neither be even nor odd. So I would say neither or neither right over here. Let's do one more example. Once again, the table defines function f. According to the table, is it even, odd, or neither?"}, {"video_title": "Even and odd functions Tables Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So we very clearly see, just looking at that data point, that this can neither be even nor odd. So I would say neither or neither right over here. Let's do one more example. Once again, the table defines function f. According to the table, is it even, odd, or neither? Pause the video again and try to answer it. All right, so actually, let's just start over here. So we have f of four is equal to negative eight."}, {"video_title": "Even and odd functions Tables Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Once again, the table defines function f. According to the table, is it even, odd, or neither? Pause the video again and try to answer it. All right, so actually, let's just start over here. So we have f of four is equal to negative eight. What is f of negative four? And the whole idea here is I wanna say, okay, if f of x is equal to something, what is f of negative x? Well, they luckily give us f of negative four."}, {"video_title": "Even and odd functions Tables Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So we have f of four is equal to negative eight. What is f of negative four? And the whole idea here is I wanna say, okay, if f of x is equal to something, what is f of negative x? Well, they luckily give us f of negative four. It is equal to eight. So it looks like it's not equal to f of x. It's equal to the negative of f of x."}, {"video_title": "Even and odd functions Tables Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Well, they luckily give us f of negative four. It is equal to eight. So it looks like it's not equal to f of x. It's equal to the negative of f of x. This is equal to the negative of f of four. So on that data point alone, at least that data point satisfies it being odd. It's equal to the negative of f of x."}, {"video_title": "Even and odd functions Tables Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "It's equal to the negative of f of x. This is equal to the negative of f of four. So on that data point alone, at least that data point satisfies it being odd. It's equal to the negative of f of x. But now let's try the other points just to make sure. So f of one is equal to five. What is f of negative one?"}, {"video_title": "Even and odd functions Tables Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "It's equal to the negative of f of x. But now let's try the other points just to make sure. So f of one is equal to five. What is f of negative one? Well, it is equal to negative five. Once again, f of negative x is equal to the negative of f of x. So that checks out."}, {"video_title": "Even and odd functions Tables Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "What is f of negative one? Well, it is equal to negative five. Once again, f of negative x is equal to the negative of f of x. So that checks out. And then f of zero, well, f of zero is, of course, equal to zero. But, of course, if you say, what is the negative of f of, if you say, what, f of negative of zero, well, that's still f of zero. And then if you were to take the negative of zero, that's still zero."}, {"video_title": "Even and odd functions Tables Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So that checks out. And then f of zero, well, f of zero is, of course, equal to zero. But, of course, if you say, what is the negative of f of, if you say, what, f of negative of zero, well, that's still f of zero. And then if you were to take the negative of zero, that's still zero. So you could view this, this is consistent still with being odd. This you could view as the negative of f of negative zero, which, of course, is still going to be zero. So this one is looking pretty good that it is odd."}, {"video_title": "Example 1 Multiplying binomials to get a difference of squares Algebra I Khan Academy.mp3", "Sentence": "So you could use FOIL. You could just straight up use the distributive property here. But the whole point of this problem, I'm guessing, is to see whether you recognize a pattern here. This is of the form a plus b times a minus b, where here a is 2x and b is 8. We have 2x plus 8 and then 2x minus 8. a plus b, a minus b. What I want to do is I'm just going to multiply this out for us and then just see what happens. Whenever you have this pattern, what the product actually looks like."}, {"video_title": "Example 1 Multiplying binomials to get a difference of squares Algebra I Khan Academy.mp3", "Sentence": "This is of the form a plus b times a minus b, where here a is 2x and b is 8. We have 2x plus 8 and then 2x minus 8. a plus b, a minus b. What I want to do is I'm just going to multiply this out for us and then just see what happens. Whenever you have this pattern, what the product actually looks like. So if you were to multiply this out, we can distribute the a plus b. We can distribute this whole thing, the distribute the whole a plus b on the a and then distribute it on the b. And I could have done this with this problem right here and it would have taken us less time to just solve it."}, {"video_title": "Example 1 Multiplying binomials to get a difference of squares Algebra I Khan Academy.mp3", "Sentence": "Whenever you have this pattern, what the product actually looks like. So if you were to multiply this out, we can distribute the a plus b. We can distribute this whole thing, the distribute the whole a plus b on the a and then distribute it on the b. And I could have done this with this problem right here and it would have taken us less time to just solve it. But I want to find out the general pattern here. So a plus b times a. So we have a times a plus b."}, {"video_title": "Example 1 Multiplying binomials to get a difference of squares Algebra I Khan Academy.mp3", "Sentence": "And I could have done this with this problem right here and it would have taken us less time to just solve it. But I want to find out the general pattern here. So a plus b times a. So we have a times a plus b. That's this times this. And then a plus b times negative b. That's negative b times a plus b."}, {"video_title": "Example 1 Multiplying binomials to get a difference of squares Algebra I Khan Academy.mp3", "Sentence": "So we have a times a plus b. That's this times this. And then a plus b times negative b. That's negative b times a plus b. So I've done distributive property once. Now I can do it again. I can distribute the a onto the a and this b."}, {"video_title": "Example 1 Multiplying binomials to get a difference of squares Algebra I Khan Academy.mp3", "Sentence": "That's negative b times a plus b. So I've done distributive property once. Now I can do it again. I can distribute the a onto the a and this b. And it gives me a squared. a times a is a squared. Plus a times b, which is ab."}, {"video_title": "Example 1 Multiplying binomials to get a difference of squares Algebra I Khan Academy.mp3", "Sentence": "I can distribute the a onto the a and this b. And it gives me a squared. a times a is a squared. Plus a times b, which is ab. And now I can do it with the negative b. Negative b times a is negative ab or negative ba, same thing. And negative b times b is negative b squared."}, {"video_title": "Example 1 Multiplying binomials to get a difference of squares Algebra I Khan Academy.mp3", "Sentence": "Plus a times b, which is ab. And now I can do it with the negative b. Negative b times a is negative ab or negative ba, same thing. And negative b times b is negative b squared. Now what does this simplify to? Well, I have an ab and then I'm subtracting an ab. So these two guys cancel out."}, {"video_title": "Example 1 Multiplying binomials to get a difference of squares Algebra I Khan Academy.mp3", "Sentence": "And negative b times b is negative b squared. Now what does this simplify to? Well, I have an ab and then I'm subtracting an ab. So these two guys cancel out. And I am just left with a squared minus b squared. So the general pattern, and this is a good one to just kind of know super fast, is that a plus b times a minus b is always going to be a squared minus b squared. So we have an a plus b times an a minus b."}, {"video_title": "Example 1 Multiplying binomials to get a difference of squares Algebra I Khan Academy.mp3", "Sentence": "So these two guys cancel out. And I am just left with a squared minus b squared. So the general pattern, and this is a good one to just kind of know super fast, is that a plus b times a minus b is always going to be a squared minus b squared. So we have an a plus b times an a minus b. So this product is going to be a squared. So it's going to be 2x squared minus b squared minus 8 squared. 2x squared, that's the same thing as 2 squared times x squared or 4x squared."}, {"video_title": "Combining mixtures example.mp3", "Sentence": "We're told a partially filled tank holds 30 liters of gasoline with an 18% concentration of ethanol. A fuel station is selling gasoline with a 25% concentration of ethanol. What volume in liters of the fuel station gasoline would we need to add to the tank to get gasoline with a 20% concentration of ethanol? Pause this video and see if you can figure this out. All right, now let's work through this together. So let's first of all just remind ourselves how concentration relates to total volume to the volume of the ingredient. The way that we calculate concentration is that it is equal to the volume of the ingredient, which is in this case, it is ethanol, over the total volume, over total volume."}, {"video_title": "Combining mixtures example.mp3", "Sentence": "Pause this video and see if you can figure this out. All right, now let's work through this together. So let's first of all just remind ourselves how concentration relates to total volume to the volume of the ingredient. The way that we calculate concentration is that it is equal to the volume of the ingredient, which is in this case, it is ethanol, over the total volume, over total volume. Now this is already interesting because this first sentence tells us a lot. It tells us our concentration, it tells us our total volume. And so if we know two parts of this, in theory, we could figure out the third part."}, {"video_title": "Combining mixtures example.mp3", "Sentence": "The way that we calculate concentration is that it is equal to the volume of the ingredient, which is in this case, it is ethanol, over the total volume, over total volume. Now this is already interesting because this first sentence tells us a lot. It tells us our concentration, it tells us our total volume. And so if we know two parts of this, in theory, we could figure out the third part. Let's try that out. We know we're dealing with an 18% concentration. That's going to be equal to, they haven't told us our volume of ingredient, we just know that the ingredient is ethanol, volume of ethanol, over the total volume they have told us, 30 liters."}, {"video_title": "Combining mixtures example.mp3", "Sentence": "And so if we know two parts of this, in theory, we could figure out the third part. Let's try that out. We know we're dealing with an 18% concentration. That's going to be equal to, they haven't told us our volume of ingredient, we just know that the ingredient is ethanol, volume of ethanol, over the total volume they have told us, 30 liters. So if we multiply both sides by 30 liters, that's going to give us the volume of ethanol because those two cancel. And what we get is 18% of 30, let's see, 18 times three is 54. So this is going to be 5.4 liters is equal to our volume of ethanol."}, {"video_title": "Combining mixtures example.mp3", "Sentence": "That's going to be equal to, they haven't told us our volume of ingredient, we just know that the ingredient is ethanol, volume of ethanol, over the total volume they have told us, 30 liters. So if we multiply both sides by 30 liters, that's going to give us the volume of ethanol because those two cancel. And what we get is 18% of 30, let's see, 18 times three is 54. So this is going to be 5.4 liters is equal to our volume of ethanol. Not only will this hopefully make it a little bit clearer how these three relate, but this is also likely to be useful information for the rest of the problem. But now let's go to where we're trying to get to. We're trying to find a volume in liters of the fuel station gasoline we would need in order to have this concentration."}, {"video_title": "Combining mixtures example.mp3", "Sentence": "So this is going to be 5.4 liters is equal to our volume of ethanol. Not only will this hopefully make it a little bit clearer how these three relate, but this is also likely to be useful information for the rest of the problem. But now let's go to where we're trying to get to. We're trying to find a volume in liters of the fuel station gasoline we would need in order to have this concentration. So let's set V equal to that. And we're trying to get a 20% concentration. So what we could write is our 20% concentration is going to be equal to our new volume of ethanol."}, {"video_title": "Combining mixtures example.mp3", "Sentence": "We're trying to find a volume in liters of the fuel station gasoline we would need in order to have this concentration. So let's set V equal to that. And we're trying to get a 20% concentration. So what we could write is our 20% concentration is going to be equal to our new volume of ethanol. Actually, let me write that out. So it's going to be new volume of ethanol divided by our new total volume, our new total volume. Now what's going to be our new total volume?"}, {"video_title": "Combining mixtures example.mp3", "Sentence": "So what we could write is our 20% concentration is going to be equal to our new volume of ethanol. Actually, let me write that out. So it's going to be new volume of ethanol divided by our new total volume, our new total volume. Now what's going to be our new total volume? We're starting with 30 liters and then we're adding V liters to it. So our new total volume is going to be the 30 liters we started with plus the V liters that we're adding. And what's our new volume of ethanol?"}, {"video_title": "Combining mixtures example.mp3", "Sentence": "Now what's going to be our new total volume? We're starting with 30 liters and then we're adding V liters to it. So our new total volume is going to be the 30 liters we started with plus the V liters that we're adding. And what's our new volume of ethanol? Well, it's going to be the ethanol that we started with, the 30 liters times 18%, which is 5.4 liters, 5.4 liters plus the volume of ethanol we're adding. Well, to figure out the volume of ethanol we're adding, we just have to multiply the volume we're adding times the concentration of that volume. So it's going to be 25%, that's the concentration of the gas that we're adding, times V plus 0.25 V. And now we have an equation to solve for V. And the best thing that I can think to do is let's start by multiplying both sides of this times 30 plus V. I'm also going to multiply this side times 30 plus V as well."}, {"video_title": "Combining mixtures example.mp3", "Sentence": "And what's our new volume of ethanol? Well, it's going to be the ethanol that we started with, the 30 liters times 18%, which is 5.4 liters, 5.4 liters plus the volume of ethanol we're adding. Well, to figure out the volume of ethanol we're adding, we just have to multiply the volume we're adding times the concentration of that volume. So it's going to be 25%, that's the concentration of the gas that we're adding, times V plus 0.25 V. And now we have an equation to solve for V. And the best thing that I can think to do is let's start by multiplying both sides of this times 30 plus V. I'm also going to multiply this side times 30 plus V as well. These two characters cancel and we are going to be left with, on the left-hand side, this over here, 20% of that is going to be, if I distribute the 20%, 20% of 30 is six, and then it's going to be plus, I'll write it as a decimal, 0.2 V. I'm just distributing the 20% over this expression here. And then that is going to be equal to, on the right-hand side, I just have the numerator here because the 30 plus V cancels with the 30 plus V. I have 5.4 plus 0.25 V. And now let's see, my V coefficient is larger on the right. So what I could do is try to subtract the 0.2 V from both sides."}, {"video_title": "Combining mixtures example.mp3", "Sentence": "So it's going to be 25%, that's the concentration of the gas that we're adding, times V plus 0.25 V. And now we have an equation to solve for V. And the best thing that I can think to do is let's start by multiplying both sides of this times 30 plus V. I'm also going to multiply this side times 30 plus V as well. These two characters cancel and we are going to be left with, on the left-hand side, this over here, 20% of that is going to be, if I distribute the 20%, 20% of 30 is six, and then it's going to be plus, I'll write it as a decimal, 0.2 V. I'm just distributing the 20% over this expression here. And then that is going to be equal to, on the right-hand side, I just have the numerator here because the 30 plus V cancels with the 30 plus V. I have 5.4 plus 0.25 V. And now let's see, my V coefficient is larger on the right. So what I could do is try to subtract the 0.2 V from both sides. So I isolate the Vs on the right. So let me do that. So minus 0.2 V, minus 0.2 V. And then, actually, I'll just do one step at a time."}, {"video_title": "Combining mixtures example.mp3", "Sentence": "So what I could do is try to subtract the 0.2 V from both sides. So I isolate the Vs on the right. So let me do that. So minus 0.2 V, minus 0.2 V. And then, actually, I'll just do one step at a time. So that's going to get me, on the left-hand side, six is equal to 5.4 plus, if I subtract here, this is 0.05 V. Now I could subtract 5.4 from both sides. And what I'm going to get is six is, or actually 0.6, I have to be careful, is going to be equal to 0.05 V. And now to solve for V, I can just divide both sides by 0.05, 0.05. That's going to get me, this is the same thing as 60 divided by five."}, {"video_title": "Combining mixtures example.mp3", "Sentence": "So minus 0.2 V, minus 0.2 V. And then, actually, I'll just do one step at a time. So that's going to get me, on the left-hand side, six is equal to 5.4 plus, if I subtract here, this is 0.05 V. Now I could subtract 5.4 from both sides. And what I'm going to get is six is, or actually 0.6, I have to be careful, is going to be equal to 0.05 V. And now to solve for V, I can just divide both sides by 0.05, 0.05. That's going to get me, this is the same thing as 60 divided by five. It gets me that V is equal to 12 liters. And we are done. And if you want, you can verify the new concentration when I add 12 liters of this concentration to the 30 liters of that concentration."}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "And then we could graph it right over here. And to graph a circle, you have to know where its center is, and you have to know what its radius is. So let me see if I can change that. And you have to know what its radius is. So what we need to do is put this in some form where we can pick out its center and its radius. Let me get my little scratch pad out and see if we can do that. So this is that same equation."}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "And you have to know what its radius is. So what we need to do is put this in some form where we can pick out its center and its radius. Let me get my little scratch pad out and see if we can do that. So this is that same equation. And what I essentially want to do is I want to complete the square in terms of x and complete the square in terms of y to put it into a form that we can recognize. So first, let's take all of the x terms. So let's take all of the x terms."}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "So this is that same equation. And what I essentially want to do is I want to complete the square in terms of x and complete the square in terms of y to put it into a form that we can recognize. So first, let's take all of the x terms. So let's take all of the x terms. So you have x squared and 4x on the left-hand side. So I could rewrite this as x squared plus 4x. And I'm going to put some parentheses around here because I want to complete the square."}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "So let's take all of the x terms. So you have x squared and 4x on the left-hand side. So I could rewrite this as x squared plus 4x. And I'm going to put some parentheses around here because I want to complete the square. And then I have my y terms. I'll circle those in. Well, the red looks too much like the purple."}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "And I'm going to put some parentheses around here because I want to complete the square. And then I have my y terms. I'll circle those in. Well, the red looks too much like the purple. I'll circle those in blue. y squared and negative 4y. So we have plus y squared minus 4y."}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "Well, the red looks too much like the purple. I'll circle those in blue. y squared and negative 4y. So we have plus y squared minus 4y. And then we have a minus 17. And I'll just do that in a neutral color. So minus 17 is equal to 0."}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "So we have plus y squared minus 4y. And then we have a minus 17. And I'll just do that in a neutral color. So minus 17 is equal to 0. Now, what I want to do is make each of these purple expressions perfect squares. So how could I do that here? Well, this would be a perfect square if I took half of this 4 and I squared it."}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "So minus 17 is equal to 0. Now, what I want to do is make each of these purple expressions perfect squares. So how could I do that here? Well, this would be a perfect square if I took half of this 4 and I squared it. So if I made this plus 4, then this entire expression would be x plus 2 squared. And you can verify that if you like. If you need review on completing the square, there's plenty of videos on Khan Academy on that."}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "Well, this would be a perfect square if I took half of this 4 and I squared it. So if I made this plus 4, then this entire expression would be x plus 2 squared. And you can verify that if you like. If you need review on completing the square, there's plenty of videos on Khan Academy on that. All we did is we took half of this coefficient and then squared it to get 4. Half of 4 is 2 squared to get 4. And that comes straight out of the idea."}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "If you need review on completing the square, there's plenty of videos on Khan Academy on that. All we did is we took half of this coefficient and then squared it to get 4. Half of 4 is 2 squared to get 4. And that comes straight out of the idea. If you take x plus 2 and square it, it's going to be x squared plus twice the product of 2 and x plus 2 squared. Now, we can't just willy-nilly add a 4 here. We had an equality before."}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "And that comes straight out of the idea. If you take x plus 2 and square it, it's going to be x squared plus twice the product of 2 and x plus 2 squared. Now, we can't just willy-nilly add a 4 here. We had an equality before. Just adding a 4, it wouldn't be equal anymore. So if we want to maintain the equality, we have to add 4 on the right-hand side as well. Now, let's do the same thing for the y's."}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "We had an equality before. Just adding a 4, it wouldn't be equal anymore. So if we want to maintain the equality, we have to add 4 on the right-hand side as well. Now, let's do the same thing for the y's. Half of this coefficient right over here is a negative 2. If we square a negative 2, it becomes a positive 4. We can't just do that on the left-hand side."}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "Now, let's do the same thing for the y's. Half of this coefficient right over here is a negative 2. If we square a negative 2, it becomes a positive 4. We can't just do that on the left-hand side. We have to do that on the right-hand side as well. Now, what we have in blue becomes y minus 2 squared. And of course, we have the minus 17."}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "We can't just do that on the left-hand side. We have to do that on the right-hand side as well. Now, what we have in blue becomes y minus 2 squared. And of course, we have the minus 17. But why don't we add 17 to both sides as well to get rid of this minus 17 here? So let's add 17 on the left and add 17 on the right. So on the left, we're just left with these two expressions."}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "And of course, we have the minus 17. But why don't we add 17 to both sides as well to get rid of this minus 17 here? So let's add 17 on the left and add 17 on the right. So on the left, we're just left with these two expressions. And on the right, we have 4 plus 4 plus 17. Well, that's 8 plus 17, which is equal to 25. Now, this is a form that we recognize."}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "So on the left, we're just left with these two expressions. And on the right, we have 4 plus 4 plus 17. Well, that's 8 plus 17, which is equal to 25. Now, this is a form that we recognize. If you have the form x minus a squared plus y minus b squared is equal to r squared, we know that the center is at the point a comma b, essentially the point that makes both of these equal to 0, and that the radius is going to be r. So if we look over here, what is our a? We have to be careful here. Our a isn't 2."}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "Now, this is a form that we recognize. If you have the form x minus a squared plus y minus b squared is equal to r squared, we know that the center is at the point a comma b, essentially the point that makes both of these equal to 0, and that the radius is going to be r. So if we look over here, what is our a? We have to be careful here. Our a isn't 2. Our a is negative 2. x minus negative 2 is equal to 2. So the x-coordinate of our center is going to be negative 2. And the y-coordinate of our center is going to be 2."}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "Our a isn't 2. Our a is negative 2. x minus negative 2 is equal to 2. So the x-coordinate of our center is going to be negative 2. And the y-coordinate of our center is going to be 2. Remember, we care about the x value that makes this 0 and the y value that makes this 0. So the center is negative 2 comma 2. And this is the radius squared."}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "And the y-coordinate of our center is going to be 2. Remember, we care about the x value that makes this 0 and the y value that makes this 0. So the center is negative 2 comma 2. And this is the radius squared. So the radius is equal to 5. So let's go back to the exercise and actually plot this. So it's negative 2 comma 2."}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "And this is the radius squared. So the radius is equal to 5. So let's go back to the exercise and actually plot this. So it's negative 2 comma 2. So our center is negative 2 comma 2. So that's right over there. x is negative 2. y is positive 2."}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "So it's negative 2 comma 2. So our center is negative 2 comma 2. So that's right over there. x is negative 2. y is positive 2. And the radius is 5. So see, this would be 1, 2, 3, 4, 5. So we have to go a little bit wider than this."}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "x is negative 2. y is positive 2. And the radius is 5. So see, this would be 1, 2, 3, 4, 5. So we have to go a little bit wider than this. My pen is having trouble. There you go. 1, 2, 3, 4, 5."}, {"video_title": "Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "we're asked to solve the equation 3 plus the principal square root of 5x plus 6 is equal to 12. And so the general strategy to solve this type of equation is to isolate the radical sign on one side of the equation and then you can square it to essentially get the radical sign to go away. But you have to be very careful there because when you square radical signs you actually lose the information that you're taking the principal square root, not the negative square root or not the plus or minus square root. You're only taking the positive square root. And so when we get our final answer we do have to check and make sure that it gels with taking the principal square root. So let's try, let's see what I'm talking about. So the first thing I want to do is I want to isolate this on one side of the equation."}, {"video_title": "Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "You're only taking the positive square root. And so when we get our final answer we do have to check and make sure that it gels with taking the principal square root. So let's try, let's see what I'm talking about. So the first thing I want to do is I want to isolate this on one side of the equation. The best way to isolate that is to get rid of this 3. And the best way to get rid of the 3 is to subtract 3 from the left-hand side. And of course if I do it on the left-hand side I also have to do it on the right-hand side."}, {"video_title": "Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "So the first thing I want to do is I want to isolate this on one side of the equation. The best way to isolate that is to get rid of this 3. And the best way to get rid of the 3 is to subtract 3 from the left-hand side. And of course if I do it on the left-hand side I also have to do it on the right-hand side. Otherwise I would lose the ability to say that they're equal. And so the left-hand side right over here simplifies to the principal square root of 5x plus 6. And this is equal to 12 minus 3."}, {"video_title": "Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "And of course if I do it on the left-hand side I also have to do it on the right-hand side. Otherwise I would lose the ability to say that they're equal. And so the left-hand side right over here simplifies to the principal square root of 5x plus 6. And this is equal to 12 minus 3. This is equal to 9. And now we can square both sides of this equation. So we could square the principal square root of 5x plus 6 and we can square 9."}, {"video_title": "Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "And this is equal to 12 minus 3. This is equal to 9. And now we can square both sides of this equation. So we could square the principal square root of 5x plus 6 and we can square 9. When you do this, when you square this you get 5x plus 6. If you square the square root of 5x plus 6 you're going to get 5x plus 6. And this is where we actually lost some information."}, {"video_title": "Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "So we could square the principal square root of 5x plus 6 and we can square 9. When you do this, when you square this you get 5x plus 6. If you square the square root of 5x plus 6 you're going to get 5x plus 6. And this is where we actually lost some information. Because we would have also gotten this if we squared the negative square root of 5x plus 6. And so that's why we have to be careful with the answers we get. And actually make sure it works when the original equation was the principal square root."}, {"video_title": "Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "And this is where we actually lost some information. Because we would have also gotten this if we squared the negative square root of 5x plus 6. And so that's why we have to be careful with the answers we get. And actually make sure it works when the original equation was the principal square root. So we get 5x plus 6 on the left-hand side and on the right-hand side we get 81. And now this is just a straight up linear equation. We want to isolate the x terms."}, {"video_title": "Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "And actually make sure it works when the original equation was the principal square root. So we get 5x plus 6 on the left-hand side and on the right-hand side we get 81. And now this is just a straight up linear equation. We want to isolate the x terms. Let's subtract 6 from both sides. On the left-hand side we have 5x and on the right-hand side we have 75. And then we can divide both sides by 5."}, {"video_title": "Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "We want to isolate the x terms. Let's subtract 6 from both sides. On the left-hand side we have 5x and on the right-hand side we have 75. And then we can divide both sides by 5. Divide both sides by 5. We get x is equal to 15. 5 times 10 is 50."}, {"video_title": "Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "And then we can divide both sides by 5. Divide both sides by 5. We get x is equal to 15. 5 times 10 is 50. 5 times 5 is 25. This is 75. So we get x is equal to 15."}, {"video_title": "Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "5 times 10 is 50. 5 times 5 is 25. This is 75. So we get x is equal to 15. But we need to make sure that this actually works for our original equation. Maybe this would have worked if this was the negative square root. So we need to make sure it actually works for the positive square root, for the principal square root."}, {"video_title": "Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "So we get x is equal to 15. But we need to make sure that this actually works for our original equation. Maybe this would have worked if this was the negative square root. So we need to make sure it actually works for the positive square root, for the principal square root. So let's apply it to our original equation. So we get 3 plus the principal square root of 5 times 15. So 75 plus 6."}, {"video_title": "Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "So we need to make sure it actually works for the positive square root, for the principal square root. So let's apply it to our original equation. So we get 3 plus the principal square root of 5 times 15. So 75 plus 6. So I just took 5 times 15 over here. I put our solution in. It should be equal to 12."}, {"video_title": "Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "So 75 plus 6. So I just took 5 times 15 over here. I put our solution in. It should be equal to 12. Or we get 3 plus the square root of 75 plus 6 is 81. It needs to be equal to 12. And this is the principal root of 81."}, {"video_title": "Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "It should be equal to 12. Or we get 3 plus the square root of 75 plus 6 is 81. It needs to be equal to 12. And this is the principal root of 81. So it's positive 9. So it's 3 plus 9 needs to be equal to 12, which is absolutely true. So we can feel pretty good about this answer."}, {"video_title": "Graphing piecewise function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "It's defined as essentially different lines. You see this right over here, even with all the decimals and the negative signs, this is essentially a line. It's defined by this line over this interval for x, this line over this interval of x, and this line over this interval of x. I want to see if we can graph it. I encourage you, especially if you have some graph paper, to see if you could graph this on your own first before I work through it. So let's think about this first interval. If when negative 10 is less than or equal to x, which is less than negative two, then our function is defined by negative 0.125x plus 4.75. So this is going to be a line, a downward-sloping line."}, {"video_title": "Graphing piecewise function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "I encourage you, especially if you have some graph paper, to see if you could graph this on your own first before I work through it. So let's think about this first interval. If when negative 10 is less than or equal to x, which is less than negative two, then our function is defined by negative 0.125x plus 4.75. So this is going to be a line, a downward-sloping line. And the easiest way I can think about graphing it is let's just plot the endpoints here and then draw the line. So when x is equal to 10, so when, or sorry, when x is equal to negative 10, so we would have negative zero, actually let me write it this way. Let me do it over here where I do the, so we're gonna have negative 0.125 times negative 10 plus 4.75."}, {"video_title": "Graphing piecewise function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So this is going to be a line, a downward-sloping line. And the easiest way I can think about graphing it is let's just plot the endpoints here and then draw the line. So when x is equal to 10, so when, or sorry, when x is equal to negative 10, so we would have negative zero, actually let me write it this way. Let me do it over here where I do the, so we're gonna have negative 0.125 times negative 10 plus 4.75. That is going to be equal to, let's see, the negative times the negative is a positive, and then 10 times this is going to be, it's going to be 1.25 plus 4.75. That is going to be equal to six. So we're going to have the point negative 10 comma six."}, {"video_title": "Graphing piecewise function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Let me do it over here where I do the, so we're gonna have negative 0.125 times negative 10 plus 4.75. That is going to be equal to, let's see, the negative times the negative is a positive, and then 10 times this is going to be, it's going to be 1.25 plus 4.75. That is going to be equal to six. So we're going to have the point negative 10 comma six. And that point, and it includes, so x is defined there, it's less than or equal to. And then we go all the way to negative two. So when x is equal to negative two, we have negative 0.125 times negative two plus 4.75 is equal to, see, negative times negative is positive."}, {"video_title": "Graphing piecewise function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So we're going to have the point negative 10 comma six. And that point, and it includes, so x is defined there, it's less than or equal to. And then we go all the way to negative two. So when x is equal to negative two, we have negative 0.125 times negative two plus 4.75 is equal to, see, negative times negative is positive. Two times this is going to be, is going to be positive 0.25 plus 4.75. It's going to be equal to positive five. Now, we might be tempted, we might be tempted to just circle in this dot over here."}, {"video_title": "Graphing piecewise function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So when x is equal to negative two, we have negative 0.125 times negative two plus 4.75 is equal to, see, negative times negative is positive. Two times this is going to be, is going to be positive 0.25 plus 4.75. It's going to be equal to positive five. Now, we might be tempted, we might be tempted to just circle in this dot over here. But remember, this interval does not include negative two. It's up to it including, it's up to negative two, not including. So I'm gonna put a little open circle there, and then I'm gonna draw the line."}, {"video_title": "Graphing piecewise function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Now, we might be tempted, we might be tempted to just circle in this dot over here. But remember, this interval does not include negative two. It's up to it including, it's up to negative two, not including. So I'm gonna put a little open circle there, and then I'm gonna draw the line. And then I'm gonna draw, and I'm gonna draw the line. I am going to draw my best attempt, my best attempt at the line. Now let's do the next interval."}, {"video_title": "Graphing piecewise function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So I'm gonna put a little open circle there, and then I'm gonna draw the line. And then I'm gonna draw, and I'm gonna draw the line. I am going to draw my best attempt, my best attempt at the line. Now let's do the next interval. The next interval, this one's a lot more straightforward. We start at x equals negative two. When x equals negative two, negative two plus seven is, negative two plus seven is five."}, {"video_title": "Graphing piecewise function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Now let's do the next interval. The next interval, this one's a lot more straightforward. We start at x equals negative two. When x equals negative two, negative two plus seven is, negative two plus seven is five. So negative two, so negative two comma five. So it actually includes that point right over there. So we're actually able to fill it in."}, {"video_title": "Graphing piecewise function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "When x equals negative two, negative two plus seven is, negative two plus seven is five. So negative two, so negative two comma five. So it actually includes that point right over there. So we're actually able to fill it in. And then when x is negative one, negative one plus seven is going to be positive six. Positive six, but we're not including x equals negative one, up to and including. So it's going to be, it's going to be right over here."}, {"video_title": "Graphing piecewise function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So we're actually able to fill it in. And then when x is negative one, negative one plus seven is going to be positive six. Positive six, but we're not including x equals negative one, up to and including. So it's going to be, it's going to be right over here. When x is negative one, we are approaching, or as x approaches negative one, we're approaching negative one plus seven is six. So that's that interval right over there. And now let's look at this last interval."}, {"video_title": "Graphing piecewise function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So it's going to be, it's going to be right over here. When x is negative one, we are approaching, or as x approaches negative one, we're approaching negative one plus seven is six. So that's that interval right over there. And now let's look at this last interval. This last interval, when x is negative one, you're going to have, well this is just going to be positive 12 over 11, because we're multiplying it by negative one, plus 54 over 11, which is equal to 66 over 11, which is equal to positive six. So we're able to fill in that right over there. And then when x is equal to 10, you have negative 120 over 11."}, {"video_title": "Graphing piecewise function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And now let's look at this last interval. This last interval, when x is negative one, you're going to have, well this is just going to be positive 12 over 11, because we're multiplying it by negative one, plus 54 over 11, which is equal to 66 over 11, which is equal to positive six. So we're able to fill in that right over there. And then when x is equal to 10, you have negative 120 over 11. I just multiplied this times 10. 12 times 10 is 120, and we have the negative, plus 54 over 11. So this is the same thing."}, {"video_title": "Graphing piecewise function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And then when x is equal to 10, you have negative 120 over 11. I just multiplied this times 10. 12 times 10 is 120, and we have the negative, plus 54 over 11. So this is the same thing. This is going to be, what is this? This is negative 66 over 11. Is that right?"}, {"video_title": "Graphing piecewise function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So this is the same thing. This is going to be, what is this? This is negative 66 over 11. Is that right? Let's see, if you, yeah, that is negative 66 over 11, which is equal to negative six. So when x is equal to 10, our function is equal to negative negative six. And so this one actually doesn't have any jumps in it."}, {"video_title": "Graphing piecewise function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Is that right? Let's see, if you, yeah, that is negative 66 over 11, which is equal to negative six. So when x is equal to 10, our function is equal to negative negative six. And so this one actually doesn't have any jumps in it. It could've, but we see. So there we have it. We have graphed this function that has been defined in a piecewise way."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "Galileo was well into his 70s, Descartes died at, what, this is only 54 years old. And he's probably most known in popular culture for this quote right over here, a very philosophical quote, I think, therefore I am. But I also wanted to throw in, and this isn't that related to algebra, but I just thought it was a really neat quote, probably his least famous quote, this one right over here. And I like it just because it's very practical, it makes you realize that these great minds, these pillars of philosophy and mathematics, at the end of the day, they really were just human beings. And he said, you just keep pushing, you just keep pushing. I made every mistake that could be made, but I just kept pushing, which I think is very, very good life advice. Now, he did many things in philosophy and mathematics, but the reason why I'm including him here as we build our foundations of algebra is that he is the individual most responsible for a very strong connection between algebra and geometry."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "And I like it just because it's very practical, it makes you realize that these great minds, these pillars of philosophy and mathematics, at the end of the day, they really were just human beings. And he said, you just keep pushing, you just keep pushing. I made every mistake that could be made, but I just kept pushing, which I think is very, very good life advice. Now, he did many things in philosophy and mathematics, but the reason why I'm including him here as we build our foundations of algebra is that he is the individual most responsible for a very strong connection between algebra and geometry. So on the left over here, you have the world of algebra, and we've discussed it a little bit. You have equations that deal with symbols, and these symbols are essentially, they can take on values. So you could have something like y is equal to 2x, y is equal to 2x minus 1."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "Now, he did many things in philosophy and mathematics, but the reason why I'm including him here as we build our foundations of algebra is that he is the individual most responsible for a very strong connection between algebra and geometry. So on the left over here, you have the world of algebra, and we've discussed it a little bit. You have equations that deal with symbols, and these symbols are essentially, they can take on values. So you could have something like y is equal to 2x, y is equal to 2x minus 1. This gives us a relationship between whatever x is and whatever y is. And we can even set up a table here and pick values for x and see what the values of y would be. And I could just pick random values for x and then figure out what y is, but I'll pick relatively straightforward values just so that the math doesn't get too complicated."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "So you could have something like y is equal to 2x, y is equal to 2x minus 1. This gives us a relationship between whatever x is and whatever y is. And we can even set up a table here and pick values for x and see what the values of y would be. And I could just pick random values for x and then figure out what y is, but I'll pick relatively straightforward values just so that the math doesn't get too complicated. So for example, if x is negative 2, then y is going to be 2 times negative 2 minus 1, which is negative 4 minus 1, which is negative 5. If x is negative 1, then y is going to be 2 times negative 1 minus 1, which is equal to, this is negative 2 minus 1, which is negative 3. If x is equal to 0, then y is going to be 2 times 0 minus 1."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "And I could just pick random values for x and then figure out what y is, but I'll pick relatively straightforward values just so that the math doesn't get too complicated. So for example, if x is negative 2, then y is going to be 2 times negative 2 minus 1, which is negative 4 minus 1, which is negative 5. If x is negative 1, then y is going to be 2 times negative 1 minus 1, which is equal to, this is negative 2 minus 1, which is negative 3. If x is equal to 0, then y is going to be 2 times 0 minus 1. 2 times 0 is 0 minus 1, is just negative 1. I'll do a couple more. If x is 1, and I could have picked any values here."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "If x is equal to 0, then y is going to be 2 times 0 minus 1. 2 times 0 is 0 minus 1, is just negative 1. I'll do a couple more. If x is 1, and I could have picked any values here. I could have said, well, what happens if x is the negative square root of 2? Or what happens if x is negative 5 halves or positive 6 sevenths? But I'm just picking these numbers because it makes the math a lot easier when I try to figure out what y is going to be."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "If x is 1, and I could have picked any values here. I could have said, well, what happens if x is the negative square root of 2? Or what happens if x is negative 5 halves or positive 6 sevenths? But I'm just picking these numbers because it makes the math a lot easier when I try to figure out what y is going to be. But when x is 1, y is going to be 2 times 1 minus 1. 2 times 1 is 2 minus 1 is 1. And I'll do one more."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "But I'm just picking these numbers because it makes the math a lot easier when I try to figure out what y is going to be. But when x is 1, y is going to be 2 times 1 minus 1. 2 times 1 is 2 minus 1 is 1. And I'll do one more. I'll do one more in a color that I have not used yet. Let's see, this purple. If x is 2, then y is going to be 2 times 2, now our x is 2, minus 1."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "And I'll do one more. I'll do one more in a color that I have not used yet. Let's see, this purple. If x is 2, then y is going to be 2 times 2, now our x is 2, minus 1. So that is 4 minus 1 is equal to 3. So fair enough. I just kind of sampled this relationship."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "If x is 2, then y is going to be 2 times 2, now our x is 2, minus 1. So that is 4 minus 1 is equal to 3. So fair enough. I just kind of sampled this relationship. I said, OK, this describes a general relationship between a variable y and a variable x. And then I just made it a little bit more concrete. I said, OK, well, then if x is one of these variables, for each of these values of x, what would be the corresponding value of y?"}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "I just kind of sampled this relationship. I said, OK, this describes a general relationship between a variable y and a variable x. And then I just made it a little bit more concrete. I said, OK, well, then if x is one of these variables, for each of these values of x, what would be the corresponding value of y? And what Descartes realized is that you could visualize this. One, you could visualize these individual points, but that could also help you in general to visualize this relationship. And so what he essentially did is he bridged the worlds of this kind of very abstract symbolic algebra and that and geometry, which was concerned with shapes and sizes and angles."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "I said, OK, well, then if x is one of these variables, for each of these values of x, what would be the corresponding value of y? And what Descartes realized is that you could visualize this. One, you could visualize these individual points, but that could also help you in general to visualize this relationship. And so what he essentially did is he bridged the worlds of this kind of very abstract symbolic algebra and that and geometry, which was concerned with shapes and sizes and angles. So over here, you have the world of geometry. And obviously, there are people in history, maybe many people who history may have forgotten who might have dabbled in this. But before Descartes, it's generally viewed that geometry was Euclidean geometry."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "And so what he essentially did is he bridged the worlds of this kind of very abstract symbolic algebra and that and geometry, which was concerned with shapes and sizes and angles. So over here, you have the world of geometry. And obviously, there are people in history, maybe many people who history may have forgotten who might have dabbled in this. But before Descartes, it's generally viewed that geometry was Euclidean geometry. And that's essentially the geometry that you study in a geometry class in eighth or ninth grade or 10th grade in a traditional high school curriculum. And that's the geometry of studying the relationships between triangles and their angles and the relationships between circles. And you have radii, and then you have triangles inscribed in circles and all the rest."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "But before Descartes, it's generally viewed that geometry was Euclidean geometry. And that's essentially the geometry that you study in a geometry class in eighth or ninth grade or 10th grade in a traditional high school curriculum. And that's the geometry of studying the relationships between triangles and their angles and the relationships between circles. And you have radii, and then you have triangles inscribed in circles and all the rest. And we go into some depth in that in the geometry playlist. But Descartes says, well, I think I can represent this visually the same way that Euclid was studying these triangles and these circles. He said, well, why don't I, if we view a piece of paper, if we think about a two-dimensional plane, you could view a piece of paper as kind of a section of a two-dimensional plane."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "And you have radii, and then you have triangles inscribed in circles and all the rest. And we go into some depth in that in the geometry playlist. But Descartes says, well, I think I can represent this visually the same way that Euclid was studying these triangles and these circles. He said, well, why don't I, if we view a piece of paper, if we think about a two-dimensional plane, you could view a piece of paper as kind of a section of a two-dimensional plane. We call it two dimensions because there's two directions that you could go in. There's the up-down direction. That's one direction."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "He said, well, why don't I, if we view a piece of paper, if we think about a two-dimensional plane, you could view a piece of paper as kind of a section of a two-dimensional plane. We call it two dimensions because there's two directions that you could go in. There's the up-down direction. That's one direction. So let me draw that. I'll do it in blue because we're trying to visualize things. So I'll do it in the geometry color."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "That's one direction. So let me draw that. I'll do it in blue because we're trying to visualize things. So I'll do it in the geometry color. So you have the up-down direction. And you have the left-right direction. That's why it's called a two-dimensional plane."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "So I'll do it in the geometry color. So you have the up-down direction. And you have the left-right direction. That's why it's called a two-dimensional plane. If we were dealing in three dimensions, you would have an in-out dimension. And it's very easy to do two dimensions on the screen because the screen is two-dimensional. And he says, well, there are two variables here, and they have this relationship."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "That's why it's called a two-dimensional plane. If we were dealing in three dimensions, you would have an in-out dimension. And it's very easy to do two dimensions on the screen because the screen is two-dimensional. And he says, well, there are two variables here, and they have this relationship. So why don't I associate each of these variables with one of these dimensions over here? And by convention, let's make the y variable, which is really the dependent of variable. The way we did it, it depends on what x is."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "And he says, well, there are two variables here, and they have this relationship. So why don't I associate each of these variables with one of these dimensions over here? And by convention, let's make the y variable, which is really the dependent of variable. The way we did it, it depends on what x is. Let's put that on the vertical axis. And let's put our independent variable, the one where I just randomly picked values for it to see what y would become. Let's put that on the horizontal axis."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "The way we did it, it depends on what x is. Let's put that on the vertical axis. And let's put our independent variable, the one where I just randomly picked values for it to see what y would become. Let's put that on the horizontal axis. And it actually was Descartes who came up with the convention of using x's and y's, and we'll see later z's in algebra so extensively as unknown variables or the variables that you're manipulating. But he says, well, if we think about it this way, if we number these dimensions, so let's say that in the x direction, let's make this right over here is negative 3. Let's make this negative 2."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "Let's put that on the horizontal axis. And it actually was Descartes who came up with the convention of using x's and y's, and we'll see later z's in algebra so extensively as unknown variables or the variables that you're manipulating. But he says, well, if we think about it this way, if we number these dimensions, so let's say that in the x direction, let's make this right over here is negative 3. Let's make this negative 2. This is negative 1. This is 0. Now I'm just numbering the x direction, the left-right direction."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "Let's make this negative 2. This is negative 1. This is 0. Now I'm just numbering the x direction, the left-right direction. Now this is positive 1. This is positive 2. This is positive 3."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "Now I'm just numbering the x direction, the left-right direction. Now this is positive 1. This is positive 2. This is positive 3. And we can do the same in the y direction. So let's see. So this could be, let's say this is negative 5, negative 4, negative 3, negative, actually, let me do it a little bit neater than that."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "This is positive 3. And we can do the same in the y direction. So let's see. So this could be, let's say this is negative 5, negative 4, negative 3, negative, actually, let me do it a little bit neater than that. Let me clean this up a little bit. So let me erase this and extend this down a little bit so I can go all the way down to negative 5 without making it look too messy. So let's go all the way down here."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "So this could be, let's say this is negative 5, negative 4, negative 3, negative, actually, let me do it a little bit neater than that. Let me clean this up a little bit. So let me erase this and extend this down a little bit so I can go all the way down to negative 5 without making it look too messy. So let's go all the way down here. And so we can number it. This is 1. This is 2."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "So let's go all the way down here. And so we can number it. This is 1. This is 2. This is 3. And then this could be negative 1, negative 2. And these are all just conventions."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "This is 2. This is 3. And then this could be negative 1, negative 2. And these are all just conventions. It could have been labeled the other way. We could have decided to put the x there and the y there and make this the positive direction, make this the negative direction. But this is just a convention that people adopted starting with Descartes."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "And these are all just conventions. It could have been labeled the other way. We could have decided to put the x there and the y there and make this the positive direction, make this the negative direction. But this is just a convention that people adopted starting with Descartes. Negative 2, negative 3, negative 4, and negative 5. And he says, well, I think I can associate each of these pairs of values with a point in two dimensions. I can take the x-coordinate."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "But this is just a convention that people adopted starting with Descartes. Negative 2, negative 3, negative 4, and negative 5. And he says, well, I think I can associate each of these pairs of values with a point in two dimensions. I can take the x-coordinate. I can take the x value right over here. And I say, OK, that's negative 2. That would be right over there along the left-right direction."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "I can take the x-coordinate. I can take the x value right over here. And I say, OK, that's negative 2. That would be right over there along the left-right direction. I'm going to the left because it's negative. And that's associated with negative 5 in the vertical direction. So I say the y value is negative 5."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "That would be right over there along the left-right direction. I'm going to the left because it's negative. And that's associated with negative 5 in the vertical direction. So I say the y value is negative 5. And so if I go 2 to the left and 5 down, I get to this point right over there. So he says, these two values, negative 2, negative 5, I can associate it with this point in this plane right over here, in this two-dimensional plane. So I'll say that point has the coordinates."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "So I say the y value is negative 5. And so if I go 2 to the left and 5 down, I get to this point right over there. So he says, these two values, negative 2, negative 5, I can associate it with this point in this plane right over here, in this two-dimensional plane. So I'll say that point has the coordinates. Tells me where to find that point, negative 2, negative 5. And these coordinates, they're called Cartesian coordinates, named for Rene Descartes. He's the guy that came up with these."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "So I'll say that point has the coordinates. Tells me where to find that point, negative 2, negative 5. And these coordinates, they're called Cartesian coordinates, named for Rene Descartes. He's the guy that came up with these. He's associating all of a sudden these relationships with points on a coordinate plane. And then he says, well, OK, let's do another one. There's this other relationship where I have when x is equal to negative 1, y is equal to negative 3."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "He's the guy that came up with these. He's associating all of a sudden these relationships with points on a coordinate plane. And then he says, well, OK, let's do another one. There's this other relationship where I have when x is equal to negative 1, y is equal to negative 3. So x is negative 1, y is negative 3. That's that point right over there. And the convention is, once again, when you list the coordinates, you list the x-coordinate, then the y-coordinate."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "There's this other relationship where I have when x is equal to negative 1, y is equal to negative 3. So x is negative 1, y is negative 3. That's that point right over there. And the convention is, once again, when you list the coordinates, you list the x-coordinate, then the y-coordinate. And that's just what people decided to do. Negative 1, negative 3, that would be that point right over there. And then you have the point when x is 0, y is negative 1."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "And the convention is, once again, when you list the coordinates, you list the x-coordinate, then the y-coordinate. And that's just what people decided to do. Negative 1, negative 3, that would be that point right over there. And then you have the point when x is 0, y is negative 1. When x is 0, right over here, which means I don't go to the left or the right, y is negative 1, which means I go 1 down. So that's that point right over there, 0, negative 1, right over there. And I could keep doing this."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "And then you have the point when x is 0, y is negative 1. When x is 0, right over here, which means I don't go to the left or the right, y is negative 1, which means I go 1 down. So that's that point right over there, 0, negative 1, right over there. And I could keep doing this. When x is 1, y is 1. When x is 2, y is 3. Actually, let me do that in that same purple color."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "And I could keep doing this. When x is 1, y is 1. When x is 2, y is 3. Actually, let me do that in that same purple color. When x is 2, y is 3. 2, 3. And then this one right over here in orange was 1, 1."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "Actually, let me do that in that same purple color. When x is 2, y is 3. 2, 3. And then this one right over here in orange was 1, 1. And this is neat by itself. I essentially just sampled possible x's. But what he realizes, not only do you sample these possible x's, but if you just kept sampling x's, if you tried sampling all the x's in between, you'd actually end up plotting out a line."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "And then this one right over here in orange was 1, 1. And this is neat by itself. I essentially just sampled possible x's. But what he realizes, not only do you sample these possible x's, but if you just kept sampling x's, if you tried sampling all the x's in between, you'd actually end up plotting out a line. So if you were to do every possible x, you would end up getting a line that looks something like that right over there. And if you pick any x and find any y, it really represents a point on this line. Or another way to think about it, any point on this line represents a solution to this equation right over here."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "But what he realizes, not only do you sample these possible x's, but if you just kept sampling x's, if you tried sampling all the x's in between, you'd actually end up plotting out a line. So if you were to do every possible x, you would end up getting a line that looks something like that right over there. And if you pick any x and find any y, it really represents a point on this line. Or another way to think about it, any point on this line represents a solution to this equation right over here. So if you have this point right over here, which looks like it's about x is 1 and 1 half, y is 2. So let me write that. 1.5, 1.5, 2."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "Or another way to think about it, any point on this line represents a solution to this equation right over here. So if you have this point right over here, which looks like it's about x is 1 and 1 half, y is 2. So let me write that. 1.5, 1.5, 2. That is a solution to this equation. When x is 1.5, 2 times 1.5 is 3, minus 1 is 2. That is right over there."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "1.5, 1.5, 2. That is a solution to this equation. When x is 1.5, 2 times 1.5 is 3, minus 1 is 2. That is right over there. So all of a sudden, he was able to bridge this gap or this relationship between algebra and geometry. We can now visualize all of the x and y pairs that satisfy this equation right over here. And so he is responsible for making this bridge."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "That is right over there. So all of a sudden, he was able to bridge this gap or this relationship between algebra and geometry. We can now visualize all of the x and y pairs that satisfy this equation right over here. And so he is responsible for making this bridge. And that's why the coordinates that we use to specify these points are called Cartesian coordinates. And as we'll see, the first type of equations we will study are equations of this form over here. And in a traditional algebra curriculum, they're called linear equations."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "And so he is responsible for making this bridge. And that's why the coordinates that we use to specify these points are called Cartesian coordinates. And as we'll see, the first type of equations we will study are equations of this form over here. And in a traditional algebra curriculum, they're called linear equations. Linear equations. And you might be saying, well, OK, this is an equation. I see that this is equal to that."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "And in a traditional algebra curriculum, they're called linear equations. Linear equations. And you might be saying, well, OK, this is an equation. I see that this is equal to that. But what's so linear about them? What makes them look like a line? And to realize why they are linear, you have to make this jump that Rene Descartes made."}, {"video_title": "Introduction to the coordinate plane Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "I see that this is equal to that. But what's so linear about them? What makes them look like a line? And to realize why they are linear, you have to make this jump that Rene Descartes made. Because if you were to plot this using Cartesian coordinates on a Euclidean plane, you will get a line. And in the future, we'll see that there's other types of equations where you won't get a line. You'll get a curve or something kind of crazy or funky."}, {"video_title": "Finding intercepts from an equation Algebra I Khan Academy.mp3", "Sentence": "So we have to find the intercepts. And then use the intercepts to graph this line on the coordinate plane. So then graph the line. So whenever someone talks about intercepts, they're talking about where you're intersecting the x and the y axes. So let me label my axes here. So this is the x-axis, and that is the y-axis there. And when I intersect the x-axis, what's going on?"}, {"video_title": "Finding intercepts from an equation Algebra I Khan Academy.mp3", "Sentence": "So whenever someone talks about intercepts, they're talking about where you're intersecting the x and the y axes. So let me label my axes here. So this is the x-axis, and that is the y-axis there. And when I intersect the x-axis, what's going on? What is my y value when I'm at the x-axis? Well, my y value is 0. I'm not above or below the x-axis."}, {"video_title": "Finding intercepts from an equation Algebra I Khan Academy.mp3", "Sentence": "And when I intersect the x-axis, what's going on? What is my y value when I'm at the x-axis? Well, my y value is 0. I'm not above or below the x-axis. So the x-intercept is when y is equal to 0. And then by that same argument, what's the y-intercept? Well, if I'm somewhere along the y-axis, what's my x value?"}, {"video_title": "Finding intercepts from an equation Algebra I Khan Academy.mp3", "Sentence": "I'm not above or below the x-axis. So the x-intercept is when y is equal to 0. And then by that same argument, what's the y-intercept? Well, if I'm somewhere along the y-axis, what's my x value? Well, I'm not to the right or the left, so my x value has to be 0. So the y-intercept occurs when x is equal to 0. So to figure out the intercepts, let's set y equal to 0 in this equation and solve for x."}, {"video_title": "Finding intercepts from an equation Algebra I Khan Academy.mp3", "Sentence": "Well, if I'm somewhere along the y-axis, what's my x value? Well, I'm not to the right or the left, so my x value has to be 0. So the y-intercept occurs when x is equal to 0. So to figure out the intercepts, let's set y equal to 0 in this equation and solve for x. And then let's set x is equal to 0 and then solve for y. So when y is equal to 0, what does this equation become? I'll do it in orange."}, {"video_title": "Finding intercepts from an equation Algebra I Khan Academy.mp3", "Sentence": "So to figure out the intercepts, let's set y equal to 0 in this equation and solve for x. And then let's set x is equal to 0 and then solve for y. So when y is equal to 0, what does this equation become? I'll do it in orange. You get negative 5x plus 4y. Well, we're saying y is 0, so 4 times 0 is equal to 20. 4 times 0 is just 0, so we can just not write that."}, {"video_title": "Finding intercepts from an equation Algebra I Khan Academy.mp3", "Sentence": "I'll do it in orange. You get negative 5x plus 4y. Well, we're saying y is 0, so 4 times 0 is equal to 20. 4 times 0 is just 0, so we can just not write that. So let me just rewrite it. So we have negative 5x is equal to 20. We can divide both sides of this equation by negative 5."}, {"video_title": "Finding intercepts from an equation Algebra I Khan Academy.mp3", "Sentence": "4 times 0 is just 0, so we can just not write that. So let me just rewrite it. So we have negative 5x is equal to 20. We can divide both sides of this equation by negative 5. The negative 5's cancel out. That was the whole point behind dividing by negative 5. And we get x is equal to 20 divided by negative 5 is negative 4."}, {"video_title": "Finding intercepts from an equation Algebra I Khan Academy.mp3", "Sentence": "We can divide both sides of this equation by negative 5. The negative 5's cancel out. That was the whole point behind dividing by negative 5. And we get x is equal to 20 divided by negative 5 is negative 4. So when y is equal to 0, we saw that right there, x is equal to negative 4. Or if we wanted to plot that point, we always put the x-coordinate for it first. So that would be the point negative 4 comma 0."}, {"video_title": "Finding intercepts from an equation Algebra I Khan Academy.mp3", "Sentence": "And we get x is equal to 20 divided by negative 5 is negative 4. So when y is equal to 0, we saw that right there, x is equal to negative 4. Or if we wanted to plot that point, we always put the x-coordinate for it first. So that would be the point negative 4 comma 0. So let me graph that. So if we go 1, 2, 3, 4, that's negative 4. And then the y value is just 0, so that point is right over there."}, {"video_title": "Finding intercepts from an equation Algebra I Khan Academy.mp3", "Sentence": "So that would be the point negative 4 comma 0. So let me graph that. So if we go 1, 2, 3, 4, that's negative 4. And then the y value is just 0, so that point is right over there. That is the x-intercept. y is 0, x is negative 4. Notice we're intersecting the x-axis."}, {"video_title": "Finding intercepts from an equation Algebra I Khan Academy.mp3", "Sentence": "And then the y value is just 0, so that point is right over there. That is the x-intercept. y is 0, x is negative 4. Notice we're intersecting the x-axis. Now let's do the exact same thing for the y-intercept. Let's set x equal to 0. So if we set x is equal to 0, we have negative 5 times 0 plus 4y is equal to 20."}, {"video_title": "Finding intercepts from an equation Algebra I Khan Academy.mp3", "Sentence": "Notice we're intersecting the x-axis. Now let's do the exact same thing for the y-intercept. Let's set x equal to 0. So if we set x is equal to 0, we have negative 5 times 0 plus 4y is equal to 20. Well, anything times 0 is 0, so we can just put that out of the way. And remember, this was setting x is equal to 0. We're doing the y-intercept now."}, {"video_title": "Finding intercepts from an equation Algebra I Khan Academy.mp3", "Sentence": "So if we set x is equal to 0, we have negative 5 times 0 plus 4y is equal to 20. Well, anything times 0 is 0, so we can just put that out of the way. And remember, this was setting x is equal to 0. We're doing the y-intercept now. So this just simplifies to 4y is equal to 20. We can divide both sides of this equation by 4 to get rid of this 4 right there. And you get y is equal to 20 over 4, which is 5."}, {"video_title": "Finding intercepts from an equation Algebra I Khan Academy.mp3", "Sentence": "We're doing the y-intercept now. So this just simplifies to 4y is equal to 20. We can divide both sides of this equation by 4 to get rid of this 4 right there. And you get y is equal to 20 over 4, which is 5. So when x is equal to 0, y is equal to 5. So the point 0, 5 is on the graph for this line. So 0, 5, 0, x is 0, and y is 1, 2, 3, 4, 5 right over there."}, {"video_title": "Finding intercepts from an equation Algebra I Khan Academy.mp3", "Sentence": "And you get y is equal to 20 over 4, which is 5. So when x is equal to 0, y is equal to 5. So the point 0, 5 is on the graph for this line. So 0, 5, 0, x is 0, and y is 1, 2, 3, 4, 5 right over there. And notice, when x is 0, we're right on the y-axis. This is our y-intercept right over there. And if we graph the line, all you need is two points to graph any line."}, {"video_title": "Finding intercepts from an equation Algebra I Khan Academy.mp3", "Sentence": "So 0, 5, 0, x is 0, and y is 1, 2, 3, 4, 5 right over there. And notice, when x is 0, we're right on the y-axis. This is our y-intercept right over there. And if we graph the line, all you need is two points to graph any line. So we just have to connect the dots. And that is our line. So let me connect the dots, try my best to draw as straight of a line as I can."}, {"video_title": "Finding intercepts from an equation Algebra I Khan Academy.mp3", "Sentence": "And if we graph the line, all you need is two points to graph any line. So we just have to connect the dots. And that is our line. So let me connect the dots, try my best to draw as straight of a line as I can. Well, I could do a better job than that. To draw as straight of a line as I can. And that's the graph of this equation using the x-intercept and the y-intercept."}, {"video_title": "Finding the x-intercept of a line Algebra I Khan Academy.mp3", "Sentence": "Or it's the x value where our graph actually intersects the x-axis. Notice right over here our y value is exactly zero. We're sitting on the x-axis. So let's think about what this x value must be. Well just looking at it from, just trying to eyeball a little bit, it looks like it's a little over two. It's between two and three. It looks like it's less than two and a half."}, {"video_title": "Finding the x-intercept of a line Algebra I Khan Academy.mp3", "Sentence": "So let's think about what this x value must be. Well just looking at it from, just trying to eyeball a little bit, it looks like it's a little over two. It's between two and three. It looks like it's less than two and a half. But we don't know the exact value. So let's go turn to the equation to figure out the exact value. So we essentially have to figure out what x value when y is equal to zero will have this equation be true."}, {"video_title": "Finding the x-intercept of a line Algebra I Khan Academy.mp3", "Sentence": "It looks like it's less than two and a half. But we don't know the exact value. So let's go turn to the equation to figure out the exact value. So we essentially have to figure out what x value when y is equal to zero will have this equation be true. So we could just say two times zero plus 3x is equal to seven. Well two times zero is just gonna be zero. So we have 3x is equal to seven."}, {"video_title": "Finding the x-intercept of a line Algebra I Khan Academy.mp3", "Sentence": "So we essentially have to figure out what x value when y is equal to zero will have this equation be true. So we could just say two times zero plus 3x is equal to seven. Well two times zero is just gonna be zero. So we have 3x is equal to seven. And then we can divide both sides by three to solve for x. And we get x is equal to seven over three. Now does that look like 7 3rds?"}, {"video_title": "Finding the x-intercept of a line Algebra I Khan Academy.mp3", "Sentence": "So we have 3x is equal to seven. And then we can divide both sides by three to solve for x. And we get x is equal to seven over three. Now does that look like 7 3rds? Well we just have to remind ourselves that seven over three is the same thing as six over three plus one over three. And six over three is two. So this is the same thing as two and 1 3rd."}, {"video_title": "Finding the x-intercept of a line Algebra I Khan Academy.mp3", "Sentence": "Now does that look like 7 3rds? Well we just have to remind ourselves that seven over three is the same thing as six over three plus one over three. And six over three is two. So this is the same thing as two and 1 3rd. Another way you could think about it is three goes into seven two times. And then you have a remainder of one. So you still gotta divide that one by three."}, {"video_title": "Finding the x-intercept of a line Algebra I Khan Academy.mp3", "Sentence": "So this is the same thing as two and 1 3rd. Another way you could think about it is three goes into seven two times. And then you have a remainder of one. So you still gotta divide that one by three. So it's two full times and then a 1 3rd. So this looks like two and 1 3rd. And so that's its x intercept."}, {"video_title": "Finding the x-intercept of a line Algebra I Khan Academy.mp3", "Sentence": "So you still gotta divide that one by three. So it's two full times and then a 1 3rd. So this looks like two and 1 3rd. And so that's its x intercept. Seven 7 3rds. If I was doing this on the exercise on Khan Academy, it's always a little easier to type in the improper fraction. So I would put in 7 3rds."}, {"video_title": "Identifying quadratic patterns Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "So when they're talking about patterns, they're really saying, hey, can we say that some of these can generally form a pattern that matches what we have here and then we can use that pattern to factor it into one of these forms. What do I mean by that? Well, let's just imagine something like u plus v squared. We've squared binomials in the past. This is going to be equal to u squared plus two times the product of these terms, so two u v, and then plus v squared. Now, when you look at this polynomial right over here, it actually has this form if you look at it carefully. How could it have this form?"}, {"video_title": "Identifying quadratic patterns Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "We've squared binomials in the past. This is going to be equal to u squared plus two times the product of these terms, so two u v, and then plus v squared. Now, when you look at this polynomial right over here, it actually has this form if you look at it carefully. How could it have this form? Well, if we view u squared as nine x to the eighth, then that means that u is, and let me write it as a capital U, u is equal to three x to the fourth because notice, if you square this, you're gonna get nine x to the eighth. So this right over here is u squared. And if we said that v squared is equal to y squared, so if this is capital V squared, then that means that v is equal to y."}, {"video_title": "Identifying quadratic patterns Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "How could it have this form? Well, if we view u squared as nine x to the eighth, then that means that u is, and let me write it as a capital U, u is equal to three x to the fourth because notice, if you square this, you're gonna get nine x to the eighth. So this right over here is u squared. And if we said that v squared is equal to y squared, so if this is capital V squared, then that means that v is equal to y. And then this would have to be two times u v, is it? Well, see, if I multiply u times v, I get three x to the fourth y, and then two times that is indeed six x to the fourth y, so this right over here is two u v. So notice, this polynomial, this higher-degree polynomial can be expressed in this pattern, which means it can be factored this way. So when they say which pattern can we use to factor this expression, well, I would use a pattern for u plus v squared, so I would go with that choice right over there."}, {"video_title": "Identifying quadratic patterns Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "And if we said that v squared is equal to y squared, so if this is capital V squared, then that means that v is equal to y. And then this would have to be two times u v, is it? Well, see, if I multiply u times v, I get three x to the fourth y, and then two times that is indeed six x to the fourth y, so this right over here is two u v. So notice, this polynomial, this higher-degree polynomial can be expressed in this pattern, which means it can be factored this way. So when they say which pattern can we use to factor this expression, well, I would use a pattern for u plus v squared, so I would go with that choice right over there. Let's do a few more examples. So here, once again, we're told the same thing. We're given a different expression, and they're asking us what pattern can we use to factor the expression."}, {"video_title": "Identifying quadratic patterns Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "So when they say which pattern can we use to factor this expression, well, I would use a pattern for u plus v squared, so I would go with that choice right over there. Let's do a few more examples. So here, once again, we're told the same thing. We're given a different expression, and they're asking us what pattern can we use to factor the expression. So I have these three terms here. It looks like maybe I could use, I can see a perfect square here. Let's see if that works."}, {"video_title": "Identifying quadratic patterns Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "We're given a different expression, and they're asking us what pattern can we use to factor the expression. So I have these three terms here. It looks like maybe I could use, I can see a perfect square here. Let's see if that works. If this is u squared, if this is u squared, then that means that u is going to be equal to two x to the third power, and if this is v squared, then that means that v is equal to five. Now, is this equal to two times u v? Well, let's see, two times u v would be equal to, well, you're not gonna have any y in it, so this is not going to be two u v, so this actually is not fitting the perfect square pattern, so we could rule this out, and both of these are perfect squares of some form."}, {"video_title": "Identifying quadratic patterns Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "Let's see if that works. If this is u squared, if this is u squared, then that means that u is going to be equal to two x to the third power, and if this is v squared, then that means that v is equal to five. Now, is this equal to two times u v? Well, let's see, two times u v would be equal to, well, you're not gonna have any y in it, so this is not going to be two u v, so this actually is not fitting the perfect square pattern, so we could rule this out, and both of these are perfect squares of some form. One just has a, I guess you'd say, adding v, the other one is subtracting v. This right over here, if I were to multiply this out, this is going to be equal to, this is a difference of squares, and we've seen this before. This is u squared minus v squared, so you wouldn't have a three-term polynomial like that, so we could rule that one out, so I would pick that we can't use any of the patterns. Let's do yet another example, and I encourage you, pause the video and see if you can work this one out on your own."}, {"video_title": "Identifying quadratic patterns Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "Well, let's see, two times u v would be equal to, well, you're not gonna have any y in it, so this is not going to be two u v, so this actually is not fitting the perfect square pattern, so we could rule this out, and both of these are perfect squares of some form. One just has a, I guess you'd say, adding v, the other one is subtracting v. This right over here, if I were to multiply this out, this is going to be equal to, this is a difference of squares, and we've seen this before. This is u squared minus v squared, so you wouldn't have a three-term polynomial like that, so we could rule that one out, so I would pick that we can't use any of the patterns. Let's do yet another example, and I encourage you, pause the video and see if you can work this one out on your own. So, same idea. They wanna factor the following expression, and this one essentially has two terms. We have a term here, and we have a term here."}, {"video_title": "Identifying quadratic patterns Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "Let's do yet another example, and I encourage you, pause the video and see if you can work this one out on your own. So, same idea. They wanna factor the following expression, and this one essentially has two terms. We have a term here, and we have a term here. They are both, they both look like they are the square of something, and we have a difference of squares, so this is making me feel pretty good about this pattern, but let's see if that works out. Remember, u plus v times u minus v is equal to u squared minus v squared, so if this is equal to u squared, then that means that capital U is equal to six x squared. That works, and if this is equal to v squared, well, that means that v is equal to y plus three, so this is fitting this pattern right over here, and they're just asking us to say, what pattern can we use to factor the expression?"}, {"video_title": "Identifying quadratic patterns Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "We have a term here, and we have a term here. They are both, they both look like they are the square of something, and we have a difference of squares, so this is making me feel pretty good about this pattern, but let's see if that works out. Remember, u plus v times u minus v is equal to u squared minus v squared, so if this is equal to u squared, then that means that capital U is equal to six x squared. That works, and if this is equal to v squared, well, that means that v is equal to y plus three, so this is fitting this pattern right over here, and they're just asking us to say, what pattern can we use to factor the expression? They're not asking us to actually factor it, so we'll just pick this choice, but once you identify the pattern, it's actually pretty straightforward to factor it, because if you say this is just going to factor into u plus v times u minus v, well, u plus v is going to be six x squared plus v plus y plus three times u minus v. U is six x squared. Minus v is minus, we could write minus y plus three, or we could distribute the negative sign, but either way, this might make it a little bit clearer what we just did. We used the pattern to factor this higher degree polynomial, which is essentially just the difference of squares."}, {"video_title": "Identifying quadratic patterns Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "That works, and if this is equal to v squared, well, that means that v is equal to y plus three, so this is fitting this pattern right over here, and they're just asking us to say, what pattern can we use to factor the expression? They're not asking us to actually factor it, so we'll just pick this choice, but once you identify the pattern, it's actually pretty straightforward to factor it, because if you say this is just going to factor into u plus v times u minus v, well, u plus v is going to be six x squared plus v plus y plus three times u minus v. U is six x squared. Minus v is minus, we could write minus y plus three, or we could distribute the negative sign, but either way, this might make it a little bit clearer what we just did. We used the pattern to factor this higher degree polynomial, which is essentially just the difference of squares. Let's do one last example. So here, once again, we want to factor an expression. Which pattern can we use?"}, {"video_title": "Identifying quadratic patterns Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "We used the pattern to factor this higher degree polynomial, which is essentially just the difference of squares. Let's do one last example. So here, once again, we want to factor an expression. Which pattern can we use? Pause the video. All right, so we have two terms here, so it looks like it might be a difference of squares. If we set u is equal to seven, then this would be u u squared."}, {"video_title": "Identifying quadratic patterns Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "Which pattern can we use? Pause the video. All right, so we have two terms here, so it looks like it might be a difference of squares. If we set u is equal to seven, then this would be u u squared. But then what can we square to get 10x to the third power? Remember, we want to have integer exponents here. And the square root of 10x to the third power, if I were to take the square root of 10x to the third power, it'd be something a little bit involved, like the square root of 10 times x times the square root of x to the third power, and I'm not going to get an integer exponent here."}, {"video_title": "Checking ordered pair solutions to equations example 1 Algebra I Khan Academy.mp3", "Sentence": "Which of the ordered pairs is a solution of the following equation? Negative three x minus y is equal to six. What we have to remind ourselves is when we're given an ordered pair, the first number is the x coordinate and the second number is the y coordinate, or the y value. So when they tell us the ordered pair negative four comma four, they're saying, hey look, if x is equal to negative four and y is equal to positive four, does that satisfy this equation? And what we can do is we can just try that out. So we have negative three, and everywhere where we see an x, everywhere where we see an x, we can replace it with negative four. So it's negative three times negative four minus, minus, and everywhere we see, and everywhere we see a, and everywhere we see a y, we can replace it with positive four."}, {"video_title": "Checking ordered pair solutions to equations example 1 Algebra I Khan Academy.mp3", "Sentence": "So when they tell us the ordered pair negative four comma four, they're saying, hey look, if x is equal to negative four and y is equal to positive four, does that satisfy this equation? And what we can do is we can just try that out. So we have negative three, and everywhere where we see an x, everywhere where we see an x, we can replace it with negative four. So it's negative three times negative four minus, minus, and everywhere we see, and everywhere we see a, and everywhere we see a y, we can replace it with positive four. We replace it with positive four. So negative three times x minus y, which is four, needs to be equal to six. Needs to be equal to six."}, {"video_title": "Checking ordered pair solutions to equations example 1 Algebra I Khan Academy.mp3", "Sentence": "So it's negative three times negative four minus, minus, and everywhere we see, and everywhere we see a, and everywhere we see a y, we can replace it with positive four. We replace it with positive four. So negative three times x minus y, which is four, needs to be equal to six. Needs to be equal to six. Now is this indeed the case? Negative three times negative four is positive 12. Positive 12 minus four, positive 12 minus four is equal to eight."}, {"video_title": "Checking ordered pair solutions to equations example 1 Algebra I Khan Academy.mp3", "Sentence": "Needs to be equal to six. Now is this indeed the case? Negative three times negative four is positive 12. Positive 12 minus four, positive 12 minus four is equal to eight. It's not equal to six. Is not equal, is not equal to six. So this one does not work out."}, {"video_title": "Checking ordered pair solutions to equations example 1 Algebra I Khan Academy.mp3", "Sentence": "Positive 12 minus four, positive 12 minus four is equal to eight. It's not equal to six. Is not equal, is not equal to six. So this one does not work out. So let's see, negative three comma three. So we can do the same thing here. Let's see what happens when x is equal to negative three and y is equal to positive three."}, {"video_title": "Checking ordered pair solutions to equations example 1 Algebra I Khan Academy.mp3", "Sentence": "So this one does not work out. So let's see, negative three comma three. So we can do the same thing here. Let's see what happens when x is equal to negative three and y is equal to positive three. So we substitute back in. We get negative three, negative three times x, which now we're gonna try out x being equal to negative three minus y, minus y, y is positive three here, minus y, let me do that y color, that blue, minus y now needs to be equal to, now needs to be equal, just like before, it needs to be equal to six. So negative three times negative three, that's going to be positive nine."}, {"video_title": "Checking ordered pair solutions to equations example 1 Algebra I Khan Academy.mp3", "Sentence": "Let's see what happens when x is equal to negative three and y is equal to positive three. So we substitute back in. We get negative three, negative three times x, which now we're gonna try out x being equal to negative three minus y, minus y, y is positive three here, minus y, let me do that y color, that blue, minus y now needs to be equal to, now needs to be equal, just like before, it needs to be equal to six. So negative three times negative three, that's going to be positive nine. Nine minus three is indeed equal to six. Nine minus three is indeed equal to six. Nine minus three is six, that is equal to six."}, {"video_title": "Solving a quadratic equation by factoring Algebra II Khan Academy.mp3", "Sentence": "We're asked to solve for s. And we have s squared minus 2s minus 35 is equal to 0. Now, if this is the first time that you've seen this type of what's essentially a quadratic equation, you might be tempted to try to solve for s using traditional algebraic means, but the best way to solve this, especially when it's explicitly equal to 0, is to factor the left-hand side. And then think about the fact that those binomials that you factor into, that they have to be equal to 0. So let's just do that. So how can we factor this? We've seen it several ways. I'll show you kind of the standard way we've been doing it, by grouping."}, {"video_title": "Solving a quadratic equation by factoring Algebra II Khan Academy.mp3", "Sentence": "So let's just do that. So how can we factor this? We've seen it several ways. I'll show you kind of the standard way we've been doing it, by grouping. And then there's a little bit of a shortcut when you have a 1 as a coefficient over here. So when you do something by grouping, when you factor by grouping, you think about two numbers whose sum is going to be equal to negative 2. So you think about two numbers whose sum, a plus b, is equal to negative 2, and whose product is going to be equal to negative 35."}, {"video_title": "Solving a quadratic equation by factoring Algebra II Khan Academy.mp3", "Sentence": "I'll show you kind of the standard way we've been doing it, by grouping. And then there's a little bit of a shortcut when you have a 1 as a coefficient over here. So when you do something by grouping, when you factor by grouping, you think about two numbers whose sum is going to be equal to negative 2. So you think about two numbers whose sum, a plus b, is equal to negative 2, and whose product is going to be equal to negative 35. Since their product is a negative number, one has to be positive, one has to be negative. And so if you think about it, ones that are about two apart, you have 5 and negative 7. I think that'll work."}, {"video_title": "Solving a quadratic equation by factoring Algebra II Khan Academy.mp3", "Sentence": "So you think about two numbers whose sum, a plus b, is equal to negative 2, and whose product is going to be equal to negative 35. Since their product is a negative number, one has to be positive, one has to be negative. And so if you think about it, ones that are about two apart, you have 5 and negative 7. I think that'll work. 5 plus negative 7 is equal to negative 2. So to factor by grouping, you split this middle term into a let me write it this way, we have s squared, and then this middle term right here, I'll do it in pink, this middle term right there, I can write it as plus 5s minus 7s, and then we have the minus 35. And of course, all of that is equal to 0."}, {"video_title": "Solving a quadratic equation by factoring Algebra II Khan Academy.mp3", "Sentence": "I think that'll work. 5 plus negative 7 is equal to negative 2. So to factor by grouping, you split this middle term into a let me write it this way, we have s squared, and then this middle term right here, I'll do it in pink, this middle term right there, I can write it as plus 5s minus 7s, and then we have the minus 35. And of course, all of that is equal to 0. Now we call it factoring by grouping because we group it. So we can group these first two terms. And these first two terms, they have a common factor of s. So let's factor that out."}, {"video_title": "Solving a quadratic equation by factoring Algebra II Khan Academy.mp3", "Sentence": "And of course, all of that is equal to 0. Now we call it factoring by grouping because we group it. So we can group these first two terms. And these first two terms, they have a common factor of s. So let's factor that out. You have s times s plus 5. That's the same thing as s squared plus 5s. Now in these second two terms right here, you have a common factor of negative 7."}, {"video_title": "Solving a quadratic equation by factoring Algebra II Khan Academy.mp3", "Sentence": "And these first two terms, they have a common factor of s. So let's factor that out. You have s times s plus 5. That's the same thing as s squared plus 5s. Now in these second two terms right here, you have a common factor of negative 7. So let's factor that out. So you have negative 7 times s plus 5. And of course, all of that is equal to 0."}, {"video_title": "Solving a quadratic equation by factoring Algebra II Khan Academy.mp3", "Sentence": "Now in these second two terms right here, you have a common factor of negative 7. So let's factor that out. So you have negative 7 times s plus 5. And of course, all of that is equal to 0. Now we have two terms here where both of them have s plus 5 as a factor. Both of them have this s plus 5 as a factor. So we can factor that out."}, {"video_title": "Solving a quadratic equation by factoring Algebra II Khan Academy.mp3", "Sentence": "And of course, all of that is equal to 0. Now we have two terms here where both of them have s plus 5 as a factor. Both of them have this s plus 5 as a factor. So we can factor that out. So let's do that. So you have s plus 5 times this s times this s right here. s plus 5 times s would give you this term."}, {"video_title": "Solving a quadratic equation by factoring Algebra II Khan Academy.mp3", "Sentence": "So we can factor that out. So let's do that. So you have s plus 5 times this s times this s right here. s plus 5 times s would give you this term. And then you have minus that 7 right there. I undistributed the s plus 5. And then this is going to be equal to 0."}, {"video_title": "Solving a quadratic equation by factoring Algebra II Khan Academy.mp3", "Sentence": "s plus 5 times s would give you this term. And then you have minus that 7 right there. I undistributed the s plus 5. And then this is going to be equal to 0. Now that we've factored it, we just have to think a little bit about what happens when you take the product of two numbers. I mean, s plus 5 is a number. s minus 7 is another number."}, {"video_title": "Solving a quadratic equation by factoring Algebra II Khan Academy.mp3", "Sentence": "And then this is going to be equal to 0. Now that we've factored it, we just have to think a little bit about what happens when you take the product of two numbers. I mean, s plus 5 is a number. s minus 7 is another number. And we're saying that the product of those two numbers is equal to 0. If I were told you that I had two numbers, if I told you that I had the numbers a times b and that they equal to 0, what do we know about either a or b or both of them? Well, at least one of them has to be equal to 0, or both of them have to be equal to 0."}, {"video_title": "Solving a quadratic equation by factoring Algebra II Khan Academy.mp3", "Sentence": "s minus 7 is another number. And we're saying that the product of those two numbers is equal to 0. If I were told you that I had two numbers, if I told you that I had the numbers a times b and that they equal to 0, what do we know about either a or b or both of them? Well, at least one of them has to be equal to 0, or both of them have to be equal to 0. So the fact that this number times that number is equal to 0 tells us that either s plus 5 is equal to 0, or s minus 7 is equal to 0. And so you have these two equations. Actually, we could say and or."}, {"video_title": "Solving a quadratic equation by factoring Algebra II Khan Academy.mp3", "Sentence": "Well, at least one of them has to be equal to 0, or both of them have to be equal to 0. So the fact that this number times that number is equal to 0 tells us that either s plus 5 is equal to 0, or s minus 7 is equal to 0. And so you have these two equations. Actually, we could say and or. It could be or and either way. Both of them could be equal to 0. So let's see how we can solve for this."}, {"video_title": "Solving a quadratic equation by factoring Algebra II Khan Academy.mp3", "Sentence": "Actually, we could say and or. It could be or and either way. Both of them could be equal to 0. So let's see how we can solve for this. Well, we could just subtract 5 from both sides of this equation right there. And so you get on the left-hand side, you have s is equal to negative 5. That is one solution to the equation."}, {"video_title": "Solving a quadratic equation by factoring Algebra II Khan Academy.mp3", "Sentence": "So let's see how we can solve for this. Well, we could just subtract 5 from both sides of this equation right there. And so you get on the left-hand side, you have s is equal to negative 5. That is one solution to the equation. Or you have, let's see, you can add 7 to both sides of that equation. And you get s is equal to 7. So if s is equal to negative 5 or s is equal to 7, then we have satisfied this equation."}, {"video_title": "Solving a quadratic equation by factoring Algebra II Khan Academy.mp3", "Sentence": "That is one solution to the equation. Or you have, let's see, you can add 7 to both sides of that equation. And you get s is equal to 7. So if s is equal to negative 5 or s is equal to 7, then we have satisfied this equation. We can even verify it. If you make s equal to negative 5, you have positive 25 plus 10, which is 35, minus 35. That does equal 0."}, {"video_title": "Solving a quadratic equation by factoring Algebra II Khan Academy.mp3", "Sentence": "So if s is equal to negative 5 or s is equal to 7, then we have satisfied this equation. We can even verify it. If you make s equal to negative 5, you have positive 25 plus 10, which is 35, minus 35. That does equal 0. If you have 7, 49 minus 14 minus 35 does equal 0. So we've solved for s. Now I mentioned there's an easier way to do it. And when you have something like this, where you have 1 as the leading coefficient, you don't have to do this two-step factoring."}, {"video_title": "Solving a quadratic equation by factoring Algebra II Khan Academy.mp3", "Sentence": "That does equal 0. If you have 7, 49 minus 14 minus 35 does equal 0. So we've solved for s. Now I mentioned there's an easier way to do it. And when you have something like this, where you have 1 as the leading coefficient, you don't have to do this two-step factoring. Let me just show you an example. If I just have x plus a times x plus b, what does that equal to? x times x is x squared."}, {"video_title": "Solving a quadratic equation by factoring Algebra II Khan Academy.mp3", "Sentence": "And when you have something like this, where you have 1 as the leading coefficient, you don't have to do this two-step factoring. Let me just show you an example. If I just have x plus a times x plus b, what does that equal to? x times x is x squared. x times b is bx. a times x is plus ax. a times b is ab."}, {"video_title": "Solving a quadratic equation by factoring Algebra II Khan Academy.mp3", "Sentence": "x times x is x squared. x times b is bx. a times x is plus ax. a times b is ab. So you get x squared plus, these two can be added, plus a plus bx plus ab. And that's the pattern that we have right here. We have 1 as a leading coefficient here."}, {"video_title": "Solving a quadratic equation by factoring Algebra II Khan Academy.mp3", "Sentence": "a times b is ab. So you get x squared plus, these two can be added, plus a plus bx plus ab. And that's the pattern that we have right here. We have 1 as a leading coefficient here. We have 1 as a leading coefficient here. So once we have our two numbers that add up to negative 2, that's our a plus b. And we have our product that gets to negative 35."}, {"video_title": "Solving a quadratic equation by factoring Algebra II Khan Academy.mp3", "Sentence": "We have 1 as a leading coefficient here. We have 1 as a leading coefficient here. So once we have our two numbers that add up to negative 2, that's our a plus b. And we have our product that gets to negative 35. Then we can straight just factor it into the product of those two things. Or the product of the binomials where those will be the a's and the b's. So we figured out it's 5 and negative 7."}, {"video_title": "Solving a quadratic equation by factoring Algebra II Khan Academy.mp3", "Sentence": "And we have our product that gets to negative 35. Then we can straight just factor it into the product of those two things. Or the product of the binomials where those will be the a's and the b's. So we figured out it's 5 and negative 7. 5 plus negative 7 is negative 2. 5 times negative 7 is negative 35. So we could have just straight factored at this point to, well actually this was a case of s. So we could have factored straight to the case of s plus 5 times s minus 7."}, {"video_title": "Solving a quadratic equation by factoring Algebra II Khan Academy.mp3", "Sentence": "So we figured out it's 5 and negative 7. 5 plus negative 7 is negative 2. 5 times negative 7 is negative 35. So we could have just straight factored at this point to, well actually this was a case of s. So we could have factored straight to the case of s plus 5 times s minus 7. We could have done that straight away and we've gotten to that right there. And of course, that whole thing was equal to 0. So that would have been a little bit of a shortcut, but factoring by grouping, it's a completely appropriate way to do it as well."}, {"video_title": "Graphing a shifted and stretched absolute value function.mp3", "Sentence": "So let's do this through a series of transformations. So the next thing I wanna graph, let's see if we can graph y, y is equal to the absolute value of x plus three. Now in previous videos, we have talked about it. If you replace your x with an x plus three, this is going to shift your graph to the left by three. You could view this as the same thing as y is equal to the absolute value of x minus negative three. And whatever you are subtracting from this x, that is how much you are shifting it. So you're going to shift it three to the left."}, {"video_title": "Graphing a shifted and stretched absolute value function.mp3", "Sentence": "If you replace your x with an x plus three, this is going to shift your graph to the left by three. You could view this as the same thing as y is equal to the absolute value of x minus negative three. And whatever you are subtracting from this x, that is how much you are shifting it. So you're going to shift it three to the left. And we're gonna do that right now, and then we're just gonna confirm that it matches up, that it makes sense. So let's first graph that. I'll get my ruler tool here."}, {"video_title": "Graphing a shifted and stretched absolute value function.mp3", "Sentence": "So you're going to shift it three to the left. And we're gonna do that right now, and then we're just gonna confirm that it matches up, that it makes sense. So let's first graph that. I'll get my ruler tool here. So if we shift three to the left, it's gonna look something like, it's gonna look something like this. So on that, when whatever we have inside the absolute value sign is positive, we're going to get essentially this slope of one. And whenever we have inside the absolute value sign is negative, we're gonna have a slope of essentially negative one."}, {"video_title": "Graphing a shifted and stretched absolute value function.mp3", "Sentence": "I'll get my ruler tool here. So if we shift three to the left, it's gonna look something like, it's gonna look something like this. So on that, when whatever we have inside the absolute value sign is positive, we're going to get essentially this slope of one. And whenever we have inside the absolute value sign is negative, we're gonna have a slope of essentially negative one. So it's gonna look, it's going to look like that. And let's confirm whether this actually makes sense. Below x equals negative three, for x values less than negative three, what we're gonna have here is this inside of the absolute value sign is going to be negative, and so then we're gonna take its opposite value, and so that makes sense."}, {"video_title": "Graphing a shifted and stretched absolute value function.mp3", "Sentence": "And whenever we have inside the absolute value sign is negative, we're gonna have a slope of essentially negative one. So it's gonna look, it's going to look like that. And let's confirm whether this actually makes sense. Below x equals negative three, for x values less than negative three, what we're gonna have here is this inside of the absolute value sign is going to be negative, and so then we're gonna take its opposite value, and so that makes sense. That's why you have this downward line right over here. Now for x is greater than negative three, when you add three to it, you're going to get a positive value, and so that's why you have this upward sloping line right over here. And at x equals negative three, you have zero inside the absolute value sign, just as if you didn't shift it, you would have had zero inside the absolute value sign at x equals zero."}, {"video_title": "Graphing a shifted and stretched absolute value function.mp3", "Sentence": "Below x equals negative three, for x values less than negative three, what we're gonna have here is this inside of the absolute value sign is going to be negative, and so then we're gonna take its opposite value, and so that makes sense. That's why you have this downward line right over here. Now for x is greater than negative three, when you add three to it, you're going to get a positive value, and so that's why you have this upward sloping line right over here. And at x equals negative three, you have zero inside the absolute value sign, just as if you didn't shift it, you would have had zero inside the absolute value sign at x equals zero. So hopefully that makes a little bit more intuitive sense of why if you replace x, if you replace x with x plus three, and this isn't just true of absolute value functions, this is true of any function, if you replace x with x plus three, you're going to shift three to the left. All right, now let's keep building. Now let's see if we can graph y is equal to two times the absolute value of x plus three."}, {"video_title": "Graphing a shifted and stretched absolute value function.mp3", "Sentence": "And at x equals negative three, you have zero inside the absolute value sign, just as if you didn't shift it, you would have had zero inside the absolute value sign at x equals zero. So hopefully that makes a little bit more intuitive sense of why if you replace x, if you replace x with x plus three, and this isn't just true of absolute value functions, this is true of any function, if you replace x with x plus three, you're going to shift three to the left. All right, now let's keep building. Now let's see if we can graph y is equal to two times the absolute value of x plus three. So what this is essentially going to do is multiply the slopes by a factor of two. It's going to stretch it vertically by a factor of two. So for x values greater than negative three, instead of having a slope of one, you're gonna have a slope of two."}, {"video_title": "Graphing a shifted and stretched absolute value function.mp3", "Sentence": "Now let's see if we can graph y is equal to two times the absolute value of x plus three. So what this is essentially going to do is multiply the slopes by a factor of two. It's going to stretch it vertically by a factor of two. So for x values greater than negative three, instead of having a slope of one, you're gonna have a slope of two. So let me get my ruler out again and see if I can draw that. So let me put that there. And then, so here instead of a slope of one, I'm gonna have a slope of two."}, {"video_title": "Graphing a shifted and stretched absolute value function.mp3", "Sentence": "So for x values greater than negative three, instead of having a slope of one, you're gonna have a slope of two. So let me get my ruler out again and see if I can draw that. So let me put that there. And then, so here instead of a slope of one, I'm gonna have a slope of two. Let me draw that. It's gonna look like that right over there. And then instead of having a slope of negative one for values less than x equals negative three, I'm gonna have a slope of negative two."}, {"video_title": "Graphing a shifted and stretched absolute value function.mp3", "Sentence": "And then, so here instead of a slope of one, I'm gonna have a slope of two. Let me draw that. It's gonna look like that right over there. And then instead of having a slope of negative one for values less than x equals negative three, I'm gonna have a slope of negative two. Let me draw that right over there. So that is the graph of y is equal to two times the absolute value of x plus three. And now to get to the f of x that we care about, we now need to add this two."}, {"video_title": "Graphing a shifted and stretched absolute value function.mp3", "Sentence": "And then instead of having a slope of negative one for values less than x equals negative three, I'm gonna have a slope of negative two. Let me draw that right over there. So that is the graph of y is equal to two times the absolute value of x plus three. And now to get to the f of x that we care about, we now need to add this two. So now I wanna graph, and we're in the home stretch, I wanna graph y is equal to two times the absolute value of x plus three plus two. Well, whatever y value I was getting for this orange function, I now wanna add two to it. So this is just gonna shift it up vertically by two."}, {"video_title": "Graphing a shifted and stretched absolute value function.mp3", "Sentence": "And now to get to the f of x that we care about, we now need to add this two. So now I wanna graph, and we're in the home stretch, I wanna graph y is equal to two times the absolute value of x plus three plus two. Well, whatever y value I was getting for this orange function, I now wanna add two to it. So this is just gonna shift it up vertically by two. So instead of, so this is gonna be shifted up by two. This is going to be shifted, every point is going to be shifted up by two. Or you can think about shifting up the entire graph by two."}, {"video_title": "Graphing a shifted and stretched absolute value function.mp3", "Sentence": "So this is just gonna shift it up vertically by two. So instead of, so this is gonna be shifted up by two. This is going to be shifted, every point is going to be shifted up by two. Or you can think about shifting up the entire graph by two. Here in the orange function, whatever y value I got for the black function, I'm gonna have to get two more than that. And so it's going to look, it's going to look like this. So let me see, I'm shifting it up by two."}, {"video_title": "Graphing a shifted and stretched absolute value function.mp3", "Sentence": "Or you can think about shifting up the entire graph by two. Here in the orange function, whatever y value I got for the black function, I'm gonna have to get two more than that. And so it's going to look, it's going to look like this. So let me see, I'm shifting it up by two. So for x less than negative three, it'll look like that. And for x greater than negative three, it is going to look like, it is going to look like that. And there you have it."}, {"video_title": "Graphing a shifted and stretched absolute value function.mp3", "Sentence": "So let me see, I'm shifting it up by two. So for x less than negative three, it'll look like that. And for x greater than negative three, it is going to look like, it is going to look like that. And there you have it. This is the graph of y, or f of x, is equal to two times the absolute value of x plus three plus two. And you could have done it in different ways. You could have shifted up two first."}, {"video_title": "Graphing a shifted and stretched absolute value function.mp3", "Sentence": "And there you have it. This is the graph of y, or f of x, is equal to two times the absolute value of x plus three plus two. And you could have done it in different ways. You could have shifted up two first. Then you could have multiplied by a factor of two. And then you could have shifted, and then, so you could have moved up two first. Then you could have multiplied by a factor of two."}, {"video_title": "Graphing a shifted and stretched absolute value function.mp3", "Sentence": "You could have shifted up two first. Then you could have multiplied by a factor of two. And then you could have shifted, and then, so you could have moved up two first. Then you could have multiplied by a factor of two. Then you could have shifted left by three. Or you could have multiplied by a factor of two first, shifted up two, and then shifted over. So there's multiple, there's three transformations going up here."}, {"video_title": "Graphing a shifted and stretched absolute value function.mp3", "Sentence": "Then you could have multiplied by a factor of two. Then you could have shifted left by three. Or you could have multiplied by a factor of two first, shifted up two, and then shifted over. So there's multiple, there's three transformations going up here. This is an, this is a, let me color them all. So this right over here tells me, this over here says, hey, shift left. Shift left by three."}, {"video_title": "Graphing a shifted and stretched absolute value function.mp3", "Sentence": "So there's multiple, there's three transformations going up here. This is an, this is a, let me color them all. So this right over here tells me, this over here says, hey, shift left. Shift left by three. This told us stretch vertically by two, or essentially multiply the slope by two. Stretch vert by two. And then that last piece says shift up by two."}, {"video_title": "Interpretting exponential expression.mp3", "Sentence": "What does two represent in this expression? So pause this video and see if you can figure it out on your own. Alright, so let's look at the expression right over here. We could write it as defining a function. So we could say leaves as a function of time is equal to five times two to the t power. And so we could try this out a little bit. If we say, well what is L of zero?"}, {"video_title": "Interpretting exponential expression.mp3", "Sentence": "We could write it as defining a function. So we could say leaves as a function of time is equal to five times two to the t power. And so we could try this out a little bit. If we say, well what is L of zero? That would be t equals zero. That's when we're zero weeks after it was planted. So this is right when it was planted."}, {"video_title": "Interpretting exponential expression.mp3", "Sentence": "If we say, well what is L of zero? That would be t equals zero. That's when we're zero weeks after it was planted. So this is right when it was planted. Well that's five times two to the zero, which is just, two to the zero is just one, so it's equal to five. And so when you see an exponential expression or an exponential function like this, that is why this number out here is often referred to as your initial value. Initial value."}, {"video_title": "Interpretting exponential expression.mp3", "Sentence": "So this is right when it was planted. Well that's five times two to the zero, which is just, two to the zero is just one, so it's equal to five. And so when you see an exponential expression or an exponential function like this, that is why this number out here is often referred to as your initial value. Initial value. And so let's explore this a little bit more. What is L of one? What happens after one week?"}, {"video_title": "Interpretting exponential expression.mp3", "Sentence": "Initial value. And so let's explore this a little bit more. What is L of one? What happens after one week? Well that's gonna be five times two to the first power, or five times two. So going from when it was planted to the first week, we are multiplying by two, the number of leaves doubles. Well what happens after two weeks?"}, {"video_title": "Interpretting exponential expression.mp3", "Sentence": "What happens after one week? Well that's gonna be five times two to the first power, or five times two. So going from when it was planted to the first week, we are multiplying by two, the number of leaves doubles. Well what happens after two weeks? The number of leaves after two weeks? Well that's gonna be five times two to the second power. Well that's the number that you had in the first week times two."}, {"video_title": "Interpretting exponential expression.mp3", "Sentence": "Well what happens after two weeks? The number of leaves after two weeks? Well that's gonna be five times two to the second power. Well that's the number that you had in the first week times two. So it looks like every week we are doubling, we are multiplying by two. And that's why this number right over here, which is what the question is about, the two, this is often referred to as the common ratio. Common ratio."}, {"video_title": "Interpretting exponential expression.mp3", "Sentence": "Well that's the number that you had in the first week times two. So it looks like every week we are doubling, we are multiplying by two. And that's why this number right over here, which is what the question is about, the two, this is often referred to as the common ratio. Common ratio. Because between any two successive weeks, the ratio between say week two and week one is two. Week two is double week one. Week one is double week zero."}, {"video_title": "Interpretting exponential expression.mp3", "Sentence": "Common ratio. Because between any two successive weeks, the ratio between say week two and week one is two. Week two is double week one. Week one is double week zero. So let's see which of these choices actually match up to that. There were initially two leaves in the plant. Well we know that there weren't two leaves in the plant, our initial value was five."}, {"video_title": "Interpretting exponential expression.mp3", "Sentence": "Week one is double week zero. So let's see which of these choices actually match up to that. There were initially two leaves in the plant. Well we know that there weren't two leaves in the plant, our initial value was five. So let me cross that one out. The number of leaves is multiplied by two each week. Well that's exactly what we just described, so I like that choice."}, {"video_title": "Interpretting exponential expression.mp3", "Sentence": "Well we know that there weren't two leaves in the plant, our initial value was five. So let me cross that one out. The number of leaves is multiplied by two each week. Well that's exactly what we just described, so I like that choice. Let's look at the last one just for good measure. The plant was planted two weeks ago. Well no, they don't tell us anything about that."}, {"video_title": "Multiplying binomials area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "So I have this big rectangle here that's divided into four smaller rectangles. So what I want to do is I want to express the area of this larger rectangle, and I want to do it two ways. The first way I want to express it as the product of two binomials, and then I want to express it as a trinomial. So let's think about this a little bit. So one way to say, well look, the height of this larger rectangle from here to here, we see that that distance is x, and then from here to here it's two. So the entire height right over here, the entire height right over here is going to be x plus two. So the height is x plus two, and what's the width?"}, {"video_title": "Multiplying binomials area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "So let's think about this a little bit. So one way to say, well look, the height of this larger rectangle from here to here, we see that that distance is x, and then from here to here it's two. So the entire height right over here, the entire height right over here is going to be x plus two. So the height is x plus two, and what's the width? Well, the width is, we go from there to there is x, and then from there to there is three. So the entire width is x plus three, x plus three. So just like that, I've expressed the area of the entire rectangle, and it's the product of two binomials."}, {"video_title": "Multiplying binomials area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "So the height is x plus two, and what's the width? Well, the width is, we go from there to there is x, and then from there to there is three. So the entire width is x plus three, x plus three. So just like that, I've expressed the area of the entire rectangle, and it's the product of two binomials. But now let's express it as a trinomial. Well, to do that, we can break down the larger area into the areas of each of these smaller rectangles. So what's the area of this purple rectangle right over here?"}, {"video_title": "Multiplying binomials area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "So just like that, I've expressed the area of the entire rectangle, and it's the product of two binomials. But now let's express it as a trinomial. Well, to do that, we can break down the larger area into the areas of each of these smaller rectangles. So what's the area of this purple rectangle right over here? Well, the purple rectangle, its height is x, and its width is x, so its area is x squared. Let me write that, that's x squared. What's the area of this yellow rectangle?"}, {"video_title": "Multiplying binomials area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "So what's the area of this purple rectangle right over here? Well, the purple rectangle, its height is x, and its width is x, so its area is x squared. Let me write that, that's x squared. What's the area of this yellow rectangle? Well, its height is x, same height as right over here. Its height is x, and its width is three. So it's gonna be x times three, or three x."}, {"video_title": "Multiplying binomials area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "What's the area of this yellow rectangle? Well, its height is x, same height as right over here. Its height is x, and its width is three. So it's gonna be x times three, or three x. It'll have an area of three x. So that area is three x. If we're summing up the area of the entire thing, this would be plus three x."}, {"video_title": "Multiplying binomials area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "So it's gonna be x times three, or three x. It'll have an area of three x. So that area is three x. If we're summing up the area of the entire thing, this would be plus three x. So this expression right over here, that's the area of this purple region plus the area of this yellow region. And then we can move on to this green region. What's the area going to be here?"}, {"video_title": "Multiplying binomials area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "If we're summing up the area of the entire thing, this would be plus three x. So this expression right over here, that's the area of this purple region plus the area of this yellow region. And then we can move on to this green region. What's the area going to be here? Well, the height is two, and the width is x. So multiplying height times width is gonna be two times x. And we can just add that, plus two times x."}, {"video_title": "Multiplying binomials area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "What's the area going to be here? Well, the height is two, and the width is x. So multiplying height times width is gonna be two times x. And we can just add that, plus two times x. And then finally, this little gray box here, its height is two, we see that right over there. Its height is two, and its width is three. We see it right over there."}, {"video_title": "Multiplying binomials area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "And we can just add that, plus two times x. And then finally, this little gray box here, its height is two, we see that right over there. Its height is two, and its width is three. We see it right over there. So it has an area of six, two times three. So plus six, and you might say, well, this isn't a trinomial. This has four terms right over here."}, {"video_title": "Multiplying binomials area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "We see it right over there. So it has an area of six, two times three. So plus six, and you might say, well, this isn't a trinomial. This has four terms right over here. But you might notice that we can add, that we can add these middle two terms. Three x plus two x. I have three x's, and I add two x's to that, I'm gonna have five x's. So this entire thing simplifies to x squared, x squared, plus five x, plus six, plus six."}, {"video_title": "Multiplying binomials area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "This has four terms right over here. But you might notice that we can add, that we can add these middle two terms. Three x plus two x. I have three x's, and I add two x's to that, I'm gonna have five x's. So this entire thing simplifies to x squared, x squared, plus five x, plus six, plus six. So this and this are two ways of expressing the area, so they're going to be equal. And that makes sense, because if you multiply it out, these binomials, and simplify it, you would get this trinomial. We could do that really fast."}, {"video_title": "Multiplying binomials area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "So this entire thing simplifies to x squared, x squared, plus five x, plus six, plus six. So this and this are two ways of expressing the area, so they're going to be equal. And that makes sense, because if you multiply it out, these binomials, and simplify it, you would get this trinomial. We could do that really fast. You multiply the x times the x. Actually, let me do it in the same colors. You multiply, you multiply the x times the x, you get the x squared."}, {"video_title": "Multiplying binomials area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "We could do that really fast. You multiply the x times the x. Actually, let me do it in the same colors. You multiply, you multiply the x times the x, you get the x squared. You multiply this x times the three, you get your three x. You multiply, you multiply the two times the x, you get your two x. And then you multiply the two times the three, and you get your six."}, {"video_title": "Introduction to factoring higher degree polynomials Algebra 2 Khan Academy (2).mp3", "Sentence": "When we first learned algebra together, we started factoring polynomials, especially quadratics. We recognized that an expression like x squared could be written as x times x. We also recognized that a polynomial like three x squared plus four x, that in this situation, both terms had the common factor of x, and you could factor that out. And so you could rewrite this as x times three x plus four. And we also learned to do fancier things. We learned to factor things like x squared plus seven x plus 12. We were able to say, hey, what two numbers would add up to seven, and if I were to multiply them, I'd get 12, and in those early videos, we show why that works, and say, well, three and four, so maybe this can be factored as x plus three times x plus four."}, {"video_title": "Introduction to factoring higher degree polynomials Algebra 2 Khan Academy (2).mp3", "Sentence": "And so you could rewrite this as x times three x plus four. And we also learned to do fancier things. We learned to factor things like x squared plus seven x plus 12. We were able to say, hey, what two numbers would add up to seven, and if I were to multiply them, I'd get 12, and in those early videos, we show why that works, and say, well, three and four, so maybe this can be factored as x plus three times x plus four. If this is unfamiliar to you, I encourage you to go review that in some of the introductory factoring quadratics on Khan Academy. It should be a review at this point in your journey. We also looked at things like differences of squares, x squared minus nine."}, {"video_title": "Introduction to factoring higher degree polynomials Algebra 2 Khan Academy (2).mp3", "Sentence": "We were able to say, hey, what two numbers would add up to seven, and if I were to multiply them, I'd get 12, and in those early videos, we show why that works, and say, well, three and four, so maybe this can be factored as x plus three times x plus four. If this is unfamiliar to you, I encourage you to go review that in some of the introductory factoring quadratics on Khan Academy. It should be a review at this point in your journey. We also looked at things like differences of squares, x squared minus nine. Say, hey, that's x squared minus three squared, so we could factor that as x plus three times x minus three, and we looked at other types of quadratics. Now, as we go deeper into our algebra journeys, we're going to build on this to factor higher degree polynomials, third degree, fourth degree, fifth degree, which will be very useful in your mathematical careers, but we're going to start doing it by really looking at some of the structure, some of the patterns that we've seen in introductory algebra. For example, let's say someone walks up to you on the street and says, can you factor x to the third plus seven x squared plus 12x?"}, {"video_title": "Introduction to factoring higher degree polynomials Algebra 2 Khan Academy (2).mp3", "Sentence": "We also looked at things like differences of squares, x squared minus nine. Say, hey, that's x squared minus three squared, so we could factor that as x plus three times x minus three, and we looked at other types of quadratics. Now, as we go deeper into our algebra journeys, we're going to build on this to factor higher degree polynomials, third degree, fourth degree, fifth degree, which will be very useful in your mathematical careers, but we're going to start doing it by really looking at some of the structure, some of the patterns that we've seen in introductory algebra. For example, let's say someone walks up to you on the street and says, can you factor x to the third plus seven x squared plus 12x? Well, at first you might say, oh, this is a third degree polynomial, that seems kind of intimidating, until you realize, hey, all of these terms have the common factor x, so if I factor that out, then it becomes x times x squared plus seven x plus 12, and then this is exactly what we saw over here, so we could rewrite all of this as x times x plus three times x plus four, so we're going to see that we might be able to do some simple factoring like this and even factoring multiple times. We might also start to appreciate structure that brings us back to some of what we saw in our introductory algebra, so for example, you might see something like this, where once again, someone walks up to you on the street and says, hey, you factor this, a to the fourth power plus seven a squared plus 12, and at first, you're like, wow, there's a fourth power here, what do I do, until you say, well, what if I were to rewrite this as a squared squared plus seven a squared plus 12, and now this a squared is looking an awful lot like this x over here. If this were an x, then this would be x squared."}, {"video_title": "Introduction to factoring higher degree polynomials Algebra 2 Khan Academy (2).mp3", "Sentence": "For example, let's say someone walks up to you on the street and says, can you factor x to the third plus seven x squared plus 12x? Well, at first you might say, oh, this is a third degree polynomial, that seems kind of intimidating, until you realize, hey, all of these terms have the common factor x, so if I factor that out, then it becomes x times x squared plus seven x plus 12, and then this is exactly what we saw over here, so we could rewrite all of this as x times x plus three times x plus four, so we're going to see that we might be able to do some simple factoring like this and even factoring multiple times. We might also start to appreciate structure that brings us back to some of what we saw in our introductory algebra, so for example, you might see something like this, where once again, someone walks up to you on the street and says, hey, you factor this, a to the fourth power plus seven a squared plus 12, and at first, you're like, wow, there's a fourth power here, what do I do, until you say, well, what if I were to rewrite this as a squared squared plus seven a squared plus 12, and now this a squared is looking an awful lot like this x over here. If this were an x, then this would be x squared. If this were an x, then this would just be an x, and then these expressions would be the same, so when I factor it, everywhere I see an x, I could replace with an a squared, so I could factor this out, really looking at the same structure we have here as a squared plus three times a squared plus four. Now, I'm going really fast through this. This is really the introductory video, the overview video."}, {"video_title": "Introduction to factoring higher degree polynomials Algebra 2 Khan Academy (2).mp3", "Sentence": "If this were an x, then this would be x squared. If this were an x, then this would just be an x, and then these expressions would be the same, so when I factor it, everywhere I see an x, I could replace with an a squared, so I could factor this out, really looking at the same structure we have here as a squared plus three times a squared plus four. Now, I'm going really fast through this. This is really the introductory video, the overview video. Don't worry if this is a little bit much too fast. This is really just to give you a sense of things. Later in this unit, we're going to dig deeper into each of these cases, but just to give you a sense of where we're going, I'll give you another example that builds off of what you likely saw in your introductory algebra learning, so building off of the structure here, if someone were to walk up to you, again, a lot of people are walking up to you, and say factor four x to the sixth minus nine y to the fourth Well, at first, this looks quite intimidating until you realize that, hey, I could write both of these as squares."}, {"video_title": "Introduction to factoring higher degree polynomials Algebra 2 Khan Academy (2).mp3", "Sentence": "This is really the introductory video, the overview video. Don't worry if this is a little bit much too fast. This is really just to give you a sense of things. Later in this unit, we're going to dig deeper into each of these cases, but just to give you a sense of where we're going, I'll give you another example that builds off of what you likely saw in your introductory algebra learning, so building off of the structure here, if someone were to walk up to you, again, a lot of people are walking up to you, and say factor four x to the sixth minus nine y to the fourth Well, at first, this looks quite intimidating until you realize that, hey, I could write both of these as squares. I could write this first one as two x to the third squared minus, and I could write this second term as three y squared squared, and now, this is just a difference of squares, so it'd be two x to the third plus three y squared times two x to the third minus three y squared. We'll also see things like this where we're going to be factoring multiple times, so once again, someone walks up to you on the street, and they say you're a very popular person. Someone walks up to you and says, up to you on the street and says, factor x to the fourth minus y to the fourth."}, {"video_title": "Introduction to factoring higher degree polynomials Algebra 2 Khan Academy (2).mp3", "Sentence": "Later in this unit, we're going to dig deeper into each of these cases, but just to give you a sense of where we're going, I'll give you another example that builds off of what you likely saw in your introductory algebra learning, so building off of the structure here, if someone were to walk up to you, again, a lot of people are walking up to you, and say factor four x to the sixth minus nine y to the fourth Well, at first, this looks quite intimidating until you realize that, hey, I could write both of these as squares. I could write this first one as two x to the third squared minus, and I could write this second term as three y squared squared, and now, this is just a difference of squares, so it'd be two x to the third plus three y squared times two x to the third minus three y squared. We'll also see things like this where we're going to be factoring multiple times, so once again, someone walks up to you on the street, and they say you're a very popular person. Someone walks up to you and says, up to you on the street and says, factor x to the fourth minus y to the fourth. Well, based on what we just saw, you could realize that this is the same thing as x squared squared minus y squared squared, and you say, okay, this is a difference of squares just like this was a difference of squares, so it's going to be the sum of x squared and y squared, x squared plus y squared, times the difference of them, x squared minus y squared. Now, this is fun because this is two a difference of squares, so we can rewrite this whole thing as I'll rewrite this first part, x squared, x squared plus y squared, and then we could factor this as a difference of squares just as we factored this up here, and we get x plus y times x minus y, so I'll leave you there. I've just bombarded you with a bunch of information, but this is really just to get you warmed up."}, {"video_title": "Plotting points of logarithmic function Logarithms Algebra II Khan Academy.mp3", "Sentence": "The three points plotted below are on the graph of y is equal to b to the x power. Based only on these three points, plot the three corresponding points that must be on the graph of y is equal to log base b of x by clicking on the graph. So I've actually copied and pasted this problem on my little scratch pad so I can mark it up a little bit. So what is this first function? This first function is telling us x and this is y is equal to b to the x power. So when x is equal to zero, y is equal to one. When x is equal to zero, y is equal to one."}, {"video_title": "Plotting points of logarithmic function Logarithms Algebra II Khan Academy.mp3", "Sentence": "So what is this first function? This first function is telling us x and this is y is equal to b to the x power. So when x is equal to zero, y is equal to one. When x is equal to zero, y is equal to one. That's this point right over here. When x is equal to one, b to the first power is equal to four. y is equal to four."}, {"video_title": "Plotting points of logarithmic function Logarithms Algebra II Khan Academy.mp3", "Sentence": "When x is equal to zero, y is equal to one. That's this point right over here. When x is equal to one, b to the first power is equal to four. y is equal to four. So another way of thinking of this, y or four is equal to b to the first power. And actually we can deduce then that b must be four. So that's this point right over there."}, {"video_title": "Plotting points of logarithmic function Logarithms Algebra II Khan Academy.mp3", "Sentence": "y is equal to four. So another way of thinking of this, y or four is equal to b to the first power. And actually we can deduce then that b must be four. So that's this point right over there. And then this point is telling us that b to the second power is equal to sixteen. So when x is equal to two, b to the second power, y is equal to sixteen. Now we want to plot the three corresponding points on this function."}, {"video_title": "Plotting points of logarithmic function Logarithms Algebra II Khan Academy.mp3", "Sentence": "So that's this point right over there. And then this point is telling us that b to the second power is equal to sixteen. So when x is equal to two, b to the second power, y is equal to sixteen. Now we want to plot the three corresponding points on this function. So let me draw another table here. So now it's essentially the inverse function where this is going to be x and we want to calculate y is equal to log base b of x. And so what are the possibilities here?"}, {"video_title": "Plotting points of logarithmic function Logarithms Algebra II Khan Academy.mp3", "Sentence": "Now we want to plot the three corresponding points on this function. So let me draw another table here. So now it's essentially the inverse function where this is going to be x and we want to calculate y is equal to log base b of x. And so what are the possibilities here? So what I want to do is think, let's take these values because these are the ones, these are essentially inverse functions. Log is the inverse of exponents. So if we take the points one, four, and sixteen."}, {"video_title": "Plotting points of logarithmic function Logarithms Algebra II Khan Academy.mp3", "Sentence": "And so what are the possibilities here? So what I want to do is think, let's take these values because these are the ones, these are essentially inverse functions. Log is the inverse of exponents. So if we take the points one, four, and sixteen. So what is y going to be here? y is going to be log base b of one. So this is saying what power I need to raise b to to get to one."}, {"video_title": "Plotting points of logarithmic function Logarithms Algebra II Khan Academy.mp3", "Sentence": "So if we take the points one, four, and sixteen. So what is y going to be here? y is going to be log base b of one. So this is saying what power I need to raise b to to get to one. Well, if we assume that b is non-zero, and that's a reasonable assumption because b to different powers are non-zero, this is going to be zero for any non-zero b. So this is going to be zero right over here. So we have the point one, zero."}, {"video_title": "Plotting points of logarithmic function Logarithms Algebra II Khan Academy.mp3", "Sentence": "So this is saying what power I need to raise b to to get to one. Well, if we assume that b is non-zero, and that's a reasonable assumption because b to different powers are non-zero, this is going to be zero for any non-zero b. So this is going to be zero right over here. So we have the point one, zero. So it's that point over there. And notice, this point corresponds to this point. We've essentially swapped the x's and y's."}, {"video_title": "Plotting points of logarithmic function Logarithms Algebra II Khan Academy.mp3", "Sentence": "So we have the point one, zero. So it's that point over there. And notice, this point corresponds to this point. We've essentially swapped the x's and y's. And in general, when you're taking an inverse, you're going to reflect over the line y is equal to x. And this is clearly a reflection over that line. So let's look over here."}, {"video_title": "Plotting points of logarithmic function Logarithms Algebra II Khan Academy.mp3", "Sentence": "We've essentially swapped the x's and y's. And in general, when you're taking an inverse, you're going to reflect over the line y is equal to x. And this is clearly a reflection over that line. So let's look over here. When x is equal to four, what is log base b of four? What is the power I need to raise b to to get to four? Well, we see right over here, b to the first power is equal to four."}, {"video_title": "Plotting points of logarithmic function Logarithms Algebra II Khan Academy.mp3", "Sentence": "So let's look over here. When x is equal to four, what is log base b of four? What is the power I need to raise b to to get to four? Well, we see right over here, b to the first power is equal to four. We already figured that out. When I take b to the first power is equal to four. So this right over here is going to be equal to one."}, {"video_title": "Plotting points of logarithmic function Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, we see right over here, b to the first power is equal to four. We already figured that out. When I take b to the first power is equal to four. So this right over here is going to be equal to one. So when x is equal to four, y is equal to one. And notice, once again, it is a reflection over the line y is equal to x. So when x is equal to 16, then y is equal to log base b of 16."}, {"video_title": "Plotting points of logarithmic function Logarithms Algebra II Khan Academy.mp3", "Sentence": "So this right over here is going to be equal to one. So when x is equal to four, y is equal to one. And notice, once again, it is a reflection over the line y is equal to x. So when x is equal to 16, then y is equal to log base b of 16. The power I need to raise b to to get to 16. Well, we already know. If we take b squared, we get to 16."}, {"video_title": "Plotting points of logarithmic function Logarithms Algebra II Khan Academy.mp3", "Sentence": "So when x is equal to 16, then y is equal to log base b of 16. The power I need to raise b to to get to 16. Well, we already know. If we take b squared, we get to 16. So this is equal to two. So when x is equal to 16, y is equal to two. Notice, we've essentially just swapped the x and y values for each of these points."}, {"video_title": "Plotting points of logarithmic function Logarithms Algebra II Khan Academy.mp3", "Sentence": "If we take b squared, we get to 16. So this is equal to two. So when x is equal to 16, y is equal to two. Notice, we've essentially just swapped the x and y values for each of these points. This is y, this is a reflection over the line y is equal to x. Now let's actually do that on the actual interface. And the whole reason is to give you this appreciation that these are inverse functions of each other."}, {"video_title": "Plotting points of logarithmic function Logarithms Algebra II Khan Academy.mp3", "Sentence": "Notice, we've essentially just swapped the x and y values for each of these points. This is y, this is a reflection over the line y is equal to x. Now let's actually do that on the actual interface. And the whole reason is to give you this appreciation that these are inverse functions of each other. So let's plot the points. So that point corresponding to that point. So x zero, y one corresponds to x one, y zero."}, {"video_title": "Plotting points of logarithmic function Logarithms Algebra II Khan Academy.mp3", "Sentence": "And the whole reason is to give you this appreciation that these are inverse functions of each other. So let's plot the points. So that point corresponding to that point. So x zero, y one corresponds to x one, y zero. Here x is one, y is four. That corresponds to x four, y one. Here x is two, y is 16."}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "Use the change of base formula to find log base 5 of 100 to the nearest thousandth. So the change of base formula is a useful formula, especially when you're going to use a calculator, because most calculators don't allow you to arbitrarily change the base of your logarithm. They have functions for log base e, which is a natural logarithm, and log base 10. So you generally have to change your base, and that's what the change of base formula is. And if we have time, I'll tell you why it makes a lot of sense or how we can derive it. So the change of base formula just tells us that log... Let me do some colors here. Log base a of b is the same thing, is the exact same thing as log base x, where x is an arbitrary base, of b over log base... that same base, base x, base x over a."}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "So you generally have to change your base, and that's what the change of base formula is. And if we have time, I'll tell you why it makes a lot of sense or how we can derive it. So the change of base formula just tells us that log... Let me do some colors here. Log base a of b is the same thing, is the exact same thing as log base x, where x is an arbitrary base, of b over log base... that same base, base x, base x over a. And the reason why this is useful is that we can change our base. Here our base is a, and we can change it to base x. So if our calculator has a certain base x function, we can convert to that base."}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "Log base a of b is the same thing, is the exact same thing as log base x, where x is an arbitrary base, of b over log base... that same base, base x, base x over a. And the reason why this is useful is that we can change our base. Here our base is a, and we can change it to base x. So if our calculator has a certain base x function, we can convert to that base. It's usually e or base 10. Base 10 is an easy way to go. And in general, if you just see someone write a logarithm like this, if they just write log of x, they're implying this implies log base 10 of x."}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "So if our calculator has a certain base x function, we can convert to that base. It's usually e or base 10. Base 10 is an easy way to go. And in general, if you just see someone write a logarithm like this, if they just write log of x, they're implying this implies log base 10 of x. If someone writes natural log of x, they are implying log base e of x. And e is obviously the number 2.71 that keeps going on and on and on forever. Now let's apply it to this problem."}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "And in general, if you just see someone write a logarithm like this, if they just write log of x, they're implying this implies log base 10 of x. If someone writes natural log of x, they are implying log base e of x. And e is obviously the number 2.71 that keeps going on and on and on forever. Now let's apply it to this problem. We have...we need to figure out the logarithm, and I'll use the colors, base 5 of 100. So this property, this change of base formula, tells us that this is the exact same thing as log, I'll make x 10, log base 10 of 100 divided by log base 10 of 5. And actually, we don't even need a calculator to evaluate this top part."}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "Now let's apply it to this problem. We have...we need to figure out the logarithm, and I'll use the colors, base 5 of 100. So this property, this change of base formula, tells us that this is the exact same thing as log, I'll make x 10, log base 10 of 100 divided by log base 10 of 5. And actually, we don't even need a calculator to evaluate this top part. Log base 10 of 100, what power do I have to raise 10 to to get to 100? The second power. So this numerator is just equal to 2, so it simplifies to 2 over log base 10 of 5."}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "And actually, we don't even need a calculator to evaluate this top part. Log base 10 of 100, what power do I have to raise 10 to to get to 100? The second power. So this numerator is just equal to 2, so it simplifies to 2 over log base 10 of 5. And we can now use our calculator because the log function on a calculator is log base 10. So let's get our calculator out. And we're going to get... we want, let me clear this, 2 divided by... when someone just writes log, they mean base 10."}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "So this numerator is just equal to 2, so it simplifies to 2 over log base 10 of 5. And we can now use our calculator because the log function on a calculator is log base 10. So let's get our calculator out. And we're going to get... we want, let me clear this, 2 divided by... when someone just writes log, they mean base 10. If they press ln, that means base e. So log, without any other information, is log base 10. So this is log base 10 of 5 is equal to 2 point... and they want it to the nearest thousandth. So 2.861."}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "And we're going to get... we want, let me clear this, 2 divided by... when someone just writes log, they mean base 10. If they press ln, that means base e. So log, without any other information, is log base 10. So this is log base 10 of 5 is equal to 2 point... and they want it to the nearest thousandth. So 2.861. So this is approximately equal to 2.861. And we can verify it, because in theory, if I raise 5 to this power, I should get 100. And it kind of makes sense, because 5 to the second power is 25, 5 to the third power is 125, and this is in between the two, and it's closer to the third power than it is to the second power, and this number is closer to 3 than it is to 2."}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "So 2.861. So this is approximately equal to 2.861. And we can verify it, because in theory, if I raise 5 to this power, I should get 100. And it kind of makes sense, because 5 to the second power is 25, 5 to the third power is 125, and this is in between the two, and it's closer to the third power than it is to the second power, and this number is closer to 3 than it is to 2. But let's verify it. So if I take 5 to that power, if I take 5 and let me type in... well, let me just type in what we did to the nearest thousandth. 5 to the 2.861, so I'm not putting in all of the digits, what do I get?"}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "And it kind of makes sense, because 5 to the second power is 25, 5 to the third power is 125, and this is in between the two, and it's closer to the third power than it is to the second power, and this number is closer to 3 than it is to 2. But let's verify it. So if I take 5 to that power, if I take 5 and let me type in... well, let me just type in what we did to the nearest thousandth. 5 to the 2.861, so I'm not putting in all of the digits, what do I get? I get 99.94. If I put all of these digits in, it should get pretty close to 100. So that's what makes you feel good, that this is the power that I have to raise 5 to to get to 100."}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "5 to the 2.861, so I'm not putting in all of the digits, what do I get? I get 99.94. If I put all of these digits in, it should get pretty close to 100. So that's what makes you feel good, that this is the power that I have to raise 5 to to get to 100. Now, with that out of the way, let's actually think about why this property, why this thing right over here makes sense. So if I write log base A, I'll try to be fair to the colors, log base A of B, let's say I set that to be equal to... let's say I set that equal to some number. Let's call that equal to C, or I could call it E for example."}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "So that's what makes you feel good, that this is the power that I have to raise 5 to to get to 100. Now, with that out of the way, let's actually think about why this property, why this thing right over here makes sense. So if I write log base A, I'll try to be fair to the colors, log base A of B, let's say I set that to be equal to... let's say I set that equal to some number. Let's call that equal to C, or I could call it E for example. Well, I'll say that's equal to C. That means that A to the Cth power is equal to B. This is an exponential way of writing this truth, this is a logarithmic way of writing this truth. Is equal to B."}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "Let's call that equal to C, or I could call it E for example. Well, I'll say that's equal to C. That means that A to the Cth power is equal to B. This is an exponential way of writing this truth, this is a logarithmic way of writing this truth. Is equal to B. Now, we can take the logarithm of any base of both sides of this. Anything you do, if you say 10 to the what power equals this, 10 to the same power will be equal to this because these two things are equal to each other. So let's take the same logarithm of both sides of this, the logarithm with the same base."}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "Is equal to B. Now, we can take the logarithm of any base of both sides of this. Anything you do, if you say 10 to the what power equals this, 10 to the same power will be equal to this because these two things are equal to each other. So let's take the same logarithm of both sides of this, the logarithm with the same base. I'll actually do log base X to prove the general case here. So I'm going to take log base X of both sides of this. So this is log base X of A to the Cth power."}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "So let's take the same logarithm of both sides of this, the logarithm with the same base. I'll actually do log base X to prove the general case here. So I'm going to take log base X of both sides of this. So this is log base X of A to the Cth power. I try to be faithful to the colors. Is equal to log base X of B. Is equal to B."}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "So this is log base X of A to the Cth power. I try to be faithful to the colors. Is equal to log base X of B. Is equal to B. Let me close it off in orange as well. And we know from our logarithm properties, log of A to the C is the same thing as C times the logarithm of whatever base we are of A. And of course, this is going to be equal to log base X of B."}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "Is equal to B. Let me close it off in orange as well. And we know from our logarithm properties, log of A to the C is the same thing as C times the logarithm of whatever base we are of A. And of course, this is going to be equal to log base X of B. And if we wanted to solve for C, you just divide both sides by log base X of A. So you get C is equal to, and I'll stick to the color, it says log base X of B, which is this, over log base X of A. And this is what C was."}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "And of course, this is going to be equal to log base X of B. And if we wanted to solve for C, you just divide both sides by log base X of A. So you get C is equal to, and I'll stick to the color, it says log base X of B, which is this, over log base X of A. And this is what C was. C was log base A of B. Is equal to log base A of B. Let me write it this way."}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "And this is what C was. C was log base A of B. Is equal to log base A of B. Let me write it this way. Let me do the original color codes just so it becomes very clear what I'm doing. I think you know where this is going. But I want to be fair to the color."}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "Let me write it this way. Let me do the original color codes just so it becomes very clear what I'm doing. I think you know where this is going. But I want to be fair to the color. So C is equal to log base X of B over, let me scroll down a little bit, log base X, dividing both sides by that, of A. And we know from here, I can just copy and paste it, this is also equal to C. This is how we defined it. So let me copy it and then let me paste it."}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "But I want to be fair to the color. So C is equal to log base X of B over, let me scroll down a little bit, log base X, dividing both sides by that, of A. And we know from here, I can just copy and paste it, this is also equal to C. This is how we defined it. So let me copy it and then let me paste it. So this is also equal to C. And we're done. We've proven the change of base formula. Log base A of B is equal to log base X of B divided by log base X of A."}, {"video_title": "Worked examples slope-intercept intro Mathematics I High School Math Khan Academy.mp3", "Sentence": "Let's do some practice examples from our intro to slope-y-intercept exercise. What is the slope of y is equal to negative four x minus three? So you might already recognize this is in slope-intercept form. Just as a reminder, slope-intercept form is y is equal to mx plus b, where the coefficient on this x term right over here, that is our slope, that is our slope, and then this constant right over here, that is going to give you your y-intercept. So if they're saying what is the slope here, well I just need to figure out what is the coefficient on this x term. And you can see that the coefficient here is a negative four. So that is going to be our m, that is going to be our slope."}, {"video_title": "Worked examples slope-intercept intro Mathematics I High School Math Khan Academy.mp3", "Sentence": "Just as a reminder, slope-intercept form is y is equal to mx plus b, where the coefficient on this x term right over here, that is our slope, that is our slope, and then this constant right over here, that is going to give you your y-intercept. So if they're saying what is the slope here, well I just need to figure out what is the coefficient on this x term. And you can see that the coefficient here is a negative four. So that is going to be our m, that is going to be our slope. Now just as a reminder, you have to make sure that it's solved in this way, that it is solved for y. It is equal to something times x minus three. So that's our slope."}, {"video_title": "Worked examples slope-intercept intro Mathematics I High School Math Khan Academy.mp3", "Sentence": "So that is going to be our m, that is going to be our slope. Now just as a reminder, you have to make sure that it's solved in this way, that it is solved for y. It is equal to something times x minus three. So that's our slope. Let's do another one of these. So we're asked, what is the y-intercept of y is equal to negative three x minus two? So once again, we already have it in slope-intercept form."}, {"video_title": "Worked examples slope-intercept intro Mathematics I High School Math Khan Academy.mp3", "Sentence": "So that's our slope. Let's do another one of these. So we're asked, what is the y-intercept of y is equal to negative three x minus two? So once again, we already have it in slope-intercept form. It's already been solved for y. It's of the form y is equal to mx plus b, where m, our slope, is given right over here, negative three, but they're not asking for our slope. They're asking for the y-intercept."}, {"video_title": "Worked examples slope-intercept intro Mathematics I High School Math Khan Academy.mp3", "Sentence": "So once again, we already have it in slope-intercept form. It's already been solved for y. It's of the form y is equal to mx plus b, where m, our slope, is given right over here, negative three, but they're not asking for our slope. They're asking for the y-intercept. Well the y-intercept is given by b here, so b is negative two. Pay close attention to the sign here. So b is equal to negative two."}, {"video_title": "Worked examples slope-intercept intro Mathematics I High School Math Khan Academy.mp3", "Sentence": "They're asking for the y-intercept. Well the y-intercept is given by b here, so b is negative two. Pay close attention to the sign here. So b is equal to negative two. But when I look at these choices, I don't see a b is equal to negative two, so what are they talking about? Well a y-intercept is, what is the y value when x is equal to zero? And you can see that here."}, {"video_title": "Worked examples slope-intercept intro Mathematics I High School Math Khan Academy.mp3", "Sentence": "So b is equal to negative two. But when I look at these choices, I don't see a b is equal to negative two, so what are they talking about? Well a y-intercept is, what is the y value when x is equal to zero? And you can see that here. If x was equal to zero, then that term goes away and y is equal to b. So if you want to know the point where the graph described by this equation intercepts the y-axis, well it's going to be, what is y when x is equal to zero? Well when x is equal to zero, y is equal to negative two."}, {"video_title": "Worked examples slope-intercept intro Mathematics I High School Math Khan Academy.mp3", "Sentence": "And you can see that here. If x was equal to zero, then that term goes away and y is equal to b. So if you want to know the point where the graph described by this equation intercepts the y-axis, well it's going to be, what is y when x is equal to zero? Well when x is equal to zero, y is equal to negative two. And you can see that in our original equation again. If x was zero, this term would go away and y would be equal to negative two. So zero comma negative two."}, {"video_title": "Worked examples slope-intercept intro Mathematics I High School Math Khan Academy.mp3", "Sentence": "Well when x is equal to zero, y is equal to negative two. And you can see that in our original equation again. If x was zero, this term would go away and y would be equal to negative two. So zero comma negative two. So it would be that choice right over there. On Khan Academy, obviously you just have to click on that. You don't have to shade it in."}, {"video_title": "Worked examples slope-intercept intro Mathematics I High School Math Khan Academy.mp3", "Sentence": "So zero comma negative two. So it would be that choice right over there. On Khan Academy, obviously you just have to click on that. You don't have to shade it in. Let's do one more. Complete the equation of the line whose slope is five and y-intercept is zero comma four. So once again, the general form is y is equal to our slope times x, if I want to put it in slope-intercept form, plus our y-intercept."}, {"video_title": "Worked examples slope-intercept intro Mathematics I High School Math Khan Academy.mp3", "Sentence": "You don't have to shade it in. Let's do one more. Complete the equation of the line whose slope is five and y-intercept is zero comma four. So once again, the general form is y is equal to our slope times x, if I want to put it in slope-intercept form, plus our y-intercept. Well they're telling us our slope is five, whose slope is five. So we know that m is going to be five and they tell us that the y-intercept is zero four. So the y-intercept b, that is the value of y when x is equal to zero."}, {"video_title": "Worked examples slope-intercept intro Mathematics I High School Math Khan Academy.mp3", "Sentence": "So once again, the general form is y is equal to our slope times x, if I want to put it in slope-intercept form, plus our y-intercept. Well they're telling us our slope is five, whose slope is five. So we know that m is going to be five and they tell us that the y-intercept is zero four. So the y-intercept b, that is the value of y when x is equal to zero. So the value of y when x equals zero is this four right over here. So that is going to be four. So I could say y is equal to five times x plus four."}, {"video_title": "Worked examples slope-intercept intro Mathematics I High School Math Khan Academy.mp3", "Sentence": "So the y-intercept b, that is the value of y when x is equal to zero. So the value of y when x equals zero is this four right over here. So that is going to be four. So I could say y is equal to five times x plus four. And when you're actually entering it on Khan Academy, you would just type it in or if you're using the app, you would use it with your finger. And I always make the mistake of writing y equals and I type in y equals five x plus four. They already gave you the y equals right over there."}, {"video_title": "Basic fractional exponents Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "We already know a good bit about exponents. For example, we know if we took the number four and raised it to the third power, this is equivalent to taking three fours and multiplying them. Or you could also view it as starting with a one and then multiplying the one by four or multiplying that by four three times. But either way, this is going to result in four times four is 16, times four is 64. We also know a little bit about negative exponents. So for example, if I were to take four to the negative three power, we know this negative tells us to take the reciprocal, one over four to the third. And we already know four to the third is 64, so this is going to be 1 64th."}, {"video_title": "Basic fractional exponents Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "But either way, this is going to result in four times four is 16, times four is 64. We also know a little bit about negative exponents. So for example, if I were to take four to the negative three power, we know this negative tells us to take the reciprocal, one over four to the third. And we already know four to the third is 64, so this is going to be 1 64th. Now let's think about fractional exponents. So we're gonna think about what is four to the 1 1\u20442 power. I encourage you to pause the video and at least take a guess about what you think this is."}, {"video_title": "Basic fractional exponents Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "And we already know four to the third is 64, so this is going to be 1 64th. Now let's think about fractional exponents. So we're gonna think about what is four to the 1 1\u20442 power. I encourage you to pause the video and at least take a guess about what you think this is. So the mathematical convention here, the mathematical definition that most people use or that frankly all people use here, is that four to the 1 1\u20442 power is the exact same thing as the square root of four. And we'll talk in the future about why this is, and the reason why this is defined this way is it has all sorts of neat and elegant properties when you start manipulating the actual exponents. But what is the square root of four, especially the principal root, mean?"}, {"video_title": "Basic fractional exponents Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "I encourage you to pause the video and at least take a guess about what you think this is. So the mathematical convention here, the mathematical definition that most people use or that frankly all people use here, is that four to the 1 1\u20442 power is the exact same thing as the square root of four. And we'll talk in the future about why this is, and the reason why this is defined this way is it has all sorts of neat and elegant properties when you start manipulating the actual exponents. But what is the square root of four, especially the principal root, mean? Well, that means, well, what is a number that if I were to multiply it by itself, or if I were to have two of those numbers and I were to multiply them times each other, that same number, I'm gonna get four. Well, what times itself is equal to four? Well, that's of course equal to two."}, {"video_title": "Basic fractional exponents Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "But what is the square root of four, especially the principal root, mean? Well, that means, well, what is a number that if I were to multiply it by itself, or if I were to have two of those numbers and I were to multiply them times each other, that same number, I'm gonna get four. Well, what times itself is equal to four? Well, that's of course equal to two. And just to get a sense of why this starts to work out well, remember, we could have also written that four is equal to two squared. So you're starting to see something interesting. Four to the 1\u20442 is equal to two."}, {"video_title": "Basic fractional exponents Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "Well, that's of course equal to two. And just to get a sense of why this starts to work out well, remember, we could have also written that four is equal to two squared. So you're starting to see something interesting. Four to the 1\u20442 is equal to two. Two squared is equal to four. So let's get a couple more examples of this just so you make sure you get what's going on. I encourage you to pause it as much as necessary and try to figure it out yourself."}, {"video_title": "Basic fractional exponents Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "Four to the 1\u20442 is equal to two. Two squared is equal to four. So let's get a couple more examples of this just so you make sure you get what's going on. I encourage you to pause it as much as necessary and try to figure it out yourself. So based on what I just told you, what do you think nine to the 1\u20442 power is going to be? Well, that's just the square root of nine, the principal root of nine. That's just going to be equal to three."}, {"video_title": "Basic fractional exponents Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "I encourage you to pause it as much as necessary and try to figure it out yourself. So based on what I just told you, what do you think nine to the 1\u20442 power is going to be? Well, that's just the square root of nine, the principal root of nine. That's just going to be equal to three. And likewise, we could have also said that three squared is, or let me write it this way, that nine is equal to three squared. These are both true statements. Let's do one more like this."}, {"video_title": "Basic fractional exponents Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "That's just going to be equal to three. And likewise, we could have also said that three squared is, or let me write it this way, that nine is equal to three squared. These are both true statements. Let's do one more like this. What is 25 to the 1\u20442 going to be? Well, this is just going to be five. Five times five is 25, or you could say that 25 is equal to five squared."}, {"video_title": "Basic fractional exponents Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "Let's do one more like this. What is 25 to the 1\u20442 going to be? Well, this is just going to be five. Five times five is 25, or you could say that 25 is equal to five squared. Now, let's think about what happens when we take something to the 1\u20443 power. So let's imagine taking eight to the 1\u20443 power. So the definition here is that taking something to the 1\u20443 power is the same thing as taking the cube root of, is the same thing as taking the cube root of that number."}, {"video_title": "Basic fractional exponents Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "Five times five is 25, or you could say that 25 is equal to five squared. Now, let's think about what happens when we take something to the 1\u20443 power. So let's imagine taking eight to the 1\u20443 power. So the definition here is that taking something to the 1\u20443 power is the same thing as taking the cube root of, is the same thing as taking the cube root of that number. And the cube root is saying, well, what number, if I had three of that number and I multiply them, that I'm going to get eight. So something times something times something is eight. Well, we already know that eight is equal to 2\u2043."}, {"video_title": "Basic fractional exponents Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "So the definition here is that taking something to the 1\u20443 power is the same thing as taking the cube root of, is the same thing as taking the cube root of that number. And the cube root is saying, well, what number, if I had three of that number and I multiply them, that I'm going to get eight. So something times something times something is eight. Well, we already know that eight is equal to 2\u2043. So the cube root of eight, or eight to the 1\u20443, is just going to be equal to two. This says, hey, give me the number that if I take, if I say that number times that number times that number, I'm gonna get eight. Well, that number is two, because two to the 3\u20443 is eight."}, {"video_title": "Basic fractional exponents Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "Well, we already know that eight is equal to 2\u2043. So the cube root of eight, or eight to the 1\u20443, is just going to be equal to two. This says, hey, give me the number that if I take, if I say that number times that number times that number, I'm gonna get eight. Well, that number is two, because two to the 3\u20443 is eight. Do a few more examples of that. What is 64 to the 1\u20443 power? Well, we already know that four times four times four is 64."}, {"video_title": "Basic fractional exponents Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "Well, that number is two, because two to the 3\u20443 is eight. Do a few more examples of that. What is 64 to the 1\u20443 power? Well, we already know that four times four times four is 64. So this is going to be four. And we already wrote over here that 64 is the same thing as four to the third. I think you're starting to see a little bit of a pattern here, a little bit of symmetry here."}, {"video_title": "Basic fractional exponents Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "Well, we already know that four times four times four is 64. So this is going to be four. And we already wrote over here that 64 is the same thing as four to the third. I think you're starting to see a little bit of a pattern here, a little bit of symmetry here. And we can extend this idea to arbitrary rational exponents. So what's happening, what happens if I were to raise, let's say I had, let me think of a good number here. So let's say I have 32."}, {"video_title": "Basic fractional exponents Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "I think you're starting to see a little bit of a pattern here, a little bit of symmetry here. And we can extend this idea to arbitrary rational exponents. So what's happening, what happens if I were to raise, let's say I had, let me think of a good number here. So let's say I have 32. I have the number 32, and I raise it to the 1\u2075 power. So this says, hey, give me the number that if I were to multiply that number, or I were to repeatedly multiply that number five times, what is that, I would get 32. Well, 32 is the same thing as two times two times two times two times two."}, {"video_title": "Factoring perfect squares common factor Mathematics II High School Math Khan Academy.mp3", "Sentence": "So let's say that we've got the polynomial 16x to the third plus 24x squared plus 9x. Now what I'd like you to do is pause the video and see if you can factor this polynomial completely. Now let's work through it together. So the first thing that you might notice is that all of the terms are divisible by x. So we can actually factor out an x. So let's do that. Actually, if we look at these coefficients, it looks like, let's see, it looks like they don't have any common factors other than one."}, {"video_title": "Factoring perfect squares common factor Mathematics II High School Math Khan Academy.mp3", "Sentence": "So the first thing that you might notice is that all of the terms are divisible by x. So we can actually factor out an x. So let's do that. Actually, if we look at these coefficients, it looks like, let's see, it looks like they don't have any common factors other than one. So it looks like the largest monomial that we can factor out is just going to be an x. So let's do that. Let's factor out an x."}, {"video_title": "Factoring perfect squares common factor Mathematics II High School Math Khan Academy.mp3", "Sentence": "Actually, if we look at these coefficients, it looks like, let's see, it looks like they don't have any common factors other than one. So it looks like the largest monomial that we can factor out is just going to be an x. So let's do that. Let's factor out an x. So then this is going to be x times. When you factor out an x from 16x to the third, you're gonna be left with 16x squared and then plus 24x and then plus 9. Now this is starting to look interesting."}, {"video_title": "Factoring perfect squares common factor Mathematics II High School Math Khan Academy.mp3", "Sentence": "Let's factor out an x. So then this is going to be x times. When you factor out an x from 16x to the third, you're gonna be left with 16x squared and then plus 24x and then plus 9. Now this is starting to look interesting. So let me just rewrite it. This is gonna be x times. This part over here looks interesting because when I see the 16x squared, this looks like a perfect square."}, {"video_title": "Factoring perfect squares common factor Mathematics II High School Math Khan Academy.mp3", "Sentence": "Now this is starting to look interesting. So let me just rewrite it. This is gonna be x times. This part over here looks interesting because when I see the 16x squared, this looks like a perfect square. Let me write it out. 16x squared, that's the same thing as 4x, 4x squared. And then we have a nine over there, which is clearly a perfect square."}, {"video_title": "Factoring perfect squares common factor Mathematics II High School Math Khan Academy.mp3", "Sentence": "This part over here looks interesting because when I see the 16x squared, this looks like a perfect square. Let me write it out. 16x squared, that's the same thing as 4x, 4x squared. And then we have a nine over there, which is clearly a perfect square. That is three squared, three squared. And when we look at this 24x, we see that it is four times three times two. And so we can write it as, let me write it this way."}, {"video_title": "Factoring perfect squares common factor Mathematics II High School Math Khan Academy.mp3", "Sentence": "And then we have a nine over there, which is clearly a perfect square. That is three squared, three squared. And when we look at this 24x, we see that it is four times three times two. And so we can write it as, let me write it this way. So this is going to be plus two times four times three x. So let me make it, so two times four times three times 3x. Now why did I take the trouble?"}, {"video_title": "Factoring perfect squares common factor Mathematics II High School Math Khan Academy.mp3", "Sentence": "And so we can write it as, let me write it this way. So this is going to be plus two times four times three x. So let me make it, so two times four times three times 3x. Now why did I take the trouble? Why did I take the trouble of writing everything like this? Because we see that it fits the pattern for a perfect square. What do I mean by that?"}, {"video_title": "Factoring perfect squares common factor Mathematics II High School Math Khan Academy.mp3", "Sentence": "Now why did I take the trouble? Why did I take the trouble of writing everything like this? Because we see that it fits the pattern for a perfect square. What do I mean by that? Well in previous videos, we saw that if you have something of the form ax plus b and you were to square it, you're going to get ax squared plus two abx plus b squared. And we have that form right over here. This is the ax squared."}, {"video_title": "Factoring perfect squares common factor Mathematics II High School Math Khan Academy.mp3", "Sentence": "What do I mean by that? Well in previous videos, we saw that if you have something of the form ax plus b and you were to square it, you're going to get ax squared plus two abx plus b squared. And we have that form right over here. This is the ax squared. Let me do the same color. The ax squared, ax squared. We have the b squared."}, {"video_title": "Factoring perfect squares common factor Mathematics II High School Math Khan Academy.mp3", "Sentence": "This is the ax squared. Let me do the same color. The ax squared, ax squared. We have the b squared. You have the b squared. And then you have the two abx, two abx right over there. So this section, this entire section, we can rewrite as being, we know what a and b are."}, {"video_title": "Factoring perfect squares common factor Mathematics II High School Math Khan Academy.mp3", "Sentence": "We have the b squared. You have the b squared. And then you have the two abx, two abx right over there. So this section, this entire section, we can rewrite as being, we know what a and b are. A is four and b is three. So this is going to be ax. So 4x plus b, which we know to be three, that whole thing squared."}, {"video_title": "Factoring perfect squares common factor Mathematics II High School Math Khan Academy.mp3", "Sentence": "So this section, this entire section, we can rewrite as being, we know what a and b are. A is four and b is three. So this is going to be ax. So 4x plus b, which we know to be three, that whole thing squared. And now we can't forget this x out front. So we have that x out front and we're done. We have just factored this."}, {"video_title": "Factoring perfect squares common factor Mathematics II High School Math Khan Academy.mp3", "Sentence": "So 4x plus b, which we know to be three, that whole thing squared. And now we can't forget this x out front. So we have that x out front and we're done. We have just factored this. We have just factored this completely. We could write it as x times 4x plus three and then write out and then say times 4x plus three or we could just write x times the quantity 4x plus three squared. And so we've just factored it completely."}, {"video_title": "Factoring perfect squares common factor Mathematics II High School Math Khan Academy.mp3", "Sentence": "We have just factored this. We have just factored this completely. We could write it as x times 4x plus three and then write out and then say times 4x plus three or we could just write x times the quantity 4x plus three squared. And so we've just factored it completely. And the key realizations here is, well one, you know, what can I factor out from all of these terms? I could factor out an x from all of those terms. And then to realize that what we had left over was a perfect square, really using this pattern that we were able to see in previous videos."}, {"video_title": "Adding rational expression unlike denominators High School Math Khan Academy.mp3", "Sentence": "Now we can work through this together. So the first thing that you might have hit when you tried to do it is you realize that they have different denominators, and it's hard to add fractions when they have different denominators. You need to rewrite them so that you have a common denominator. And the easiest way to get a common denominator is you could just multiply the two denominators, especially in a case like this where they don't seem to share any factors. Both of these are about as factors as you can get, and they don't share anything in common. And so let's set up a common denominator. So this is going to be equal to, it's going to be equal to something, let's see, it's going to be equal to something over our common denominator."}, {"video_title": "Adding rational expression unlike denominators High School Math Khan Academy.mp3", "Sentence": "And the easiest way to get a common denominator is you could just multiply the two denominators, especially in a case like this where they don't seem to share any factors. Both of these are about as factors as you can get, and they don't share anything in common. And so let's set up a common denominator. So this is going to be equal to, it's going to be equal to something, let's see, it's going to be equal to something over our common denominator. Let's make it, let's make it two x, let me just add another color. So we're gonna make it two x minus three times three x plus one, times three x plus one, and then plus, plus something else over two x minus three, two x minus three times three x plus one, times three x plus one. And so to go from two x, to go from just a two x minus three here, the denominator, to a two x three times three x plus one, we multiply the denominator by three x plus one."}, {"video_title": "Adding rational expression unlike denominators High School Math Khan Academy.mp3", "Sentence": "So this is going to be equal to, it's going to be equal to something, let's see, it's going to be equal to something over our common denominator. Let's make it, let's make it two x, let me just add another color. So we're gonna make it two x minus three times three x plus one, times three x plus one, and then plus, plus something else over two x minus three, two x minus three times three x plus one, times three x plus one. And so to go from two x, to go from just a two x minus three here, the denominator, to a two x three times three x plus one, we multiply the denominator by three x plus one. So if we do that to the denominator, we don't want to change the value of the rational expression, we'd also have to do that to the numerator. So the original numerator was five x, let me do that in blue color. So the original numerator was five x, and now we're gonna multiply it by the three x plus one."}, {"video_title": "Adding rational expression unlike denominators High School Math Khan Academy.mp3", "Sentence": "And so to go from two x, to go from just a two x minus three here, the denominator, to a two x three times three x plus one, we multiply the denominator by three x plus one. So if we do that to the denominator, we don't want to change the value of the rational expression, we'd also have to do that to the numerator. So the original numerator was five x, let me do that in blue color. So the original numerator was five x, and now we're gonna multiply it by the three x plus one. So times three x plus one. Notice I didn't change the value of this expression, I multiplied it by three x plus one over three x plus one, which is one, as long as three x plus one does not equal zero. So let's do the same thing over here."}, {"video_title": "Adding rational expression unlike denominators High School Math Khan Academy.mp3", "Sentence": "So the original numerator was five x, and now we're gonna multiply it by the three x plus one. So times three x plus one. Notice I didn't change the value of this expression, I multiplied it by three x plus one over three x plus one, which is one, as long as three x plus one does not equal zero. So let's do the same thing over here. Over here I have a denominator of three x plus one, I multiply it by two x minus three, so I would take my numerator, which is negative four x squared, and I would also multiply it by two x minus three. Two x minus three. Let me put parentheses around this so it doesn't look like I'm subtracting four x squared."}, {"video_title": "Adding rational expression unlike denominators High School Math Khan Academy.mp3", "Sentence": "So let's do the same thing over here. Over here I have a denominator of three x plus one, I multiply it by two x minus three, so I would take my numerator, which is negative four x squared, and I would also multiply it by two x minus three. Two x minus three. Let me put parentheses around this so it doesn't look like I'm subtracting four x squared. And so then I can rewrite all of this business as being equal to, well in the numerator, in the numerator I'm gonna have five x times three x, which is 15 x squared, five x times one, which is plus five x, and then over here, let me do this in green, let's see I could do four x times two, negative four x times two x, which would be negative eight x squared, and then negative four x times negative three, which is plus 12 x squared, did I do that right? Negative, oh let me be very careful, negative four, my spider sense could tell that I did something shady, in fact if you want to pause the video you could see, try to figure out what I just did that's wrong. So negative four x squared times two x is negative eight x to the third power, negative eight x to the third power, and then negative four x squared times negative three is 12 x squared, and then our entire denominator, our entire denominator, we have a common denominator now, so we're able to just add everything, is two x minus three, two x minus three times three x plus one, times three x plus one, and let's see how can we simplify this?"}, {"video_title": "Adding rational expression unlike denominators High School Math Khan Academy.mp3", "Sentence": "Let me put parentheses around this so it doesn't look like I'm subtracting four x squared. And so then I can rewrite all of this business as being equal to, well in the numerator, in the numerator I'm gonna have five x times three x, which is 15 x squared, five x times one, which is plus five x, and then over here, let me do this in green, let's see I could do four x times two, negative four x times two x, which would be negative eight x squared, and then negative four x times negative three, which is plus 12 x squared, did I do that right? Negative, oh let me be very careful, negative four, my spider sense could tell that I did something shady, in fact if you want to pause the video you could see, try to figure out what I just did that's wrong. So negative four x squared times two x is negative eight x to the third power, negative eight x to the third power, and then negative four x squared times negative three is 12 x squared, and then our entire denominator, our entire denominator, we have a common denominator now, so we're able to just add everything, is two x minus three, two x minus three times three x plus one, times three x plus one, and let's see how can we simplify this? So this is all going to be equal to, let me draw, make sure we recognize it's a rational expression, and so let's see, we can look at, our highest degree term here is the negative eight x to the third, so it's negative eight x to the third power, and then we have a 15 x squared, and we also have a 12 x squared, we could add those two together to get a 27 x squared, so we've already taken care of this, we've taken care, let me do that in a green color, so we've taken care of this, taken care of those two, and we're just left with a five x, so plus five x, and then all of that is over two x minus three times three x plus one, three x plus one, and we are, and we are all done. It doesn't seem like there's any easy way to simplify this further. You could factor out an x out of the numerator, but that's not going to cancel out with anything in the denominator, and it looks like we are all done."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "And it's actually the basis for the quadratic formula. In either the next video or the video after that, I'll prove the quadratic formula using completing the square. But before we do that, we need to understand even what it's all about. And that really just builds off of what we did in the last video, where we solved quadratics using perfect squares. So let's say I have the quadratic equation x squared minus 4x is equal to 5. And I put this big space here for a reason. In the last video, we saw that these can be pretty straightforward to solve if the left-hand side is a perfect square."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "And that really just builds off of what we did in the last video, where we solved quadratics using perfect squares. So let's say I have the quadratic equation x squared minus 4x is equal to 5. And I put this big space here for a reason. In the last video, we saw that these can be pretty straightforward to solve if the left-hand side is a perfect square. So how can we make? You see, completing the square is all about making the quadratic equation into a perfect square. Engineering it, adding and subtracting from both sides so it becomes a perfect square."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "In the last video, we saw that these can be pretty straightforward to solve if the left-hand side is a perfect square. So how can we make? You see, completing the square is all about making the quadratic equation into a perfect square. Engineering it, adding and subtracting from both sides so it becomes a perfect square. So how can we do that? Well, in order for this left-hand side to be a perfect square, there has to be some number here that if I have my number squared, I get that number. And then if I have 2 times my number, I get negative 4."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "Engineering it, adding and subtracting from both sides so it becomes a perfect square. So how can we do that? Well, in order for this left-hand side to be a perfect square, there has to be some number here that if I have my number squared, I get that number. And then if I have 2 times my number, I get negative 4. Remember that. I think it'll become clear with a few examples. I want x squared minus 4x plus something to be equal to x minus a squared."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "And then if I have 2 times my number, I get negative 4. Remember that. I think it'll become clear with a few examples. I want x squared minus 4x plus something to be equal to x minus a squared. We don't know what a is just yet. But we know a couple of things. When I square things, so this is going to be x squared minus 2a plus a squared."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "I want x squared minus 4x plus something to be equal to x minus a squared. We don't know what a is just yet. But we know a couple of things. When I square things, so this is going to be x squared minus 2a plus a squared. So if you look at this pattern right here, this over here, that has to be x squared minus 2ax. This right here has to be 2ax. And this right here would have to be a squared."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "When I square things, so this is going to be x squared minus 2a plus a squared. So if you look at this pattern right here, this over here, that has to be x squared minus 2ax. This right here has to be 2ax. And this right here would have to be a squared. So this number, a, is going to be half of negative 4. a has to be negative 2, right? Because 2 times a is going to be negative 4. a is negative 2. And if a is negative 2, what is a squared?"}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "And this right here would have to be a squared. So this number, a, is going to be half of negative 4. a has to be negative 2, right? Because 2 times a is going to be negative 4. a is negative 2. And if a is negative 2, what is a squared? Well, then a squared is going to be positive 4. And this might look all complicated to you right now, but I'm showing you the rationale. You literally just look at this coefficient right here, and you say, OK, well, what's half of that coefficient?"}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "And if a is negative 2, what is a squared? Well, then a squared is going to be positive 4. And this might look all complicated to you right now, but I'm showing you the rationale. You literally just look at this coefficient right here, and you say, OK, well, what's half of that coefficient? Well, half of that coefficient is negative 2. So we could say a is equal to negative 2. Same idea there."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "You literally just look at this coefficient right here, and you say, OK, well, what's half of that coefficient? Well, half of that coefficient is negative 2. So we could say a is equal to negative 2. Same idea there. And then you square it. You square a, you get positive 4. So we add positive 4 here."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "Same idea there. And then you square it. You square a, you get positive 4. So we add positive 4 here. Now, from the very first equation we ever did, you should know that you can never do something to just one side of the equation. You can't add 4 to just one side of the equation. If x squared minus 4x was equal to 5, then when I add 4, it's not going to be equal to 5 anymore."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "So we add positive 4 here. Now, from the very first equation we ever did, you should know that you can never do something to just one side of the equation. You can't add 4 to just one side of the equation. If x squared minus 4x was equal to 5, then when I add 4, it's not going to be equal to 5 anymore. It's going to be equal to 5 plus 4. We added 4 on the left-hand side because we wanted this to be a perfect square. But if you add something to the left-hand side, you've got to add it to the right-hand side."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "If x squared minus 4x was equal to 5, then when I add 4, it's not going to be equal to 5 anymore. It's going to be equal to 5 plus 4. We added 4 on the left-hand side because we wanted this to be a perfect square. But if you add something to the left-hand side, you've got to add it to the right-hand side. And now we've gotten ourselves to a problem that's just like the problems we did in the last video. What is this left-hand side? Let me rewrite the whole thing."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "But if you add something to the left-hand side, you've got to add it to the right-hand side. And now we've gotten ourselves to a problem that's just like the problems we did in the last video. What is this left-hand side? Let me rewrite the whole thing. We have x squared minus 4x plus 4 is equal to 9 now. All we did is add 4 to both sides of the equation. But we added 4 on purpose so that this left-hand side becomes a perfect square."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "Let me rewrite the whole thing. We have x squared minus 4x plus 4 is equal to 9 now. All we did is add 4 to both sides of the equation. But we added 4 on purpose so that this left-hand side becomes a perfect square. Now, what is this? What number, when I multiply it by itself, is equal to 4? And when I add it to itself, I'm equal to negative 2."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "But we added 4 on purpose so that this left-hand side becomes a perfect square. Now, what is this? What number, when I multiply it by itself, is equal to 4? And when I add it to itself, I'm equal to negative 2. Well, we already answered that question. It's negative 2. So we get x minus 2 times x minus 2 is equal to 9."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "And when I add it to itself, I'm equal to negative 2. Well, we already answered that question. It's negative 2. So we get x minus 2 times x minus 2 is equal to 9. Or we could have even skipped this step and written x minus 2 squared is equal to 9. And then you take the square root of both sides. You get x minus 2 is equal to plus or minus 3."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "So we get x minus 2 times x minus 2 is equal to 9. Or we could have even skipped this step and written x minus 2 squared is equal to 9. And then you take the square root of both sides. You get x minus 2 is equal to plus or minus 3. Add 2 to both sides. You get x is equal to 2 plus or minus 3. That tells us that x could be equal to 2 plus 3, which is 5."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "You get x minus 2 is equal to plus or minus 3. Add 2 to both sides. You get x is equal to 2 plus or minus 3. That tells us that x could be equal to 2 plus 3, which is 5. Or x could be equal to 2 minus 3, which is negative 1. And we are done. Now, I want to be very clear."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "That tells us that x could be equal to 2 plus 3, which is 5. Or x could be equal to 2 minus 3, which is negative 1. And we are done. Now, I want to be very clear. You could have done this without completing the square. We could have started off with x squared minus 4x is equal to 5. We could have subtracted 5 from both sides and gotten x squared minus 4x minus 5 is equal to 0."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "Now, I want to be very clear. You could have done this without completing the square. We could have started off with x squared minus 4x is equal to 5. We could have subtracted 5 from both sides and gotten x squared minus 4x minus 5 is equal to 0. And you could say, hey, if I have a negative 5 times a positive 1, their product is negative 5 and their sum is negative 4. So I could say this is x minus 5 times x plus 1 is equal to 0, and then we would say that x is equal to 5 or x is equal to negative 1. And in this case, this actually probably would have been a faster way to do the problem."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "We could have subtracted 5 from both sides and gotten x squared minus 4x minus 5 is equal to 0. And you could say, hey, if I have a negative 5 times a positive 1, their product is negative 5 and their sum is negative 4. So I could say this is x minus 5 times x plus 1 is equal to 0, and then we would say that x is equal to 5 or x is equal to negative 1. And in this case, this actually probably would have been a faster way to do the problem. But the neat thing about completing the square is it will always work. It'll always work no matter what the coefficients are or no matter how crazy the problem is. And let me prove it to you."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "And in this case, this actually probably would have been a faster way to do the problem. But the neat thing about completing the square is it will always work. It'll always work no matter what the coefficients are or no matter how crazy the problem is. And let me prove it to you. Let's do one that traditionally would have been a pretty painful problem if we just tried to do it by factoring, especially if we did it using grouping or something like that. Let's say we had 10x squared minus 30x minus 8 is equal to 0. Now, right from the get-go, you could say, hey, look, we can maybe divide both sides by 2."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "And let me prove it to you. Let's do one that traditionally would have been a pretty painful problem if we just tried to do it by factoring, especially if we did it using grouping or something like that. Let's say we had 10x squared minus 30x minus 8 is equal to 0. Now, right from the get-go, you could say, hey, look, we can maybe divide both sides by 2. Yeah, that does simplify a little bit. Let's divide both sides by 2. So if you divide everything by 2, what do you get?"}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "Now, right from the get-go, you could say, hey, look, we can maybe divide both sides by 2. Yeah, that does simplify a little bit. Let's divide both sides by 2. So if you divide everything by 2, what do you get? We get 5x squared minus 15x minus 4 is equal to 0. But once again, now we have this crazy 5 in front of this coefficient. We would have to solve it by grouping, which is a reasonably painful process."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "So if you divide everything by 2, what do you get? We get 5x squared minus 15x minus 4 is equal to 0. But once again, now we have this crazy 5 in front of this coefficient. We would have to solve it by grouping, which is a reasonably painful process. But we can now go straight to completing the square. And to do that, I'm going to now divide by 5 to get a 1 leading coefficient here. And you're going to see why this is different than what we've traditionally done."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "We would have to solve it by grouping, which is a reasonably painful process. But we can now go straight to completing the square. And to do that, I'm going to now divide by 5 to get a 1 leading coefficient here. And you're going to see why this is different than what we've traditionally done. So if I divide this whole thing by 5, I could have just divided by 10 from the get-go. But I wanted to go to this step first just to show you that this really didn't give us much. Let's divide everything by 5."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "And you're going to see why this is different than what we've traditionally done. So if I divide this whole thing by 5, I could have just divided by 10 from the get-go. But I wanted to go to this step first just to show you that this really didn't give us much. Let's divide everything by 5. So if you divide everything by 5, you get x squared minus 3x minus 4 fifths is equal to 0. So you might say, hey, why did we ever do that factoring by grouping? If we could just always divide by this leading coefficient, we can get rid of that."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "Let's divide everything by 5. So if you divide everything by 5, you get x squared minus 3x minus 4 fifths is equal to 0. So you might say, hey, why did we ever do that factoring by grouping? If we could just always divide by this leading coefficient, we can get rid of that. We can always turn this into a 1 or a negative 1 if we divide by the right number. But notice, by doing that, we got this crazy 4 fifths here. So this is super hard to do just using factoring."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "If we could just always divide by this leading coefficient, we can get rid of that. We can always turn this into a 1 or a negative 1 if we divide by the right number. But notice, by doing that, we got this crazy 4 fifths here. So this is super hard to do just using factoring. You'd have to say, gee, what two numbers when I take the product is equal to negative 4 fifths? It's a fraction. And when I take their sum is equal to negative 3."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "So this is super hard to do just using factoring. You'd have to say, gee, what two numbers when I take the product is equal to negative 4 fifths? It's a fraction. And when I take their sum is equal to negative 3. This is a hard problem with factoring. This is hard using factoring. So the best thing to do is to use completing the square."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "And when I take their sum is equal to negative 3. This is a hard problem with factoring. This is hard using factoring. So the best thing to do is to use completing the square. So let's think a little bit about how we can turn this into a perfect square. And what I like to do, and you'll see this done some ways, and I'll show you both ways because you'll see teachers do it both ways. I like to get the 4 fifths on the other side."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "So the best thing to do is to use completing the square. So let's think a little bit about how we can turn this into a perfect square. And what I like to do, and you'll see this done some ways, and I'll show you both ways because you'll see teachers do it both ways. I like to get the 4 fifths on the other side. So let's add 4 fifths to both sides of this equation. You don't have to do it this way, but I like to get the 4 fifths out of the way. And then what do we get?"}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "I like to get the 4 fifths on the other side. So let's add 4 fifths to both sides of this equation. You don't have to do it this way, but I like to get the 4 fifths out of the way. And then what do we get? If we add 4 fifths to both sides of the equation, the left-hand side of the equation just becomes x squared minus 3x. No 4 fifths there. I'm going to leave a little bit of space."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "And then what do we get? If we add 4 fifths to both sides of the equation, the left-hand side of the equation just becomes x squared minus 3x. No 4 fifths there. I'm going to leave a little bit of space. And that's going to be equal to 4 fifths. Now, just like the last problem, we want to turn this left-hand side into the perfect square of a binomial. How do we do that?"}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "I'm going to leave a little bit of space. And that's going to be equal to 4 fifths. Now, just like the last problem, we want to turn this left-hand side into the perfect square of a binomial. How do we do that? Well, we say, well, what number times 2 is equal to negative 3? So some number times 2 is negative 3. Or we essentially just take negative 3 and divide it by 2, which is negative 3 halves."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "How do we do that? Well, we say, well, what number times 2 is equal to negative 3? So some number times 2 is negative 3. Or we essentially just take negative 3 and divide it by 2, which is negative 3 halves. And then we square negative 3 halves. So in the example, we'll say a is negative 3 halves. And if we square negative 3 halves, what do we get?"}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "Or we essentially just take negative 3 and divide it by 2, which is negative 3 halves. And then we square negative 3 halves. So in the example, we'll say a is negative 3 halves. And if we square negative 3 halves, what do we get? We get plus, we get positive 9 over 4. I just took half of this coefficient, squared it, got positive 9 over 4. The whole purpose of doing that is to turn this left-hand side into a perfect square."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "And if we square negative 3 halves, what do we get? We get plus, we get positive 9 over 4. I just took half of this coefficient, squared it, got positive 9 over 4. The whole purpose of doing that is to turn this left-hand side into a perfect square. Now, anything you do to one side of the equation, you've got to do to the other side. So we added a 9 fourth here. Let's add a 9 fourths over there."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "The whole purpose of doing that is to turn this left-hand side into a perfect square. Now, anything you do to one side of the equation, you've got to do to the other side. So we added a 9 fourth here. Let's add a 9 fourths over there. What does our equation become? We get x squared minus 3x plus 9 over 4 is equal to, let's see, we get a common denominator. So 4 fifths is the same thing as 16 over 20."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "Let's add a 9 fourths over there. What does our equation become? We get x squared minus 3x plus 9 over 4 is equal to, let's see, we get a common denominator. So 4 fifths is the same thing as 16 over 20. Just multiply the numerator and denominator by 4. Plus over 20, 9 over 4 is the same thing if you multiply the numerator by 5 as 45 over 20. And so what's 16 plus 45?"}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "So 4 fifths is the same thing as 16 over 20. Just multiply the numerator and denominator by 4. Plus over 20, 9 over 4 is the same thing if you multiply the numerator by 5 as 45 over 20. And so what's 16 plus 45? You see, this is kind of getting kind of hairy, but that's the fun, I guess, of completing the square sometimes. 16 plus 45, you see that's 5561. So this is equal to 61 over 20."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "And so what's 16 plus 45? You see, this is kind of getting kind of hairy, but that's the fun, I guess, of completing the square sometimes. 16 plus 45, you see that's 5561. So this is equal to 61 over 20. So we get, well, let me just rewrite it. x squared minus 3x plus 9 over 4 is equal to 61 over 20. Crazy number."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "So this is equal to 61 over 20. So we get, well, let me just rewrite it. x squared minus 3x plus 9 over 4 is equal to 61 over 20. Crazy number. Now, this, at least on the left-hand side, is a perfect square. This is the same thing as x minus 3 halves squared. And it was by design."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "Crazy number. Now, this, at least on the left-hand side, is a perfect square. This is the same thing as x minus 3 halves squared. And it was by design. Negative 3 halves times negative 3 halves is positive 9 fourths. Negative 3 halves plus negative 3 halves is equal to negative 3. So this squared is equal to 61 over 20."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "And it was by design. Negative 3 halves times negative 3 halves is positive 9 fourths. Negative 3 halves plus negative 3 halves is equal to negative 3. So this squared is equal to 61 over 20. We can take the square root of both sides, and we get x minus 3 halves is equal to the positive or the negative square root of 61 over 20. And now we can add 3 halves to both sides of this equation, and you get x is equal to positive 3 halves plus or minus the square root of 61 over 20. And this is a crazy number, and it's hopefully obvious you would not have been able to, at least I would not have been able to, get to this number just by factoring."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "So this squared is equal to 61 over 20. We can take the square root of both sides, and we get x minus 3 halves is equal to the positive or the negative square root of 61 over 20. And now we can add 3 halves to both sides of this equation, and you get x is equal to positive 3 halves plus or minus the square root of 61 over 20. And this is a crazy number, and it's hopefully obvious you would not have been able to, at least I would not have been able to, get to this number just by factoring. And if you want their actual values, you can get your calculator out. And then let me clear all of this. Let me exit from here."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "And this is a crazy number, and it's hopefully obvious you would not have been able to, at least I would not have been able to, get to this number just by factoring. And if you want their actual values, you can get your calculator out. And then let me clear all of this. Let me exit from here. Maybe I'll clear it there. And 3 halves, let's do the plus version first. So we want to do 3 divided by 2 plus the second square root."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "Let me exit from here. Maybe I'll clear it there. And 3 halves, let's do the plus version first. So we want to do 3 divided by 2 plus the second square root. We want to pick that little yellow square root. So the square root of 61 divided by 20, which is 3.24, this crazy 3.2464. I'll just write 3.246."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "So we want to do 3 divided by 2 plus the second square root. We want to pick that little yellow square root. So the square root of 61 divided by 20, which is 3.24, this crazy 3.2464. I'll just write 3.246. So this is approximately equal to 3.246. And or, that was just the positive version. Let's do the subtraction version."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "I'll just write 3.246. So this is approximately equal to 3.246. And or, that was just the positive version. Let's do the subtraction version. So we can actually put our entry. If you do second and then entry, that we want that little yellow entry, that's why I pressed the second button, so I press enter. It puts in what we just put."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "Let's do the subtraction version. So we can actually put our entry. If you do second and then entry, that we want that little yellow entry, that's why I pressed the second button, so I press enter. It puts in what we just put. We could just change that positive, or that addition, to a subtraction. And you get negative 0.246. So you get negative 0.246."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "It puts in what we just put. We could just change that positive, or that addition, to a subtraction. And you get negative 0.246. So you get negative 0.246. And you can actually verify that these satisfy our original equation. Our original equation was up here. Let me just verify it for one of them."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "So you get negative 0.246. And you can actually verify that these satisfy our original equation. Our original equation was up here. Let me just verify it for one of them. So let's say we have, so the second answer on your graphing calculator is the last answer you use. So if you use the variable answer, that's this number right here. So if I have my answer squared, I'm using answer represents negative 0.24."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "Let me just verify it for one of them. So let's say we have, so the second answer on your graphing calculator is the last answer you use. So if you use the variable answer, that's this number right here. So if I have my answer squared, I'm using answer represents negative 0.24. Answer squared minus 3 times answer minus 4 fifths, or divided by 5, it equals this. Because they, just a little bit of explanation, this doesn't store the entire number. It goes up to some level of precision."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "So if I have my answer squared, I'm using answer represents negative 0.24. Answer squared minus 3 times answer minus 4 fifths, or divided by 5, it equals this. Because they, just a little bit of explanation, this doesn't store the entire number. It goes up to some level of precision. It stores some number of digits. So when it calculated it using this stored number right here, it got 1 times 10 to the negative 14. So that is 0.0000."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "It goes up to some level of precision. It stores some number of digits. So when it calculated it using this stored number right here, it got 1 times 10 to the negative 14. So that is 0.0000. So that's 13 zeros and then a 1. A decimal, then 13 zeros and a 1. So this is pretty much 0."}, {"video_title": "Solving quadratic equations by completing the square Algebra II Khan Academy.mp3", "Sentence": "So that is 0.0000. So that's 13 zeros and then a 1. A decimal, then 13 zeros and a 1. So this is pretty much 0. Or actually, if you got the exact answer right here, if you went through an infinite level of precision here, or maybe if you kept it in this radical form, you would get that it is indeed equal to 0. So hopefully you found that helpful. This whole notion of completing the square."}, {"video_title": "Worked example Rewriting expressions by completing the square High School Math Khan Academy.mp3", "Sentence": "Let's see if we can take this quadratic expression here, x squared plus 16x plus nine, and write it in this form. And you might be saying, hey Sal, why do I even need to worry about this? And one, it is just good algebraic practice to be able to manipulate things. But as we'll see in the future, what we're about to do is called completing the square. And it's a really valuable technique for solving quadratics, and it's actually the basis for the proof of the quadratic formula, which you'll learn in the future. So it's actually a pretty interesting technique. So how do we write this in this form?"}, {"video_title": "Worked example Rewriting expressions by completing the square High School Math Khan Academy.mp3", "Sentence": "But as we'll see in the future, what we're about to do is called completing the square. And it's a really valuable technique for solving quadratics, and it's actually the basis for the proof of the quadratic formula, which you'll learn in the future. So it's actually a pretty interesting technique. So how do we write this in this form? Well, one way to think about it is if we expanded this x plus a squared, we know if we square x plus a, it'd be x squared plus two a, x plus a squared, and then you still have that plus b right over there. So one way to think about it is, let's take this expression, this x squared plus 16x plus nine, and I'm just gonna write it with a few spaces in it. X squared plus 16x, and then plus nine, just like that."}, {"video_title": "Worked example Rewriting expressions by completing the square High School Math Khan Academy.mp3", "Sentence": "So how do we write this in this form? Well, one way to think about it is if we expanded this x plus a squared, we know if we square x plus a, it'd be x squared plus two a, x plus a squared, and then you still have that plus b right over there. So one way to think about it is, let's take this expression, this x squared plus 16x plus nine, and I'm just gonna write it with a few spaces in it. X squared plus 16x, and then plus nine, just like that. And so if we say, all right, we have an x squared here, we have an x squared here. If we say that two a x is the same thing as that, then what's a going to be? So if this is two a times x, well, that means two a is 16, or that a is equal to eight."}, {"video_title": "Worked example Rewriting expressions by completing the square High School Math Khan Academy.mp3", "Sentence": "X squared plus 16x, and then plus nine, just like that. And so if we say, all right, we have an x squared here, we have an x squared here. If we say that two a x is the same thing as that, then what's a going to be? So if this is two a times x, well, that means two a is 16, or that a is equal to eight. And so if I want to have an a squared over here, well, if a is eight, I would add an eight squared, which would be a 64. Well, I can't just add numbers willy-nilly to an expression without changing the value of an expression, so if I don't want to change the value of the expression, I still need to subtract 64. So notice, all that I have done now is I just took our original expression, and I added 64, and I subtracted 64."}, {"video_title": "Worked example Rewriting expressions by completing the square High School Math Khan Academy.mp3", "Sentence": "So if this is two a times x, well, that means two a is 16, or that a is equal to eight. And so if I want to have an a squared over here, well, if a is eight, I would add an eight squared, which would be a 64. Well, I can't just add numbers willy-nilly to an expression without changing the value of an expression, so if I don't want to change the value of the expression, I still need to subtract 64. So notice, all that I have done now is I just took our original expression, and I added 64, and I subtracted 64. So I have not changed the value of that expression. But what was valuable about me doing that is now this first part of the expression, this part right over here, it fits the pattern of a perfect square quadratic right over here. We have x squared plus two a x, where a is eight, plus a squared, 64."}, {"video_title": "Worked example Rewriting expressions by completing the square High School Math Khan Academy.mp3", "Sentence": "So notice, all that I have done now is I just took our original expression, and I added 64, and I subtracted 64. So I have not changed the value of that expression. But what was valuable about me doing that is now this first part of the expression, this part right over here, it fits the pattern of a perfect square quadratic right over here. We have x squared plus two a x, where a is eight, plus a squared, 64. Once again, how did I get 64? I took half of the 16, and I squared it to get to the 64. And so this stuff in this that I just squared off, this is going to be x plus eight squared."}, {"video_title": "Worked example Rewriting expressions by completing the square High School Math Khan Academy.mp3", "Sentence": "We have x squared plus two a x, where a is eight, plus a squared, 64. Once again, how did I get 64? I took half of the 16, and I squared it to get to the 64. And so this stuff in this that I just squared off, this is going to be x plus eight squared. X plus eight squared. Once again, I know that because a is eight, a is eight. So this is x plus eight squared."}, {"video_title": "Worked example Rewriting expressions by completing the square High School Math Khan Academy.mp3", "Sentence": "And so this stuff in this that I just squared off, this is going to be x plus eight squared. X plus eight squared. Once again, I know that because a is eight, a is eight. So this is x plus eight squared. And then all of this business on the right-hand side, what is nine minus 64? Well, 64 minus nine is 55, so this is going to be negative 55. So minus 55."}, {"video_title": "Introduction to function inverses Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Let's think about what functions really do, and then we'll think about the idea of an inverse of a function. Let's start with a pretty straightforward function. Let's have f of x is equal to 2x plus 4. And so if I take f of 2, f of 2 is going to be equal to 2 times 2 plus 4, which is 4 plus 4, which is 8. I could take f of 3, which is 2 times 3 plus 4, which is equal to 10, 6 plus 4. So let's think about it in a little bit more of an abstract sense. So there's a set of things that I can input into this function."}, {"video_title": "Introduction to function inverses Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And so if I take f of 2, f of 2 is going to be equal to 2 times 2 plus 4, which is 4 plus 4, which is 8. I could take f of 3, which is 2 times 3 plus 4, which is equal to 10, 6 plus 4. So let's think about it in a little bit more of an abstract sense. So there's a set of things that I can input into this function. You might already be familiar with that notion. It's the domain, the set of all of the things that I can input into that function. That is the domain."}, {"video_title": "Introduction to function inverses Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So there's a set of things that I can input into this function. You might already be familiar with that notion. It's the domain, the set of all of the things that I can input into that function. That is the domain. And in that domain, 2 is sitting there, 2 is there, you have 3 over there. Pretty much you could input any real number into this function. So this is going to be all reals, but we're making it a nice contained set here just to help you visualize it."}, {"video_title": "Introduction to function inverses Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "That is the domain. And in that domain, 2 is sitting there, 2 is there, you have 3 over there. Pretty much you could input any real number into this function. So this is going to be all reals, but we're making it a nice contained set here just to help you visualize it. Now, when you apply the function, let's think about what it means to take f of 2. We're inputting a number 2, and then the function is outputting the number 8. It is mapping us from 2 to 8."}, {"video_title": "Introduction to function inverses Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So this is going to be all reals, but we're making it a nice contained set here just to help you visualize it. Now, when you apply the function, let's think about what it means to take f of 2. We're inputting a number 2, and then the function is outputting the number 8. It is mapping us from 2 to 8. So let's make another set here of all of the possible values that my function can take on. All of the possible values that my function can take on, and we can call that the range. There are more formal ways to talk about this, and there's a much more rigorous discussion of this later on, especially in the linear algebra playlist."}, {"video_title": "Introduction to function inverses Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "It is mapping us from 2 to 8. So let's make another set here of all of the possible values that my function can take on. All of the possible values that my function can take on, and we can call that the range. There are more formal ways to talk about this, and there's a much more rigorous discussion of this later on, especially in the linear algebra playlist. But this is all the different values I can take on. So if I take the number 2 from our domain, I input it into the function, we're getting mapped to the number 8. So let me draw that out."}, {"video_title": "Introduction to function inverses Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "There are more formal ways to talk about this, and there's a much more rigorous discussion of this later on, especially in the linear algebra playlist. But this is all the different values I can take on. So if I take the number 2 from our domain, I input it into the function, we're getting mapped to the number 8. So let me draw that out. So we're going from 2 to the number 8 right there. And it's being done by the function. The function is doing that mapping."}, {"video_title": "Introduction to function inverses Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So let me draw that out. So we're going from 2 to the number 8 right there. And it's being done by the function. The function is doing that mapping. That function is mapping us from 2 to 8. This right here, that is equal to f of 2. Same idea."}, {"video_title": "Introduction to function inverses Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "The function is doing that mapping. That function is mapping us from 2 to 8. This right here, that is equal to f of 2. Same idea. You start with 3. 3 is being mapped by the function to 10. It's creating an association."}, {"video_title": "Introduction to function inverses Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Same idea. You start with 3. 3 is being mapped by the function to 10. It's creating an association. The function is mapping us from 3 to 10. Now, this raises an interesting question. Is there a way to get back from 8 to the 2?"}, {"video_title": "Introduction to function inverses Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "It's creating an association. The function is mapping us from 3 to 10. Now, this raises an interesting question. Is there a way to get back from 8 to the 2? Or is there a way to go back from the 10 to the 3? Or is there some other function? We can call that the inverse of f. That'll take us back."}, {"video_title": "Introduction to function inverses Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Is there a way to get back from 8 to the 2? Or is there a way to go back from the 10 to the 3? Or is there some other function? We can call that the inverse of f. That'll take us back. Is there some other function that'll take us from 10 back to 3? We'll call that the inverse of f, and we'll use that as a notation. And it'll take us back from 10 to 3."}, {"video_title": "Introduction to function inverses Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "We can call that the inverse of f. That'll take us back. Is there some other function that'll take us from 10 back to 3? We'll call that the inverse of f, and we'll use that as a notation. And it'll take us back from 10 to 3. Is there a way to do that? That same inverse of f, will it take us back from, if we apply 8 to it, will that take us back to 2? Will the inverse of f, if we apply it to 8, will it take us back to 2?"}, {"video_title": "Introduction to function inverses Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And it'll take us back from 10 to 3. Is there a way to do that? That same inverse of f, will it take us back from, if we apply 8 to it, will that take us back to 2? Will the inverse of f, if we apply it to 8, will it take us back to 2? Now, all of this seems very abstract and difficult. What you'll find is it's actually very easy to solve for this inverse of f. I think once we solve for it, it'll make it clear what I'm talking about. That the function takes you from 2 to 8."}, {"video_title": "Introduction to function inverses Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Will the inverse of f, if we apply it to 8, will it take us back to 2? Now, all of this seems very abstract and difficult. What you'll find is it's actually very easy to solve for this inverse of f. I think once we solve for it, it'll make it clear what I'm talking about. That the function takes you from 2 to 8. The inverse will take us back from 8 to 2. To think about that, let's just define, let's just say y is equal to f of x. So let's say y is equal to f of x."}, {"video_title": "Introduction to function inverses Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "That the function takes you from 2 to 8. The inverse will take us back from 8 to 2. To think about that, let's just define, let's just say y is equal to f of x. So let's say y is equal to f of x. So y is equal to f of x is equal to 2x plus 4. So I could write just y is equal to 2x plus 4. And this, once again, this is our function."}, {"video_title": "Introduction to function inverses Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So let's say y is equal to f of x. So y is equal to f of x is equal to 2x plus 4. So I could write just y is equal to 2x plus 4. And this, once again, this is our function. You give me an x, it'll give me a y. But we want to go the other way around. We want to give you a y and get an x."}, {"video_title": "Introduction to function inverses Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And this, once again, this is our function. You give me an x, it'll give me a y. But we want to go the other way around. We want to give you a y and get an x. So all we have to do is solve for x in terms of y. So let's do that. If we subtract 4 from both sides of this equation, let me switch colors."}, {"video_title": "Introduction to function inverses Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "We want to give you a y and get an x. So all we have to do is solve for x in terms of y. So let's do that. If we subtract 4 from both sides of this equation, let me switch colors. If we subtract 4 from both sides of this equation, we get y minus 4 is equal to 2x. And then if we divide both sides of this equation by 2, we get y over 2 minus 2, right? 4 divided by 2 is 2, is equal to x."}, {"video_title": "Introduction to function inverses Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "If we subtract 4 from both sides of this equation, let me switch colors. If we subtract 4 from both sides of this equation, we get y minus 4 is equal to 2x. And then if we divide both sides of this equation by 2, we get y over 2 minus 2, right? 4 divided by 2 is 2, is equal to x. So if we just want to write it that way, we can just swap the sides. We get x is equal to 1 half y, same thing as y over 2, minus 2. So what we have here is a function of y that gives us an x, which is exactly what we wanted."}, {"video_title": "Introduction to function inverses Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "4 divided by 2 is 2, is equal to x. So if we just want to write it that way, we can just swap the sides. We get x is equal to 1 half y, same thing as y over 2, minus 2. So what we have here is a function of y that gives us an x, which is exactly what we wanted. We want a function of these values that map back to an x. So we could call this, we could say that this is equal to, I'll do it in that same color, this is equal to f inverse as a function of y. Or let me just write a little bit cleaner."}, {"video_title": "Introduction to function inverses Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So what we have here is a function of y that gives us an x, which is exactly what we wanted. We want a function of these values that map back to an x. So we could call this, we could say that this is equal to, I'll do it in that same color, this is equal to f inverse as a function of y. Or let me just write a little bit cleaner. We could say f inverse as a function of y. So we could have 10 or 8. So now the range is now the domain for f inverse."}, {"video_title": "Introduction to function inverses Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Or let me just write a little bit cleaner. We could say f inverse as a function of y. So we could have 10 or 8. So now the range is now the domain for f inverse. f inverse as a function of y is equal to 1 half y minus 2. So all we did is we started with our original function, y is equal to 2x plus 4. We solved for, over here we have, we've solved for y in terms of x."}, {"video_title": "Introduction to function inverses Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So now the range is now the domain for f inverse. f inverse as a function of y is equal to 1 half y minus 2. So all we did is we started with our original function, y is equal to 2x plus 4. We solved for, over here we have, we've solved for y in terms of x. Then we just do a little bit of algebra, solve for x in terms of y. And we say that that is our inverse as a function of y, which is right over here. And then if we, you know, you could say this is, you could replace the y with an a, a b, an x, whatever you want to do."}, {"video_title": "Introduction to function inverses Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "We solved for, over here we have, we've solved for y in terms of x. Then we just do a little bit of algebra, solve for x in terms of y. And we say that that is our inverse as a function of y, which is right over here. And then if we, you know, you could say this is, you could replace the y with an a, a b, an x, whatever you want to do. So then we can just rename the y as x. So if you put an x into this function, you would get f inverse of x is equal to 1 half x minus 2. So all you do, you solve for x, and then you swap the y and the x, if you want to view it that way."}, {"video_title": "Introduction to function inverses Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And then if we, you know, you could say this is, you could replace the y with an a, a b, an x, whatever you want to do. So then we can just rename the y as x. So if you put an x into this function, you would get f inverse of x is equal to 1 half x minus 2. So all you do, you solve for x, and then you swap the y and the x, if you want to view it that way. That's the easiest way to think about it. And one thing I want to point out is what happens when you graph the function and the inverse. So let me just do a little quick and dirty graph right here."}, {"video_title": "Introduction to function inverses Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So all you do, you solve for x, and then you swap the y and the x, if you want to view it that way. That's the easiest way to think about it. And one thing I want to point out is what happens when you graph the function and the inverse. So let me just do a little quick and dirty graph right here. And then I'll do a bunch of examples of actually solving for inverses. But I really just wanted to give you the general idea. Function takes you from the domain of the range, the inverse will take you from that point back to the original value, if it exists."}, {"video_title": "Introduction to function inverses Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So let me just do a little quick and dirty graph right here. And then I'll do a bunch of examples of actually solving for inverses. But I really just wanted to give you the general idea. Function takes you from the domain of the range, the inverse will take you from that point back to the original value, if it exists. So if I were to graph these, so let me draw a little coordinate axis right here. Draw a little bit of a coordinate axis right there. This first function, 2x plus 4, its y-intercept is going to be 1, 2, 3, 4, just like that."}, {"video_title": "Introduction to function inverses Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Function takes you from the domain of the range, the inverse will take you from that point back to the original value, if it exists. So if I were to graph these, so let me draw a little coordinate axis right here. Draw a little bit of a coordinate axis right there. This first function, 2x plus 4, its y-intercept is going to be 1, 2, 3, 4, just like that. And then its slope will look like this. It has a slope of 2, so it will look something like this. That's what that function looks like."}, {"video_title": "Introduction to function inverses Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "This first function, 2x plus 4, its y-intercept is going to be 1, 2, 3, 4, just like that. And then its slope will look like this. It has a slope of 2, so it will look something like this. That's what that function looks like. What does this function look like? What does the inverse function look like as a function of x? Remember, we solved for x and then we swapped the x and the y, essentially."}, {"video_title": "Introduction to function inverses Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "That's what that function looks like. What does this function look like? What does the inverse function look like as a function of x? Remember, we solved for x and then we swapped the x and the y, essentially. We could say now that y is equal to f inverse of x. So we have a y-intercept of negative 2, 1, 2. And now the slope is 1 half."}, {"video_title": "Introduction to function inverses Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Remember, we solved for x and then we swapped the x and the y, essentially. We could say now that y is equal to f inverse of x. So we have a y-intercept of negative 2, 1, 2. And now the slope is 1 half. So the slope is 1 half. The slope looks like this. Let me see if I can draw it."}, {"video_title": "Introduction to function inverses Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And now the slope is 1 half. So the slope is 1 half. The slope looks like this. Let me see if I can draw it. The slope looks, or the line looks something like that. There's a relationship here. These look kind of related."}, {"video_title": "Introduction to function inverses Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Let me see if I can draw it. The slope looks, or the line looks something like that. There's a relationship here. These look kind of related. It looks like they're reflected about something. It will be a little bit more clear what they're reflected about if we draw the line y is equal to x. The line y equals x looks like that."}, {"video_title": "Introduction to function inverses Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "These look kind of related. It looks like they're reflected about something. It will be a little bit more clear what they're reflected about if we draw the line y is equal to x. The line y equals x looks like that. I'll do it as a dotted line. The line y equals x looks something like that. That is the line y is equal to x."}, {"video_title": "Introduction to function inverses Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "The line y equals x looks like that. I'll do it as a dotted line. The line y equals x looks something like that. That is the line y is equal to x. You can see you have the function and its inverse. They're reflected about the line y is equal to x. Hopefully that makes sense here. Because over here on this line, let's take an easy example."}, {"video_title": "Introduction to function inverses Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "That is the line y is equal to x. You can see you have the function and its inverse. They're reflected about the line y is equal to x. Hopefully that makes sense here. Because over here on this line, let's take an easy example. Our function, when you take 0, so f of 0 is equal to 4. Our function is mapping 0 to 4. The inverse function, if you take f inverse of 4, f inverse of 4 is equal to 0."}, {"video_title": "Introduction to function inverses Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Because over here on this line, let's take an easy example. Our function, when you take 0, so f of 0 is equal to 4. Our function is mapping 0 to 4. The inverse function, if you take f inverse of 4, f inverse of 4 is equal to 0. The inverse function is mapping us from 4 to 0, which is exactly what we expected. The function takes us from the x to the y world. Then we swap it."}, {"video_title": "Introduction to function inverses Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "The inverse function, if you take f inverse of 4, f inverse of 4 is equal to 0. The inverse function is mapping us from 4 to 0, which is exactly what we expected. The function takes us from the x to the y world. Then we swap it. We were swapping the x and the y when we take the inverse. That's why it's reflected around y equals x. This example that I just showed you right here, function takes you from 0 to 4."}, {"video_title": "Introduction to function inverses Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Then we swap it. We were swapping the x and the y when we take the inverse. That's why it's reflected around y equals x. This example that I just showed you right here, function takes you from 0 to 4. I should do that in the function color. The function takes you from 0 to 4. That's the function."}, {"video_title": "Introduction to function inverses Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "This example that I just showed you right here, function takes you from 0 to 4. I should do that in the function color. The function takes you from 0 to 4. That's the function. f of 0 is 4. You see that right there. It goes from 0 to 4."}, {"video_title": "Introduction to function inverses Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "That's the function. f of 0 is 4. You see that right there. It goes from 0 to 4. Then the inverse takes us back from 4 to 0. f inverse takes us back from 4 to 0. You saw that right there. When you evaluate 4 here, 1 half times 4 minus 2 is 0."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor Khan Academy.mp3", "Sentence": "So to factor this, we need to figure out what the greatest common factor of each of these terms are. So let me rewrite it. So we have 4x to the fourth y, and we have minus 8x to the third y, and then we have minus 2x squared. So in the other videos, we looked at it in terms of breaking it down to its simplest parts, but I think we have enough practice now to be able to do a little bit more of it in our head. So what is the largest number that divides into all of these? When I say number, I'm actually talking about the actual, I guess, coefficients. We have a 4, an 8, a 2."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor Khan Academy.mp3", "Sentence": "So in the other videos, we looked at it in terms of breaking it down to its simplest parts, but I think we have enough practice now to be able to do a little bit more of it in our head. So what is the largest number that divides into all of these? When I say number, I'm actually talking about the actual, I guess, coefficients. We have a 4, an 8, a 2. We don't have to worry about the negative signs just yet. And we say, well, the largest common factor of 2, 8, and 4 is 2. 2 goes into all of them, and obviously that's the largest number that can go into 2."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor Khan Academy.mp3", "Sentence": "We have a 4, an 8, a 2. We don't have to worry about the negative signs just yet. And we say, well, the largest common factor of 2, 8, and 4 is 2. 2 goes into all of them, and obviously that's the largest number that can go into 2. So that is the largest number that's going to be part of the greatest common factor. So let's write that down. So it's going to be 2."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor Khan Academy.mp3", "Sentence": "2 goes into all of them, and obviously that's the largest number that can go into 2. So that is the largest number that's going to be part of the greatest common factor. So let's write that down. So it's going to be 2. And then what's the greatest, I guess, factor, what's the greatest degree of x that's divisible into all three of these? Well, x squared goes into all three of these, and obviously that's the greatest degree of x that can be divided into this last term. So x squared is going to be the greatest common x degree in all of them, 2x squared."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor Khan Academy.mp3", "Sentence": "So it's going to be 2. And then what's the greatest, I guess, factor, what's the greatest degree of x that's divisible into all three of these? Well, x squared goes into all three of these, and obviously that's the greatest degree of x that can be divided into this last term. So x squared is going to be the greatest common x degree in all of them, 2x squared. And then what's the largest degree of y that's divisible into all of them? Well, these two guys are divisible by y, but this guy isn't. So there is no degree of y that's divisible into all of them."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor Khan Academy.mp3", "Sentence": "So x squared is going to be the greatest common x degree in all of them, 2x squared. And then what's the largest degree of y that's divisible into all of them? Well, these two guys are divisible by y, but this guy isn't. So there is no degree of y that's divisible into all of them. So the greatest common factor of all three of these guys right here is 2x squared. So what we can do now is we can think about each of these terms as a product of 2x squared and something else. And to figure out something else, we can literally undistribute the 2x squared and say this is the same thing."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor Khan Academy.mp3", "Sentence": "So there is no degree of y that's divisible into all of them. So the greatest common factor of all three of these guys right here is 2x squared. So what we can do now is we can think about each of these terms as a product of 2x squared and something else. And to figure out something else, we can literally undistribute the 2x squared and say this is the same thing. Or even before we undistribute the 2x squared, we could say, look, 4x to the fourth y is the same thing as 2x squared times 4x to the fourth y over 2x squared. Right? If you just multiply this out, you'd get 4xy."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor Khan Academy.mp3", "Sentence": "And to figure out something else, we can literally undistribute the 2x squared and say this is the same thing. Or even before we undistribute the 2x squared, we could say, look, 4x to the fourth y is the same thing as 2x squared times 4x to the fourth y over 2x squared. Right? If you just multiply this out, you'd get 4xy. Similarly, you could say that 8x to the third y, I'll put the negative out front, you could say that 8x to the third y is the same thing as 2x squared, our greatest common factor, times 8x to the third y over 2x squared. And then finally, 2x squared is the same thing as if we factor out 2x squared. So we have that negative sign out front."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor Khan Academy.mp3", "Sentence": "If you just multiply this out, you'd get 4xy. Similarly, you could say that 8x to the third y, I'll put the negative out front, you could say that 8x to the third y is the same thing as 2x squared, our greatest common factor, times 8x to the third y over 2x squared. And then finally, 2x squared is the same thing as if we factor out 2x squared. So we have that negative sign out front. We have this negative sign, 2x squared. If we factor out 2x squared, same thing as 2x squared times 2x squared over 2x squared. This is almost silly what I'm doing here, but I'm just showing you that I'm just multiplying and dividing both of these terms by 2x squared."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor Khan Academy.mp3", "Sentence": "So we have that negative sign out front. We have this negative sign, 2x squared. If we factor out 2x squared, same thing as 2x squared times 2x squared over 2x squared. This is almost silly what I'm doing here, but I'm just showing you that I'm just multiplying and dividing both of these terms by 2x squared. Multiplying and dividing. Here it's trivially simple. This just simplifies to 2x squared right there, or this 2x squared times 1."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor Khan Academy.mp3", "Sentence": "This is almost silly what I'm doing here, but I'm just showing you that I'm just multiplying and dividing both of these terms by 2x squared. Multiplying and dividing. Here it's trivially simple. This just simplifies to 2x squared right there, or this 2x squared times 1. That simplifies to 1. Maybe I should write it below. That simplifies to 1."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor Khan Academy.mp3", "Sentence": "This just simplifies to 2x squared right there, or this 2x squared times 1. That simplifies to 1. Maybe I should write it below. That simplifies to 1. But what do these simplify to? This first term over here, this simplifies to 2x squared times, now you get 4 divided by 2 is 2. X to the fourth divided by x squared is x squared."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor Khan Academy.mp3", "Sentence": "That simplifies to 1. But what do these simplify to? This first term over here, this simplifies to 2x squared times, now you get 4 divided by 2 is 2. X to the fourth divided by x squared is x squared. And then y divided by 1 is just going to be a y. So it's 2x squared times 2x squared y. And then you have minus 2x squared times 8 divided by 2 is 4."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor Khan Academy.mp3", "Sentence": "X to the fourth divided by x squared is x squared. And then y divided by 1 is just going to be a y. So it's 2x squared times 2x squared y. And then you have minus 2x squared times 8 divided by 2 is 4. X to the third divided by x squared is x. And y divided by 1, you can imagine, is just y. And then finally, of course, you have minus 2x squared times this right here simplifies to 1."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor Khan Academy.mp3", "Sentence": "And then you have minus 2x squared times 8 divided by 2 is 4. X to the third divided by x squared is x. And y divided by 1, you can imagine, is just y. And then finally, of course, you have minus 2x squared times this right here simplifies to 1. Times 1. Now, if you were to undistribute 2x squared out of the expression, you'd essentially get 2x squared times this term minus this term minus this term. If you distribute this out, if you take that out of each of the terms, you're going to get 2x squared times this 2x squared y minus 4xy."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor Khan Academy.mp3", "Sentence": "And then finally, of course, you have minus 2x squared times this right here simplifies to 1. Times 1. Now, if you were to undistribute 2x squared out of the expression, you'd essentially get 2x squared times this term minus this term minus this term. If you distribute this out, if you take that out of each of the terms, you're going to get 2x squared times this 2x squared y minus 4xy. And then you have minus 1. And we're done. We've factored the problem."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor Khan Academy.mp3", "Sentence": "If you distribute this out, if you take that out of each of the terms, you're going to get 2x squared times this 2x squared y minus 4xy. And then you have minus 1. And we're done. We've factored the problem. Now, it looks like we did a lot of steps. And the reason why I kind of went through great pains to show you exactly what we're doing is so you know exactly what we're doing. In the future, you might be able to do this a little bit quicker."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor Khan Academy.mp3", "Sentence": "We've factored the problem. Now, it looks like we did a lot of steps. And the reason why I kind of went through great pains to show you exactly what we're doing is so you know exactly what we're doing. In the future, you might be able to do this a little bit quicker. You might be able to do many of the steps in your head. You might say, okay, let me look at each of these. Well, the biggest coefficient that divides all of these is a 2."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor Khan Academy.mp3", "Sentence": "In the future, you might be able to do this a little bit quicker. You might be able to do many of the steps in your head. You might say, okay, let me look at each of these. Well, the biggest coefficient that divides all of these is a 2. So let me factor a 2 out. Well, all of these are divisible by x squared. That's the largest degree of x."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor Khan Academy.mp3", "Sentence": "Well, the biggest coefficient that divides all of these is a 2. So let me factor a 2 out. Well, all of these are divisible by x squared. That's the largest degree of x. Let me factor an x squared out. And this guy doesn't have a y, so I can't factor a y out. So let's say it's going to be 2x squared times."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor Khan Academy.mp3", "Sentence": "That's the largest degree of x. Let me factor an x squared out. And this guy doesn't have a y, so I can't factor a y out. So let's say it's going to be 2x squared times. And what's this guy divided by 2x squared? Well, 4 divided by 2 is 2. x to the fourth divided by x squared is x squared. y divided by 1."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor Khan Academy.mp3", "Sentence": "So let's say it's going to be 2x squared times. And what's this guy divided by 2x squared? Well, 4 divided by 2 is 2. x to the fourth divided by x squared is x squared. y divided by 1. There is no other y degree that we factored out, so it's just going to be y. And then you have minus 8 divided by 2 is 4. x to the third divided by x squared is x. And you have y divided by, say, 1 is just y."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor Khan Academy.mp3", "Sentence": "y divided by 1. There is no other y degree that we factored out, so it's just going to be y. And then you have minus 8 divided by 2 is 4. x to the third divided by x squared is x. And you have y divided by, say, 1 is just y. And then you have minus 2 divided by 2 is 1. x squared divided by x squared is 1. So 2x squared divided by 2x squared is just 1. So in the future, you'll do it more like this, where you kind of just factor it out in your head."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor Khan Academy.mp3", "Sentence": "And you have y divided by, say, 1 is just y. And then you have minus 2 divided by 2 is 1. x squared divided by x squared is 1. So 2x squared divided by 2x squared is just 1. So in the future, you'll do it more like this, where you kind of just factor it out in your head. But I really want you to understand what we did here. There is no magic. And to realize that there's no magic, you could just use the distributive property to multiply this out again."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3 \u00c2\u00b2 Algebra I High School Math Khan Academy.mp3", "Sentence": "And I encourage you to pause the video and work through it on your own. All right, now let's work through this together. So we just have to remember, we're squaring the entire binomial. So this thing is going to be the same thing as x plus seven times x plus seven. I'm gonna write the second x plus seven in a different color, which is going to be helpful when we actually multiply things out. When we see it like this, then we can multiply these out the way we would multiply any binomials. And I'll first do it, I guess you could say the slower way, but the more intuitive way, applying the distributive property twice, and then we'll think about maybe some shortcuts or some patterns we might be able to recognize, especially when we are squaring binomials."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3 \u00c2\u00b2 Algebra I High School Math Khan Academy.mp3", "Sentence": "So this thing is going to be the same thing as x plus seven times x plus seven. I'm gonna write the second x plus seven in a different color, which is going to be helpful when we actually multiply things out. When we see it like this, then we can multiply these out the way we would multiply any binomials. And I'll first do it, I guess you could say the slower way, but the more intuitive way, applying the distributive property twice, and then we'll think about maybe some shortcuts or some patterns we might be able to recognize, especially when we are squaring binomials. So let's start with just applying the distributive property twice. So let's distribute this yellow x plus seven over this magenta x plus seven. So we can multiply it by the x, this magenta x."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3 \u00c2\u00b2 Algebra I High School Math Khan Academy.mp3", "Sentence": "And I'll first do it, I guess you could say the slower way, but the more intuitive way, applying the distributive property twice, and then we'll think about maybe some shortcuts or some patterns we might be able to recognize, especially when we are squaring binomials. So let's start with just applying the distributive property twice. So let's distribute this yellow x plus seven over this magenta x plus seven. So we can multiply it by the x, this magenta x. So it's going to be x, let me do it in that same color. So it's going to be magenta x times x plus seven, plus magenta seven times yellow x plus seven, x plus seven. And now we can apply the distributive property again."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3 \u00c2\u00b2 Algebra I High School Math Khan Academy.mp3", "Sentence": "So we can multiply it by the x, this magenta x. So it's going to be x, let me do it in that same color. So it's going to be magenta x times x plus seven, plus magenta seven times yellow x plus seven, x plus seven. And now we can apply the distributive property again. We can take this magenta x and distribute it over the x plus seven. So x times x is x squared. X times seven is seven x."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3 \u00c2\u00b2 Algebra I High School Math Khan Academy.mp3", "Sentence": "And now we can apply the distributive property again. We can take this magenta x and distribute it over the x plus seven. So x times x is x squared. X times seven is seven x. And then we can do it again over here. This seven, let me do it in a different color. So this seven times that x is going to be plus another seven x."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3 \u00c2\u00b2 Algebra I High School Math Khan Academy.mp3", "Sentence": "X times seven is seven x. And then we can do it again over here. This seven, let me do it in a different color. So this seven times that x is going to be plus another seven x. And then the seven times the seven is going to be 49. And we're in the home stretch. We can then simplify it."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3 \u00c2\u00b2 Algebra I High School Math Khan Academy.mp3", "Sentence": "So this seven times that x is going to be plus another seven x. And then the seven times the seven is going to be 49. And we're in the home stretch. We can then simplify it. This is going to be x squared. And then these two middle terms we can add together. Seven x, seven x, let me do this in orange."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3 \u00c2\u00b2 Algebra I High School Math Khan Academy.mp3", "Sentence": "We can then simplify it. This is going to be x squared. And then these two middle terms we can add together. Seven x, seven x, let me do this in orange. Seven x plus seven x, that's going to be 14 x, plus 14 x plus 49. So plus 49. And we're done."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3 \u00c2\u00b2 Algebra I High School Math Khan Academy.mp3", "Sentence": "Seven x, seven x, let me do this in orange. Seven x plus seven x, that's going to be 14 x, plus 14 x plus 49. So plus 49. And we're done. Now the key question is do we see some patterns here? Do we see some patterns that we can generalize and that might help us square binomials a little bit faster in the future? Well when we first looked at just multiplying binomials, we saw a pattern like x plus a times x plus b is going to be equal to x squared."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3 \u00c2\u00b2 Algebra I High School Math Khan Academy.mp3", "Sentence": "And we're done. Now the key question is do we see some patterns here? Do we see some patterns that we can generalize and that might help us square binomials a little bit faster in the future? Well when we first looked at just multiplying binomials, we saw a pattern like x plus a times x plus b is going to be equal to x squared. Let me write it this way. It's going to be equal to x squared plus a plus b, x plus b squared. And so if both a and b are the same thing, we could say that x plus a times x plus a is going to be equal to x squared."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3 \u00c2\u00b2 Algebra I High School Math Khan Academy.mp3", "Sentence": "Well when we first looked at just multiplying binomials, we saw a pattern like x plus a times x plus b is going to be equal to x squared. Let me write it this way. It's going to be equal to x squared plus a plus b, x plus b squared. And so if both a and b are the same thing, we could say that x plus a times x plus a is going to be equal to x squared. And this is the case when we have a coefficient of one on both of these x's, x squared. Now in this case, a and b are both a. So this is going to be a plus a times x, or we could just say plus two a x."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3 \u00c2\u00b2 Algebra I High School Math Khan Academy.mp3", "Sentence": "And so if both a and b are the same thing, we could say that x plus a times x plus a is going to be equal to x squared. And this is the case when we have a coefficient of one on both of these x's, x squared. Now in this case, a and b are both a. So this is going to be a plus a times x, or we could just say plus two a x. Let me be clear what I just did. Instead of writing a plus b, I could just view this as a plus a times x, and then plus a squared. Or that's the same thing as x squared, plus two a x plus a squared."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3 \u00c2\u00b2 Algebra I High School Math Khan Academy.mp3", "Sentence": "So this is going to be a plus a times x, or we could just say plus two a x. Let me be clear what I just did. Instead of writing a plus b, I could just view this as a plus a times x, and then plus a squared. Or that's the same thing as x squared, plus two a x plus a squared. This is a general way of expressing a squared binomial like this. A squared binomial where the coefficients on both x's are one. We can see that's exactly what we saw over here."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3 \u00c2\u00b2 Algebra I High School Math Khan Academy.mp3", "Sentence": "Or that's the same thing as x squared, plus two a x plus a squared. This is a general way of expressing a squared binomial like this. A squared binomial where the coefficients on both x's are one. We can see that's exactly what we saw over here. In the example we did, seven is our a. So we got x squared right over there. Let me circle it."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3 \u00c2\u00b2 Algebra I High School Math Khan Academy.mp3", "Sentence": "We can see that's exactly what we saw over here. In the example we did, seven is our a. So we got x squared right over there. Let me circle it. So we have this blue x squared. That corresponds to that over there. And then seven is our a, so two a x, two times seven is 14 x."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3 \u00c2\u00b2 Algebra I High School Math Khan Academy.mp3", "Sentence": "Let me circle it. So we have this blue x squared. That corresponds to that over there. And then seven is our a, so two a x, two times seven is 14 x. Notice we have the 14 x right over there. So this 14 x corresponds to two a x. And then finally, if a is seven, a squared is 49."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3 \u00c2\u00b2 Algebra I High School Math Khan Academy.mp3", "Sentence": "And then seven is our a, so two a x, two times seven is 14 x. Notice we have the 14 x right over there. So this 14 x corresponds to two a x. And then finally, if a is seven, a squared is 49. So in general, if you're squaring a binomial, a fast way to do it is to do this pattern here. We could do another example real fast just to make sure that we've understood things. If I were to tell you what is x, I'll throw a negative in here, x minus three squared."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3 \u00c2\u00b2 Algebra I High School Math Khan Academy.mp3", "Sentence": "And then finally, if a is seven, a squared is 49. So in general, if you're squaring a binomial, a fast way to do it is to do this pattern here. We could do another example real fast just to make sure that we've understood things. If I were to tell you what is x, I'll throw a negative in here, x minus three squared. I encourage you to pause the video and think about expressing this using this pattern. Well, this is going to be, in this case, our a, we have to be careful, our a is going to be negative three. So that is our a right over there."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3 \u00c2\u00b2 Algebra I High School Math Khan Academy.mp3", "Sentence": "If I were to tell you what is x, I'll throw a negative in here, x minus three squared. I encourage you to pause the video and think about expressing this using this pattern. Well, this is going to be, in this case, our a, we have to be careful, our a is going to be negative three. So that is our a right over there. So this is going to be equal to x squared. Now, two a x, let me do it in the same colors actually, just for fun. So it's going to be x squared."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3 \u00c2\u00b2 Algebra I High School Math Khan Academy.mp3", "Sentence": "So that is our a right over there. So this is going to be equal to x squared. Now, two a x, let me do it in the same colors actually, just for fun. So it's going to be x squared. Now, what is two times a times x? A is negative three. So two times a is negative six."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3 \u00c2\u00b2 Algebra I High School Math Khan Academy.mp3", "Sentence": "So it's going to be x squared. Now, what is two times a times x? A is negative three. So two times a is negative six. So it's going to be negative six x. So minus six x, that's two times a is the coefficient, and then we have our x there. And then plus a squared."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3 \u00c2\u00b2 Algebra I High School Math Khan Academy.mp3", "Sentence": "So two times a is negative six. So it's going to be negative six x. So minus six x, that's two times a is the coefficient, and then we have our x there. And then plus a squared. Well, if a is negative three, what is negative three times negative three? It's going to be positive nine. And just like that, when we looked at this pattern, we were able to very quickly figure out what this binomial squared actually is."}, {"video_title": "Algebraic expression adding fractions Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "So let's start with some fairly straightforward ones. So let's say that I had, let's say I had A over B plus C over D. And if I actually wanted to add these things so it is just one fraction, how would I do that? Well, what we could do is we could find a common denominator. Well, over here, we don't know what B is, we don't know what D is, but we know a common denominator is just going to be B times D. That is going to be a common multiple of B and D. So we could rewrite this as two fractions with a common denominator BD. So plus BD. Actually, let me color code it a little bit. So A over B is going to be the same thing as what over BD?"}, {"video_title": "Algebraic expression adding fractions Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "Well, over here, we don't know what B is, we don't know what D is, but we know a common denominator is just going to be B times D. That is going to be a common multiple of B and D. So we could rewrite this as two fractions with a common denominator BD. So plus BD. Actually, let me color code it a little bit. So A over B is going to be the same thing as what over BD? Well, to get to BD, I multiply the denominator by D. So let me multiply the numerator by D as well. Then I haven't changed the value of the fraction. I'm just multiplying by D over D. So this is going to be A times D over B times D. Notice, I could divide the numerator and the denominator by D, and I'm going to get back to A over B."}, {"video_title": "Algebraic expression adding fractions Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "So A over B is going to be the same thing as what over BD? Well, to get to BD, I multiply the denominator by D. So let me multiply the numerator by D as well. Then I haven't changed the value of the fraction. I'm just multiplying by D over D. So this is going to be A times D over B times D. Notice, I could divide the numerator and the denominator by D, and I'm going to get back to A over B. And then we could look at the second fraction, C over D. To go from D to BD, we multiplied by B. And so if I multiply the denominator by B, if I don't want to change the value of the fraction, I have to multiply the numerator by B as well. So let's multiply the numerator by B as well, and it's going to be BC."}, {"video_title": "Algebraic expression adding fractions Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "I'm just multiplying by D over D. So this is going to be A times D over B times D. Notice, I could divide the numerator and the denominator by D, and I'm going to get back to A over B. And then we could look at the second fraction, C over D. To go from D to BD, we multiplied by B. And so if I multiply the denominator by B, if I don't want to change the value of the fraction, I have to multiply the numerator by B as well. So let's multiply the numerator by B as well, and it's going to be BC. BC. BC over BD. This is C over D. So what I have here in magenta, this fraction is equivalent to this fraction."}, {"video_title": "Algebraic expression adding fractions Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "So let's multiply the numerator by B as well, and it's going to be BC. BC. BC over BD. This is C over D. So what I have here in magenta, this fraction is equivalent to this fraction. I just multiplied it by D over D, which we can assume is one, if we assume that D is not equal to zero. And then if we just multiply C over D times one, which is the same thing as B over B, if we assume B is not equal to zero, then this fraction and this fraction are equivalent. Now why did I go through all of this trouble?"}, {"video_title": "Algebraic expression adding fractions Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "This is C over D. So what I have here in magenta, this fraction is equivalent to this fraction. I just multiplied it by D over D, which we can assume is one, if we assume that D is not equal to zero. And then if we just multiply C over D times one, which is the same thing as B over B, if we assume B is not equal to zero, then this fraction and this fraction are equivalent. Now why did I go through all of this trouble? Well, now I have a common denominator, so I can add these two fractions. So what's this going to be? Well, common denominator is BD."}, {"video_title": "Algebraic expression adding fractions Introduction to algebra Algebra I Khan Academy.mp3", "Sentence": "Now why did I go through all of this trouble? Well, now I have a common denominator, so I can add these two fractions. So what's this going to be? Well, common denominator is BD. So let me just, so common denominator is BD. And I can just add the numerators, just like you would have done if these were numbers, if this wasn't an algebraic expression. So this is going to be, this is going to be AD plus BC, all of that over BD."}, {"video_title": "Example 5 Factoring a fourth degree polynomial using the perfect square\u201d form Khan Academy.mp3", "Sentence": "But there's something about this that might pop out at you. The thing that pops out at me at least is that 25 is a perfect square, x to the fourth is a perfect square, so 25x to the fourth is a perfect square, and 9 is also a perfect square. So maybe this is the square of some binomial, and to confirm it, this center term has to be 2 times the product of the terms that you're squaring on either end. Let me explain that a little bit better. So 25x to the fourth, that is the same thing as 5x squared squared, right? So it's a perfect square. 9 is the exact same thing as, well, it could be plus or minus 3 squared."}, {"video_title": "Example 5 Factoring a fourth degree polynomial using the perfect square\u201d form Khan Academy.mp3", "Sentence": "Let me explain that a little bit better. So 25x to the fourth, that is the same thing as 5x squared squared, right? So it's a perfect square. 9 is the exact same thing as, well, it could be plus or minus 3 squared. It could be either one. Now what is 30x squared? Well, what happens if we take 5 times plus or minus 3?"}, {"video_title": "Example 5 Factoring a fourth degree polynomial using the perfect square\u201d form Khan Academy.mp3", "Sentence": "9 is the exact same thing as, well, it could be plus or minus 3 squared. It could be either one. Now what is 30x squared? Well, what happens if we take 5 times plus or minus 3? So remember, this needs to be 2 times the product of what's inside the square, the square root of this and the square root of that. So if we take, given that there's a negative sign here and 5 is positive, we want to take the negative 3, right? That's the only way we're going to get a negative over there, so let's just try it with negative 3."}, {"video_title": "Example 5 Factoring a fourth degree polynomial using the perfect square\u201d form Khan Academy.mp3", "Sentence": "Well, what happens if we take 5 times plus or minus 3? So remember, this needs to be 2 times the product of what's inside the square, the square root of this and the square root of that. So if we take, given that there's a negative sign here and 5 is positive, we want to take the negative 3, right? That's the only way we're going to get a negative over there, so let's just try it with negative 3. So what is 2 times 5x squared times negative 3? What is this? Well, 2 times 5x squared is 10x squared times negative 3."}, {"video_title": "Example 5 Factoring a fourth degree polynomial using the perfect square\u201d form Khan Academy.mp3", "Sentence": "That's the only way we're going to get a negative over there, so let's just try it with negative 3. So what is 2 times 5x squared times negative 3? What is this? Well, 2 times 5x squared is 10x squared times negative 3. It is equal to negative 30x squared. We know that this is a perfect square, so we can just rewrite this as this is equal to 5x squared. We do it the same color."}, {"video_title": "Example 5 Factoring a fourth degree polynomial using the perfect square\u201d form Khan Academy.mp3", "Sentence": "Well, 2 times 5x squared is 10x squared times negative 3. It is equal to negative 30x squared. We know that this is a perfect square, so we can just rewrite this as this is equal to 5x squared. We do it the same color. 5x squared minus 3 times 5x squared minus 3. And we saw in the last video why this works, and if you want to verify it for yourself, multiply this out. You will get 25x to the fourth minus 30x squared plus 9."}, {"video_title": "Least common multiple of polynomials Mathematics III High School Math Khan Academy.mp3", "Sentence": "And the second one is seven z to the fourth plus 21 z to the third plus 14 z squared. Now if you're saying, well what is the, you're familiar with least common multiple of two numbers, one way to think about them is if I were to find, say, the least common multiple between, I don't know, four and six, you literally could look at all the multiples and see which ones are, which one is least. So you go four, eight, 12, 16, so on and so forth. You could do the same thing for six. You could go six, 12, 18, 24, so on and so forth. And you immediately see that they do have, they'll actually have multiple common multiples, but the least of the common multiples, you immediately see, is going to be 12. Now another way to think about it is, is to actually factor these numbers out."}, {"video_title": "Least common multiple of polynomials Mathematics III High School Math Khan Academy.mp3", "Sentence": "You could do the same thing for six. You could go six, 12, 18, 24, so on and so forth. And you immediately see that they do have, they'll actually have multiple common multiples, but the least of the common multiples, you immediately see, is going to be 12. Now another way to think about it is, is to actually factor these numbers out. We could view four, you could view four as being two times two if you look at its prime factorization and six is two times three. So you could say that the least common multiple, the LCM of four and six, is going to be, or it's going to be equal to, well it's going to have to have the factors of both of them. So it's going to have to have two fours, two times two, and it's going to have to have a two and a three."}, {"video_title": "Least common multiple of polynomials Mathematics III High School Math Khan Academy.mp3", "Sentence": "Now another way to think about it is, is to actually factor these numbers out. We could view four, you could view four as being two times two if you look at its prime factorization and six is two times three. So you could say that the least common multiple, the LCM of four and six, is going to be, or it's going to be equal to, well it's going to have to have the factors of both of them. So it's going to have to have two fours, two times two, and it's going to have to have a two and a three. Well we already have a two, in fact we have two of them, so in order to be divisible by three, we have to be, in order to be divisible by six, we have to have three as one of the factors. And so when you look at it that way, you say, hey look, we have to contain all of the factors of each of them, we have to have at least two twos, and we have to have at least one three, because the two twos take care of this one two right over there, and you see that this is also going to be equal to 12. Now when we think about it for polynomials, we're going to think about it a little bit more, it's essentially the same idea, but we're going to think about it a little bit more with the second lens, where we're going to think about the factors, and say well the least common multiple needs to contain the factors of both, but it shouldn't contain more than that."}, {"video_title": "Least common multiple of polynomials Mathematics III High School Math Khan Academy.mp3", "Sentence": "So it's going to have to have two fours, two times two, and it's going to have to have a two and a three. Well we already have a two, in fact we have two of them, so in order to be divisible by three, we have to be, in order to be divisible by six, we have to have three as one of the factors. And so when you look at it that way, you say, hey look, we have to contain all of the factors of each of them, we have to have at least two twos, and we have to have at least one three, because the two twos take care of this one two right over there, and you see that this is also going to be equal to 12. Now when we think about it for polynomials, we're going to think about it a little bit more, it's essentially the same idea, but we're going to think about it a little bit more with the second lens, where we're going to think about the factors, and say well the least common multiple needs to contain the factors of both, but it shouldn't contain more than that. You can always find a multiple of two polynomials by just multiplying them, but we don't want to find just any multiple, we want to find the least common multiple. So let's factor each of them. So this first one, three z to the third minus six z squared minus nine z."}, {"video_title": "Least common multiple of polynomials Mathematics III High School Math Khan Academy.mp3", "Sentence": "Now when we think about it for polynomials, we're going to think about it a little bit more, it's essentially the same idea, but we're going to think about it a little bit more with the second lens, where we're going to think about the factors, and say well the least common multiple needs to contain the factors of both, but it shouldn't contain more than that. You can always find a multiple of two polynomials by just multiplying them, but we don't want to find just any multiple, we want to find the least common multiple. So let's factor each of them. So this first one, three z to the third minus six z squared minus nine z. Let's see, immediately, let's see, all of these terms are divisible by three z, so let's factor out a three z. So it's three z times z squared, if you factor out a three z out of that. See, it's going to be minus two z, if you factor out a three z out of that, and then minus three."}, {"video_title": "Least common multiple of polynomials Mathematics III High School Math Khan Academy.mp3", "Sentence": "So this first one, three z to the third minus six z squared minus nine z. Let's see, immediately, let's see, all of these terms are divisible by three z, so let's factor out a three z. So it's three z times z squared, if you factor out a three z out of that. See, it's going to be minus two z, if you factor out a three z out of that, and then minus three. And notice, if you were to distribute this three z back, you would get exactly what we have up here. And so let's see, can we factor this further? Can we think of two numbers that if you multiply them we get negative three, and if we add them we get negative two, and one's going to be positive, one's going to be negative, since their product is negative."}, {"video_title": "Least common multiple of polynomials Mathematics III High School Math Khan Academy.mp3", "Sentence": "See, it's going to be minus two z, if you factor out a three z out of that, and then minus three. And notice, if you were to distribute this three z back, you would get exactly what we have up here. And so let's see, can we factor this further? Can we think of two numbers that if you multiply them we get negative three, and if we add them we get negative two, and one's going to be positive, one's going to be negative, since their product is negative. So let's see, it sounds like negative three and positive one, so we could rewrite this as three z times z plus one times z minus three. I think I've factored this first polynomial about as much as I can. One times negative three is negative three."}, {"video_title": "Least common multiple of polynomials Mathematics III High School Math Khan Academy.mp3", "Sentence": "Can we think of two numbers that if you multiply them we get negative three, and if we add them we get negative two, and one's going to be positive, one's going to be negative, since their product is negative. So let's see, it sounds like negative three and positive one, so we could rewrite this as three z times z plus one times z minus three. I think I've factored this first polynomial about as much as I can. One times negative three is negative three. One z minus three z is negative two z, so that looks good. So now let's factor this other character over here, this fourth degree polynomial. So every one of those terms look like they're divisible by seven z squared, so I could write this as seven z squared times z squared, when you factor out a seven z squared here you're just left with z squared."}, {"video_title": "Least common multiple of polynomials Mathematics III High School Math Khan Academy.mp3", "Sentence": "One times negative three is negative three. One z minus three z is negative two z, so that looks good. So now let's factor this other character over here, this fourth degree polynomial. So every one of those terms look like they're divisible by seven z squared, so I could write this as seven z squared times z squared, when you factor out a seven z squared here you're just left with z squared. And then plus, 21 divided by seven is three, z to the third divided by z squared is z, and then plus 14 divided by seven is two, z squared divided by z squared is one, so it's just going to be a two there. And so this is going to be the same thing as seven z squared and this can be factored into, let's see, two times one is two, two plus one is three, so z plus one times z plus two. Now let's think about the least common multiple."}, {"video_title": "Least common multiple of polynomials Mathematics III High School Math Khan Academy.mp3", "Sentence": "So every one of those terms look like they're divisible by seven z squared, so I could write this as seven z squared times z squared, when you factor out a seven z squared here you're just left with z squared. And then plus, 21 divided by seven is three, z to the third divided by z squared is z, and then plus 14 divided by seven is two, z squared divided by z squared is one, so it's just going to be a two there. And so this is going to be the same thing as seven z squared and this can be factored into, let's see, two times one is two, two plus one is three, so z plus one times z plus two. Now let's think about the least common multiple. We've factored each of these just the way that we, when we did the prime factorization for regular numbers, now we have factored this down to as simple expressions as we will find useful. And so the least common multiple of these two things has to contain each of these factors. So the least common multiple's got to contain a three z, it's got to contain, and let me expand it out a little bit, it's got to contain a three, it's got to contain a z, it's got to contain a z plus one, z plus one, it's, I don't have to write a little dot there, since I'm, it's got to contain a z plus one, it's got to contain a z minus three, let's see, it's got to contain a seven."}, {"video_title": "Least common multiple of polynomials Mathematics III High School Math Khan Academy.mp3", "Sentence": "Now let's think about the least common multiple. We've factored each of these just the way that we, when we did the prime factorization for regular numbers, now we have factored this down to as simple expressions as we will find useful. And so the least common multiple of these two things has to contain each of these factors. So the least common multiple's got to contain a three z, it's got to contain, and let me expand it out a little bit, it's got to contain a three, it's got to contain a z, it's got to contain a z plus one, z plus one, it's, I don't have to write a little dot there, since I'm, it's got to contain a z plus one, it's got to contain a z minus three, let's see, it's got to contain a seven. We do not have a seven here yet, so we have to include, we have to include a seven, so I'll put the seven out front with the numbers. It's got to include a seven. It's got to include a z squared."}, {"video_title": "Least common multiple of polynomials Mathematics III High School Math Khan Academy.mp3", "Sentence": "So the least common multiple's got to contain a three z, it's got to contain, and let me expand it out a little bit, it's got to contain a three, it's got to contain a z, it's got to contain a z plus one, z plus one, it's, I don't have to write a little dot there, since I'm, it's got to contain a z plus one, it's got to contain a z minus three, let's see, it's got to contain a seven. We do not have a seven here yet, so we have to include, we have to include a seven, so I'll put the seven out front with the numbers. It's got to include a seven. It's got to include a z squared. Well we only have a z right now, so let's throw in another z. So I could throw in another, I could write, I could put a z out front, or I could just make this a squared. It still contains that z, but now we contain another z, or we're multiplying by another z to have z squared."}, {"video_title": "Least common multiple of polynomials Mathematics III High School Math Khan Academy.mp3", "Sentence": "It's got to include a z squared. Well we only have a z right now, so let's throw in another z. So I could throw in another, I could write, I could put a z out front, or I could just make this a squared. It still contains that z, but now we contain another z, or we're multiplying by another z to have z squared. See, we already have a z plus one in here. We need a z plus two as well. Z plus two as well."}, {"video_title": "Least common multiple of polynomials Mathematics III High School Math Khan Academy.mp3", "Sentence": "It still contains that z, but now we contain another z, or we're multiplying by another z to have z squared. See, we already have a z plus one in here. We need a z plus two as well. Z plus two as well. And there you have it. This is the least common multiple. If I were to write it all out in a neutral color, it's going to be 21 z squared times z plus one times z minus three times z plus two."}, {"video_title": "Least common multiple of polynomials Mathematics III High School Math Khan Academy.mp3", "Sentence": "Z plus two as well. And there you have it. This is the least common multiple. If I were to write it all out in a neutral color, it's going to be 21 z squared times z plus one times z minus three times z plus two. I say two and I write six. Z plus two. And we are all done."}, {"video_title": "Least common multiple of polynomials Mathematics III High School Math Khan Academy.mp3", "Sentence": "If I were to write it all out in a neutral color, it's going to be 21 z squared times z plus one times z minus three times z plus two. I say two and I write six. Z plus two. And we are all done. And I really want you to appreciate, this is the exact same thing we're doing when we're doing, or a very similar thing than what we're doing when we're finding least common multiples of regular numbers. We're looking at their factors, and in the case of numbers, prime factors, and then we say, okay, the least common multiple has to contain, each of, has to be a superset, has to contain all of these, but we don't want to contain, you know, I could multiply this times 100. It's still going to be a common multiple of these two, but it's no longer the least common multiple."}, {"video_title": "Least common multiple of polynomials Mathematics III High School Math Khan Academy.mp3", "Sentence": "And we are all done. And I really want you to appreciate, this is the exact same thing we're doing when we're doing, or a very similar thing than what we're doing when we're finding least common multiples of regular numbers. We're looking at their factors, and in the case of numbers, prime factors, and then we say, okay, the least common multiple has to contain, each of, has to be a superset, has to contain all of these, but we don't want to contain, you know, I could multiply this times 100. It's still going to be a common multiple of these two, but it's no longer the least common multiple. Likewise, 12 is the least common multiple of four and six. If I just wanted a common multiple, I could multiply that times 100, or 1,200 would also be a multiple of four and six, but it wouldn't be the least common multiple, so we don't want to do that. Hopefully you found that interesting."}, {"video_title": "Multiplying rational expressions Precalculus Khan Academy.mp3", "Sentence": "All right, now let's work through this together. So multiplying rational expressions like this, it's very analogous to multiplying fractions. For example, if I were to multiply 6 25ths times 15 over nine, there's a few ways you could do it. You could just multiply six times 15 in the numerator and 25 times nine in the denominator. But the way that many of us approach it so that it's easier to reduce to lowest terms is to factor things, to realize that, look, six is two times three, nine is three times three, 15 is three times five, and 25 is five times five. And then you can realize in your eventual product, you're going to have a five in the numerator, five in the denominator. You can, five divided by five is one, three divided by three is one, and then three divided by three is one."}, {"video_title": "Multiplying rational expressions Precalculus Khan Academy.mp3", "Sentence": "You could just multiply six times 15 in the numerator and 25 times nine in the denominator. But the way that many of us approach it so that it's easier to reduce to lowest terms is to factor things, to realize that, look, six is two times three, nine is three times three, 15 is three times five, and 25 is five times five. And then you can realize in your eventual product, you're going to have a five in the numerator, five in the denominator. You can, five divided by five is one, three divided by three is one, and then three divided by three is one. And so all you'd be left with is that two and that five. So this is going to be equal to 2 5ths. So we'll do the analogous thing here with these rational expressions."}, {"video_title": "Multiplying rational expressions Precalculus Khan Academy.mp3", "Sentence": "You can, five divided by five is one, three divided by three is one, and then three divided by three is one. And so all you'd be left with is that two and that five. So this is going to be equal to 2 5ths. So we'll do the analogous thing here with these rational expressions. We're going to factor them, all of them, then the numerators and the denominators, and then we'll see if we can divide the numerator and the denominator by the same thing. Now, the one thing we have to make sure of as we do that is we keep track of the domain because these rational expressions here, they might have x values that make their denominators equal to zero. And so even if we reduce to lowest terms and we get rid of those expressions, in order for the expressions to be the same expression, we have to constrain the domain in the same way."}, {"video_title": "Multiplying rational expressions Precalculus Khan Academy.mp3", "Sentence": "So we'll do the analogous thing here with these rational expressions. We're going to factor them, all of them, then the numerators and the denominators, and then we'll see if we can divide the numerator and the denominator by the same thing. Now, the one thing we have to make sure of as we do that is we keep track of the domain because these rational expressions here, they might have x values that make their denominators equal to zero. And so even if we reduce to lowest terms and we get rid of those expressions, in order for the expressions to be the same expression, we have to constrain the domain in the same way. So let's get started. So this is going to be equal to, I'll just rewrite everything, x squared minus nine. How do we factor that?"}, {"video_title": "Multiplying rational expressions Precalculus Khan Academy.mp3", "Sentence": "And so even if we reduce to lowest terms and we get rid of those expressions, in order for the expressions to be the same expression, we have to constrain the domain in the same way. So let's get started. So this is going to be equal to, I'll just rewrite everything, x squared minus nine. How do we factor that? Well, that's going to be a difference of squares. We could write that as x plus three times x minus three. And then that is going to be over this business."}, {"video_title": "Multiplying rational expressions Precalculus Khan Academy.mp3", "Sentence": "How do we factor that? Well, that's going to be a difference of squares. We could write that as x plus three times x minus three. And then that is going to be over this business. And let's see, five squared is 25, negative five plus negative five is negative 10. So this is going to be x minus five times x minus five. If what I'm just doing here with the factoring is not making sense, I encourage you to review factoring on Khan Academy."}, {"video_title": "Multiplying rational expressions Precalculus Khan Academy.mp3", "Sentence": "And then that is going to be over this business. And let's see, five squared is 25, negative five plus negative five is negative 10. So this is going to be x minus five times x minus five. If what I'm just doing here with the factoring is not making sense, I encourage you to review factoring on Khan Academy. And then we multiply that times, let's see, in this numerator here, I can factor out a four. So that's going to be four times x minus five, which is going to be useful. I have an x minus five there, x minus five there."}, {"video_title": "Multiplying rational expressions Precalculus Khan Academy.mp3", "Sentence": "If what I'm just doing here with the factoring is not making sense, I encourage you to review factoring on Khan Academy. And then we multiply that times, let's see, in this numerator here, I can factor out a four. So that's going to be four times x minus five, which is going to be useful. I have an x minus five there, x minus five there. And then that's going to be over, let's see, this expression over here, two plus three is five, two times three is six. So it's going to be x plus two times x plus three. Now, before I start reducing to lowest terms, let's think about the domain here."}, {"video_title": "Multiplying rational expressions Precalculus Khan Academy.mp3", "Sentence": "I have an x minus five there, x minus five there. And then that's going to be over, let's see, this expression over here, two plus three is five, two times three is six. So it's going to be x plus two times x plus three. Now, before I start reducing to lowest terms, let's think about the domain here. And the domain is going to be constrained by things that make these denominators equal to zero. So the domain would be all real numbers except x cannot equal five. Let me write it over here."}, {"video_title": "Multiplying rational expressions Precalculus Khan Academy.mp3", "Sentence": "Now, before I start reducing to lowest terms, let's think about the domain here. And the domain is going to be constrained by things that make these denominators equal to zero. So the domain would be all real numbers except x cannot equal five. Let me write it over here. X cannot equal five, because if that happened, then this denominator would be equal to zero. X cannot be equal to negative two. X could not be equal to negative two because that would make the denominator here zero, which would make the denominator here zero."}, {"video_title": "Multiplying rational expressions Precalculus Khan Academy.mp3", "Sentence": "Let me write it over here. X cannot equal five, because if that happened, then this denominator would be equal to zero. X cannot be equal to negative two. X could not be equal to negative two because that would make the denominator here zero, which would make the denominator here zero. And x cannot be equal to negative three. So the domain is constrained in this way. And we have to carry this throughout."}, {"video_title": "Multiplying rational expressions Precalculus Khan Academy.mp3", "Sentence": "X could not be equal to negative two because that would make the denominator here zero, which would make the denominator here zero. And x cannot be equal to negative three. So the domain is constrained in this way. And we have to carry this throughout. No matter what we do to the expression, this is the constraints on our domain. With that out of the way, now we can reduce to lowest terms. So we have an x plus three in the numerator, x plus three in the denominator, x minus five in the numerator, x minus five in the denominator."}, {"video_title": "Multiplying rational expressions Precalculus Khan Academy.mp3", "Sentence": "And we have to carry this throughout. No matter what we do to the expression, this is the constraints on our domain. With that out of the way, now we can reduce to lowest terms. So we have an x plus three in the numerator, x plus three in the denominator, x minus five in the numerator, x minus five in the denominator. And I think we've gone about as far as we can. And so when we multiply the numerators, we are going to get, this business is going to be four times x minus three, four times x minus three over, we have an x minus five here, x minus five. And then we have an x plus two."}, {"video_title": "Multiplying rational expressions Precalculus Khan Academy.mp3", "Sentence": "So we have an x plus three in the numerator, x plus three in the denominator, x minus five in the numerator, x minus five in the denominator. And I think we've gone about as far as we can. And so when we multiply the numerators, we are going to get, this business is going to be four times x minus three, four times x minus three over, we have an x minus five here, x minus five. And then we have an x plus two. We have an x plus two. And we could leave it like this if you want. In some cases, people like to multiply the things out, but we're done."}, {"video_title": "Multiplying rational expressions Precalculus Khan Academy.mp3", "Sentence": "And then we have an x plus two. We have an x plus two. And we could leave it like this if you want. In some cases, people like to multiply the things out, but we're done. We've just finished multiplying these rational expressions. And we have to remind ourselves that x cannot be equal to any of these things. Now, the way that we've simplified it, we still have an x minus five right over here."}, {"video_title": "Multiplying rational expressions Precalculus Khan Academy.mp3", "Sentence": "In some cases, people like to multiply the things out, but we're done. We've just finished multiplying these rational expressions. And we have to remind ourselves that x cannot be equal to any of these things. Now, the way that we've simplified it, we still have an x minus five right over here. So it might be redundant to say that x cannot be equal to five because that's still the case in our reduced terms expression here. And that's true also of the x cannot be equal to negative two. We still have an x plus two here."}, {"video_title": "Multiplying rational expressions Precalculus Khan Academy.mp3", "Sentence": "Now, the way that we've simplified it, we still have an x minus five right over here. So it might be redundant to say that x cannot be equal to five because that's still the case in our reduced terms expression here. And that's true also of the x cannot be equal to negative two. We still have an x plus two here. So still, even in this expression, it's pretty clear that x cannot be equal to negative two. But the x cannot equal negative three isn't so obvious when you just look at this expression. But in order for this expression to be completely equivalent to the original, it has to have the same domain."}, {"video_title": "Powers of zero Exponents, radicals, and scientific notation Pre-algebra Khan Academy.mp3", "Sentence": "In this video, we're going to talk about powers of zero. And just as a little bit of a reminder, let's start with a non-zero number, just to remind ourselves what exponentiation is all about. So if I were to take two to the first power, one way to think about this is we always start with a one, and then we multiply this base that many times times that one. So here, we're only gonna have one, two. So it's gonna be one times two, which is of course equal to two. If I were to say what is two to the second power, well, that's going to be equal to one times, and now I'm gonna have two twos, so times two times two, which is equal to four. And you could keep going like that."}, {"video_title": "Powers of zero Exponents, radicals, and scientific notation Pre-algebra Khan Academy.mp3", "Sentence": "So here, we're only gonna have one, two. So it's gonna be one times two, which is of course equal to two. If I were to say what is two to the second power, well, that's going to be equal to one times, and now I'm gonna have two twos, so times two times two, which is equal to four. And you could keep going like that. Now, the reason why I have this one here, and we've done this before, is to justify, and there's many other good reasons why two to the zero power should be equal to one. But you could see if we use the same exact idea here, you start with a one, and then you multiply it by two zero times. Well, that's just going to end up with a one."}, {"video_title": "Powers of zero Exponents, radicals, and scientific notation Pre-algebra Khan Academy.mp3", "Sentence": "And you could keep going like that. Now, the reason why I have this one here, and we've done this before, is to justify, and there's many other good reasons why two to the zero power should be equal to one. But you could see if we use the same exact idea here, you start with a one, and then you multiply it by two zero times. Well, that's just going to end up with a one. So, so far, I've told you this video's about powers of zero, but I've been doing powers of two, so let's focus on zero now. So what do you think zero to the first power is going to be? Pause this video and try to figure that out."}, {"video_title": "Powers of zero Exponents, radicals, and scientific notation Pre-algebra Khan Academy.mp3", "Sentence": "Well, that's just going to end up with a one. So, so far, I've told you this video's about powers of zero, but I've been doing powers of two, so let's focus on zero now. So what do you think zero to the first power is going to be? Pause this video and try to figure that out. Well, you do the exact same idea. You start with a one, and then multiply it by zero one time. So times zero, and this is going to be equal to zero."}, {"video_title": "Powers of zero Exponents, radicals, and scientific notation Pre-algebra Khan Academy.mp3", "Sentence": "Pause this video and try to figure that out. Well, you do the exact same idea. You start with a one, and then multiply it by zero one time. So times zero, and this is going to be equal to zero. What do you think zero to the second power is going to be equal to? Pause this video and think about that. Well, it's gonna be one times zero twice."}, {"video_title": "Powers of zero Exponents, radicals, and scientific notation Pre-algebra Khan Academy.mp3", "Sentence": "So times zero, and this is going to be equal to zero. What do you think zero to the second power is going to be equal to? Pause this video and think about that. Well, it's gonna be one times zero twice. So times zero times zero, and I think you see where this is going. This is also going to be equal to zero. What do you think zero to some arbitrary positive, positive integer is going to be?"}, {"video_title": "Powers of zero Exponents, radicals, and scientific notation Pre-algebra Khan Academy.mp3", "Sentence": "Well, it's gonna be one times zero twice. So times zero times zero, and I think you see where this is going. This is also going to be equal to zero. What do you think zero to some arbitrary positive, positive integer is going to be? Well, it's going to be equal to one times zero, that positive integer number of times. So once again, it's going to be equal to zero. And in general, you can extend that zero to any positive value exponent is going to give you zero."}, {"video_title": "Powers of zero Exponents, radicals, and scientific notation Pre-algebra Khan Academy.mp3", "Sentence": "What do you think zero to some arbitrary positive, positive integer is going to be? Well, it's going to be equal to one times zero, that positive integer number of times. So once again, it's going to be equal to zero. And in general, you can extend that zero to any positive value exponent is going to give you zero. So that's pretty straightforward. But there is an interesting edge case here. What do you think zero to the zeroth power should be?"}, {"video_title": "Powers of zero Exponents, radicals, and scientific notation Pre-algebra Khan Academy.mp3", "Sentence": "And in general, you can extend that zero to any positive value exponent is going to give you zero. So that's pretty straightforward. But there is an interesting edge case here. What do you think zero to the zeroth power should be? Pause this video and think about that. Well, this is actually contested. Different people will tell you different things."}, {"video_title": "Powers of zero Exponents, radicals, and scientific notation Pre-algebra Khan Academy.mp3", "Sentence": "What do you think zero to the zeroth power should be? Pause this video and think about that. Well, this is actually contested. Different people will tell you different things. If you use the intuition behind exponentiation that we've been using in this video, you would say, all right, I would start with a one and then multiply it by zero zero times. Or in other words, I just wouldn't multiply it by zero, in which case I'm just left with the one. The zero to the zeroth power should be equal to one."}, {"video_title": "Powers of zero Exponents, radicals, and scientific notation Pre-algebra Khan Academy.mp3", "Sentence": "Different people will tell you different things. If you use the intuition behind exponentiation that we've been using in this video, you would say, all right, I would start with a one and then multiply it by zero zero times. Or in other words, I just wouldn't multiply it by zero, in which case I'm just left with the one. The zero to the zeroth power should be equal to one. Other folks would say, hey, no, I'm with a zero and that's the zeroth power, maybe it should be a zero. And that's why a lot of folks leave it undefined. Most of the time, you're going to see zero to the zeroth power either being undefined or that it is equal to one."}, {"video_title": "Simplifying with exponent properties Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "So let me actually color code these exponents just so we can keep track of them a little better. So that's the 1 half power in blue. This is the 5th root here in magenta. And let's see in green, let's think about this 3rd power. So one way to think about this 5th root is that this is the exact same thing as raising this 6 to the 1 5th power. So let's write it like that. So this part right over here we could rewrite as 6 to the 1 5th power."}, {"video_title": "Simplifying with exponent properties Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "And let's see in green, let's think about this 3rd power. So one way to think about this 5th root is that this is the exact same thing as raising this 6 to the 1 5th power. So let's write it like that. So this part right over here we could rewrite as 6 to the 1 5th power. And then that whole thing gets raised to the 3rd power. And of course we have this 6 to the 1 half power out here. 6 to the 1 half power times all of this business right over here."}, {"video_title": "Simplifying with exponent properties Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "So this part right over here we could rewrite as 6 to the 1 5th power. And then that whole thing gets raised to the 3rd power. And of course we have this 6 to the 1 half power out here. 6 to the 1 half power times all of this business right over here. Now what happens if we raise something to an exponent and then raise that whole thing to another exponent? Well we've already seen in our exponent properties that's the equivalent of raising this to the product of these two exponents. So this part right over here could be rewritten as 6 to the 3 5th power."}, {"video_title": "Simplifying with exponent properties Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "6 to the 1 half power times all of this business right over here. Now what happens if we raise something to an exponent and then raise that whole thing to another exponent? Well we've already seen in our exponent properties that's the equivalent of raising this to the product of these two exponents. So this part right over here could be rewritten as 6 to the 3 5th power. And of course we're multiplying that times 6 to the 1 half power. And if you're multiplying some base to this exponent and then the same base again to another exponent, we know that this is going to be the same thing. And actually we could put these equal signs the whole way because these all equal each other."}, {"video_title": "Simplifying with exponent properties Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "So this part right over here could be rewritten as 6 to the 3 5th power. And of course we're multiplying that times 6 to the 1 half power. And if you're multiplying some base to this exponent and then the same base again to another exponent, we know that this is going to be the same thing. And actually we could put these equal signs the whole way because these all equal each other. This is the same thing as 6 being raised to the 1 half plus 3 5th power. 1 half plus 3 over 5. Now what's 1 half plus 3 over 5?"}, {"video_title": "Simplifying with exponent properties Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "And actually we could put these equal signs the whole way because these all equal each other. This is the same thing as 6 being raised to the 1 half plus 3 5th power. 1 half plus 3 over 5. Now what's 1 half plus 3 over 5? Well we could find a common denominator. It would be 10. So that's the same thing as 6 to the 1 half."}, {"video_title": "Simplifying with exponent properties Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "Now what's 1 half plus 3 over 5? Well we could find a common denominator. It would be 10. So that's the same thing as 6 to the 1 half. We can write it as 5 over 10 plus 3 5th is the same thing as 6 over 10. 6 over 10 power which is the same thing. And we deserve a little bit of a drum roll here."}, {"video_title": "Simplifying with exponent properties Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "So that's the same thing as 6 to the 1 half. We can write it as 5 over 10 plus 3 5th is the same thing as 6 over 10. 6 over 10 power which is the same thing. And we deserve a little bit of a drum roll here. This wasn't that long of a problem. 6 to the 11 10th power. And so that looks pretty simplified to me."}, {"video_title": "Example 3 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So let me just rewrite it. So 4x squared plus 40x minus 300 is equal to 0. So just as a first step here, I don't like having this 4 out front as a coefficient on the x squared term. I'd prefer if that was a 1. So let's just divide both sides of this equation by 4. So let's just divide everything by 4. So this divided by 4, this divided by 4, that divided by 4, and the 0 divided by 4."}, {"video_title": "Example 3 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "I'd prefer if that was a 1. So let's just divide both sides of this equation by 4. So let's just divide everything by 4. So this divided by 4, this divided by 4, that divided by 4, and the 0 divided by 4. Just dividing both sides by 4. So this will simplify to x squared plus 10x. And I can obviously do that because as long as I do whatever I do to the left-hand side, I also do to the right-hand side, that will make the equality continue to be valid."}, {"video_title": "Example 3 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So this divided by 4, this divided by 4, that divided by 4, and the 0 divided by 4. Just dividing both sides by 4. So this will simplify to x squared plus 10x. And I can obviously do that because as long as I do whatever I do to the left-hand side, I also do to the right-hand side, that will make the equality continue to be valid. So that's why I can do that. So 40 divided by 4 is 10x, and then 300 divided by 4 is what? That is 75."}, {"video_title": "Example 3 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "And I can obviously do that because as long as I do whatever I do to the left-hand side, I also do to the right-hand side, that will make the equality continue to be valid. So that's why I can do that. So 40 divided by 4 is 10x, and then 300 divided by 4 is what? That is 75. Let me verify that. 400, 4 goes into 30 7 times. 7 times 4 is 28."}, {"video_title": "Example 3 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "That is 75. Let me verify that. 400, 4 goes into 30 7 times. 7 times 4 is 28. You subtract, you get a remainder of 2, bring down the 0. 4 goes into 20 5 times. 5 times 4 is 20."}, {"video_title": "Example 3 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "7 times 4 is 28. You subtract, you get a remainder of 2, bring down the 0. 4 goes into 20 5 times. 5 times 4 is 20. Subtract 0. So it goes 75 times. This is minus 75 is equal to 0."}, {"video_title": "Example 3 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "5 times 4 is 20. Subtract 0. So it goes 75 times. This is minus 75 is equal to 0. And right when you look at this, just the way it's written, you might try to factor this in some way, but it's pretty clear this is not a complete square, or this is not a perfect square trinomial. Because if you look at this term right here, this 10, half of this 10 is 5, and 5 squared is not 75. So this is not a perfect square."}, {"video_title": "Example 3 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "This is minus 75 is equal to 0. And right when you look at this, just the way it's written, you might try to factor this in some way, but it's pretty clear this is not a complete square, or this is not a perfect square trinomial. Because if you look at this term right here, this 10, half of this 10 is 5, and 5 squared is not 75. So this is not a perfect square. So what we want to do is somehow turn whatever we have on the left-hand side into a perfect square. And I'm going to start out by kind of getting this 75 out of the way. You'll sometimes see it where people leave the 75 on the left-hand side."}, {"video_title": "Example 3 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So this is not a perfect square. So what we want to do is somehow turn whatever we have on the left-hand side into a perfect square. And I'm going to start out by kind of getting this 75 out of the way. You'll sometimes see it where people leave the 75 on the left-hand side. I'm going to put it on the right-hand side, just so it kind of clears things up a little bit. So let's add 75 to both sides to get rid of the 75 from the left-hand side of the equation. And so we get x squared plus 10x, and then negative 75 plus 75, those guys cancel out, and I'm going to leave some space here because we're going to add something here to complete the square."}, {"video_title": "Example 3 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "You'll sometimes see it where people leave the 75 on the left-hand side. I'm going to put it on the right-hand side, just so it kind of clears things up a little bit. So let's add 75 to both sides to get rid of the 75 from the left-hand side of the equation. And so we get x squared plus 10x, and then negative 75 plus 75, those guys cancel out, and I'm going to leave some space here because we're going to add something here to complete the square. That is equal to 75. So all I did is add 75 to both sides of this equation. Now in this step, this is where this is really the meat of completing the square."}, {"video_title": "Example 3 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "And so we get x squared plus 10x, and then negative 75 plus 75, those guys cancel out, and I'm going to leave some space here because we're going to add something here to complete the square. That is equal to 75. So all I did is add 75 to both sides of this equation. Now in this step, this is where this is really the meat of completing the square. I want to add something to both sides of this equation. I can't add to only one side of the equation. So I want to add something to both sides of this equation so that this left-hand side becomes a perfect square."}, {"video_title": "Example 3 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Now in this step, this is where this is really the meat of completing the square. I want to add something to both sides of this equation. I can't add to only one side of the equation. So I want to add something to both sides of this equation so that this left-hand side becomes a perfect square. And the way we can do that, and we looked at, we saw this in the last video where we constructed a perfect square trinomial, is that this last term, or let's just say this, what we see on the left-hand side, not the last term, this expression on the left-hand side, it will be a perfect square if we have a term, a constant term, that is the square of half of the coefficient on the first degree term. So the coefficient here is 10. Half of 10 is 5."}, {"video_title": "Example 3 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So I want to add something to both sides of this equation so that this left-hand side becomes a perfect square. And the way we can do that, and we looked at, we saw this in the last video where we constructed a perfect square trinomial, is that this last term, or let's just say this, what we see on the left-hand side, not the last term, this expression on the left-hand side, it will be a perfect square if we have a term, a constant term, that is the square of half of the coefficient on the first degree term. So the coefficient here is 10. Half of 10 is 5. 5 squared is 25. So I'm going to add 25 to the left-hand side. And of course, in order to maintain the equality, anything I do to the left-hand side, I also have to do to the right-hand side."}, {"video_title": "Example 3 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Half of 10 is 5. 5 squared is 25. So I'm going to add 25 to the left-hand side. And of course, in order to maintain the equality, anything I do to the left-hand side, I also have to do to the right-hand side. And now we see that this is a perfect square. We say, hey, what two numbers, if I add them, I get 10, and when I multiply them, I get 25. Well, that's 5 and 5."}, {"video_title": "Example 3 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "And of course, in order to maintain the equality, anything I do to the left-hand side, I also have to do to the right-hand side. And now we see that this is a perfect square. We say, hey, what two numbers, if I add them, I get 10, and when I multiply them, I get 25. Well, that's 5 and 5. So when we factor this, what we see on the left-hand side simplifies to, this is x plus 5 squared. x plus 5 times x plus 5. And you can look at the videos on factoring if you find that confusing."}, {"video_title": "Example 3 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Well, that's 5 and 5. So when we factor this, what we see on the left-hand side simplifies to, this is x plus 5 squared. x plus 5 times x plus 5. And you can look at the videos on factoring if you find that confusing. Or you can look at the last video on constructing perfect square trinomials. I encourage you to square this and see that you get exactly this. And this will be equal to 75 plus 25, which is equal to 100."}, {"video_title": "Example 3 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "And you can look at the videos on factoring if you find that confusing. Or you can look at the last video on constructing perfect square trinomials. I encourage you to square this and see that you get exactly this. And this will be equal to 75 plus 25, which is equal to 100. And so now we're saying that something squared is equal to 100. So really this something right over here, if I say something squared is equal to 100, that means that that something is one of the square roots of 100. And we know that 100 has two square roots."}, {"video_title": "Example 3 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "And this will be equal to 75 plus 25, which is equal to 100. And so now we're saying that something squared is equal to 100. So really this something right over here, if I say something squared is equal to 100, that means that that something is one of the square roots of 100. And we know that 100 has two square roots. It has positive 10 and it has negative 10. So we could say that x plus 5, the something that we were squaring, that must be one of the square roots of 100. So that must be equal to the plus or minus square root of 100, or plus or minus 10."}, {"video_title": "Example 3 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "And we know that 100 has two square roots. It has positive 10 and it has negative 10. So we could say that x plus 5, the something that we were squaring, that must be one of the square roots of 100. So that must be equal to the plus or minus square root of 100, or plus or minus 10. Or we could separate it out. We could say that x plus 5 is equal to 10, or x plus 5 is equal to negative 10. On this side right here, I can just subtract 5 from both sides of this equation, and I would get, I'll just write it out, subtracting 5 from both sides, I get x is equal to 5."}, {"video_title": "Example 3 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So that must be equal to the plus or minus square root of 100, or plus or minus 10. Or we could separate it out. We could say that x plus 5 is equal to 10, or x plus 5 is equal to negative 10. On this side right here, I can just subtract 5 from both sides of this equation, and I would get, I'll just write it out, subtracting 5 from both sides, I get x is equal to 5. And over here I could subtract 5 from both sides again. I subtracted 5 in both cases. Subtract 5 again, and I can get x is equal to negative 15."}, {"video_title": "Example 3 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "On this side right here, I can just subtract 5 from both sides of this equation, and I would get, I'll just write it out, subtracting 5 from both sides, I get x is equal to 5. And over here I could subtract 5 from both sides again. I subtracted 5 in both cases. Subtract 5 again, and I can get x is equal to negative 15. So those are my two solutions that I got to solve this equation. We can verify that they actually work, and I'll do that in blue. So let's try with 5."}, {"video_title": "Example 3 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Subtract 5 again, and I can get x is equal to negative 15. So those are my two solutions that I got to solve this equation. We can verify that they actually work, and I'll do that in blue. So let's try with 5. I'll just do one of them. I'll leave the other one for you. I'll leave the other one for you to verify that it works."}, {"video_title": "Example 3 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So let's try with 5. I'll just do one of them. I'll leave the other one for you. I'll leave the other one for you to verify that it works. So 4 times x squared, so 4 times 25, plus 40 times 5, minus 300 needs to be equal to 0. 4 times 25 is 100. 40 times 5 is 200."}, {"video_title": "Example 3 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "I'll leave the other one for you to verify that it works. So 4 times x squared, so 4 times 25, plus 40 times 5, minus 300 needs to be equal to 0. 4 times 25 is 100. 40 times 5 is 200. We're going to subtract that 300. 100 plus 200 minus 300, that definitely equals 0. So x equals 5 worked, and I think you'll find that x equals negative 15 will also work when you substitute it into this right over here."}, {"video_title": "Factoring perfect squares missing values Mathematics II High School Math Khan Academy.mp3", "Sentence": "So what I want to figure out in this video is what is c, given the information that we have right over here, what is c going to be equal to and what is d going to be equal to? And like always, pause the video and see if you can figure it out. Well, let's work through this together. We're saying that x squared plus five x plus c can be rewritten as x plus d squared. So let me write that down. So this part, this part, x squared plus five x plus c, we're saying that that could be written as x plus d squared. This is equal to x plus d squared."}, {"video_title": "Factoring perfect squares missing values Mathematics II High School Math Khan Academy.mp3", "Sentence": "We're saying that x squared plus five x plus c can be rewritten as x plus d squared. So let me write that down. So this part, this part, x squared plus five x plus c, we're saying that that could be written as x plus d squared. This is equal to x plus d squared. Now we can rewrite x plus d squared is going to be equal to x squared plus two x, two, let me write, two dx, two dx plus d squared. If this step right over here you find strange, I encourage you to watch the videos on squaring binomials or on perfect square polynomials, either one, so you can see the pattern that this is going to be x squared plus two times the product of both of these terms plus d squared. And so when you look at it like this, you can start to pattern match a little bit."}, {"video_title": "Factoring perfect squares missing values Mathematics II High School Math Khan Academy.mp3", "Sentence": "This is equal to x plus d squared. Now we can rewrite x plus d squared is going to be equal to x squared plus two x, two, let me write, two dx, two dx plus d squared. If this step right over here you find strange, I encourage you to watch the videos on squaring binomials or on perfect square polynomials, either one, so you can see the pattern that this is going to be x squared plus two times the product of both of these terms plus d squared. And so when you look at it like this, you can start to pattern match a little bit. You can say, all right, well, five x right over here, that is going to have to be equal to two d, and then you can also say that c is going to have to be equal to d squared. So once again, we could say two d is equal to five, two d is equal to five, or that d is equal to five halves. So we've figured out what d is equal to, and now we can figure out what c is, because we know that c needs to be equal to d squared, c, let me do that orange color actually."}, {"video_title": "Factoring perfect squares missing values Mathematics II High School Math Khan Academy.mp3", "Sentence": "And so when you look at it like this, you can start to pattern match a little bit. You can say, all right, well, five x right over here, that is going to have to be equal to two d, and then you can also say that c is going to have to be equal to d squared. So once again, we could say two d is equal to five, two d is equal to five, or that d is equal to five halves. So we've figured out what d is equal to, and now we can figure out what c is, because we know that c needs to be equal to d squared, c, let me do that orange color actually. So we know that c is equal to d squared, which is the same thing as five halves squared. We just figured out what d is equal to. It's going to be five halves squared, which is going to be 25 over four."}, {"video_title": "Example 2 Finding the square of a binomial with one variable Algebra I Khan Academy.mp3", "Sentence": "Now the first thing I will show you is exactly what you should not do. There's a huge temptation. A lot of people will look and they'll say, oh, that's 7x squared plus 10 squared. This is wrong. And I'll write it in caps. This is wrong. What your brain is doing is thinking, oh, if I had 7x times 10 and I squared that, this would be 7x squared times 10 squared."}, {"video_title": "Example 2 Finding the square of a binomial with one variable Algebra I Khan Academy.mp3", "Sentence": "This is wrong. And I'll write it in caps. This is wrong. What your brain is doing is thinking, oh, if I had 7x times 10 and I squared that, this would be 7x squared times 10 squared. We aren't multiplying here. We're adding 7x to 10. So you can't just square each of these terms."}, {"video_title": "Example 2 Finding the square of a binomial with one variable Algebra I Khan Academy.mp3", "Sentence": "What your brain is doing is thinking, oh, if I had 7x times 10 and I squared that, this would be 7x squared times 10 squared. We aren't multiplying here. We're adding 7x to 10. So you can't just square each of these terms. I just wanted to highlight, this is completely wrong. And to see why it's wrong, you have to just remind yourself that 7x plus 10 squared, this is the exact same thing as 7x plus 10 times 7x plus 10. That's what it means to square something."}, {"video_title": "Example 2 Finding the square of a binomial with one variable Algebra I Khan Academy.mp3", "Sentence": "So you can't just square each of these terms. I just wanted to highlight, this is completely wrong. And to see why it's wrong, you have to just remind yourself that 7x plus 10 squared, this is the exact same thing as 7x plus 10 times 7x plus 10. That's what it means to square something. You're multiplying it by itself, I guess, twice here. If you only did it once, you'd only have 1 7x plus 10. So this is what it is."}, {"video_title": "Example 2 Finding the square of a binomial with one variable Algebra I Khan Academy.mp3", "Sentence": "That's what it means to square something. You're multiplying it by itself, I guess, twice here. If you only did it once, you'd only have 1 7x plus 10. So this is what it is. So we're really just multiplying a binomial or two binomials. They just happen to be the same one. And you could use FOIL."}, {"video_title": "Example 2 Finding the square of a binomial with one variable Algebra I Khan Academy.mp3", "Sentence": "So this is what it is. So we're really just multiplying a binomial or two binomials. They just happen to be the same one. And you could use FOIL. You could use the distributive method. But this is actually a special case when you're squaring a binomial. So let's just think about it as a special case first."}, {"video_title": "Example 2 Finding the square of a binomial with one variable Algebra I Khan Academy.mp3", "Sentence": "And you could use FOIL. You could use the distributive method. But this is actually a special case when you're squaring a binomial. So let's just think about it as a special case first. And we can apply whatever we learn to this. So we could have just done it straight here. But I want to learn the general case so you can apply it to any problem that you might see."}, {"video_title": "Example 2 Finding the square of a binomial with one variable Algebra I Khan Academy.mp3", "Sentence": "So let's just think about it as a special case first. And we can apply whatever we learn to this. So we could have just done it straight here. But I want to learn the general case so you can apply it to any problem that you might see. So if I have a plus b squared, we already realize that's not a squared plus b squared. That is a plus b times a plus b. And now we can use the distributive property."}, {"video_title": "Example 2 Finding the square of a binomial with one variable Algebra I Khan Academy.mp3", "Sentence": "But I want to learn the general case so you can apply it to any problem that you might see. So if I have a plus b squared, we already realize that's not a squared plus b squared. That is a plus b times a plus b. And now we can use the distributive property. We can distribute this a plus b times this a. So we get a times a plus b. And then we could distribute the a plus b times this b."}, {"video_title": "Example 2 Finding the square of a binomial with one variable Algebra I Khan Academy.mp3", "Sentence": "And now we can use the distributive property. We can distribute this a plus b times this a. So we get a times a plus b. And then we could distribute the a plus b times this b. Plus b times a plus b. And then we distribute this a. We get a squared plus ab."}, {"video_title": "Example 2 Finding the square of a binomial with one variable Algebra I Khan Academy.mp3", "Sentence": "And then we could distribute the a plus b times this b. Plus b times a plus b. And then we distribute this a. We get a squared plus ab. Plus b times a is another ab. And I'm just swapping the order so it's the same as this. Plus b times b, which is b squared."}, {"video_title": "Example 2 Finding the square of a binomial with one variable Algebra I Khan Academy.mp3", "Sentence": "We get a squared plus ab. Plus b times a is another ab. And I'm just swapping the order so it's the same as this. Plus b times b, which is b squared. These are the same, or these are like terms. So we can add them. One of something plus another of that something will give you two of that something."}, {"video_title": "Example 2 Finding the square of a binomial with one variable Algebra I Khan Academy.mp3", "Sentence": "Plus b times b, which is b squared. These are the same, or these are like terms. So we can add them. One of something plus another of that something will give you two of that something. Two ab. We have a squared plus 2ab plus b squared. So the pattern here, if I have a plus b squared, it's equal to a squared plus 2 times the product of these numbers plus b squared."}, {"video_title": "Example 2 Finding the square of a binomial with one variable Algebra I Khan Academy.mp3", "Sentence": "One of something plus another of that something will give you two of that something. Two ab. We have a squared plus 2ab plus b squared. So the pattern here, if I have a plus b squared, it's equal to a squared plus 2 times the product of these numbers plus b squared. So over here I have 7x plus 10 squared. So this is going to be equal to 7x squared plus 2 times the product of 7x and 10. 2 times 7x times 10 plus 10 squared."}, {"video_title": "Example 2 Finding the square of a binomial with one variable Algebra I Khan Academy.mp3", "Sentence": "So the pattern here, if I have a plus b squared, it's equal to a squared plus 2 times the product of these numbers plus b squared. So over here I have 7x plus 10 squared. So this is going to be equal to 7x squared plus 2 times the product of 7x and 10. 2 times 7x times 10 plus 10 squared. So the difference between the right answer and the wrong answer is that you have this middle term here that you might have forgotten about if you did it this way. And this comes out when you're multiplying all of the different combinations of the terms here. And if we simplify this, if we simplify it, 7x squared, that's 7 squared times x squared."}, {"video_title": "Example 2 Finding the square of a binomial with one variable Algebra I Khan Academy.mp3", "Sentence": "2 times 7x times 10 plus 10 squared. So the difference between the right answer and the wrong answer is that you have this middle term here that you might have forgotten about if you did it this way. And this comes out when you're multiplying all of the different combinations of the terms here. And if we simplify this, if we simplify it, 7x squared, that's 7 squared times x squared. So 7 squared is 49 times x squared. When you multiply this part out, 2 times 7 times 10, that's 14 times 10, which is 140. And then we have our x, no other x there."}, {"video_title": "Example 2 Finding the square of a binomial with one variable Algebra I Khan Academy.mp3", "Sentence": "And if we simplify this, if we simplify it, 7x squared, that's 7 squared times x squared. So 7 squared is 49 times x squared. When you multiply this part out, 2 times 7 times 10, that's 14 times 10, which is 140. And then we have our x, no other x there. And then plus 10 squared. So plus 100. And we are done."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Function g is defined as g of x is equal to f of negative x, also fair enough. What is the graph of g? And on Khan Academy, it's multiple choice, but I thought for the sake of this video, it'd be fun to think about what g would look like without having any choices, just sketching it out. So pause this video and try to think about it, at least in your head. All right, now let's work through this together. So we've already gone over that g of x is equal to f of negative x. So whatever the value of f is at a certain value, we would expect g to take on that value at the negative of that."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So pause this video and try to think about it, at least in your head. All right, now let's work through this together. So we've already gone over that g of x is equal to f of negative x. So whatever the value of f is at a certain value, we would expect g to take on that value at the negative of that. So for example, we can see that f of four is equal to two, so we would expect g of negative four to be equal to two. Because once again, g of negative four, we could write it over here, g of negative four is going to be equal to f of the negative of negative four, which is equal to f of four. And so we could keep going with that."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So whatever the value of f is at a certain value, we would expect g to take on that value at the negative of that. So for example, we can see that f of four is equal to two, so we would expect g of negative four to be equal to two. Because once again, g of negative four, we could write it over here, g of negative four is going to be equal to f of the negative of negative four, which is equal to f of four. And so we could keep going with that. What would g of negative two be? Well, that would be the same thing as f of two, which is zero, so it would be right over there. What would g of zero be?"}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And so we could keep going with that. What would g of negative two be? Well, that would be the same thing as f of two, which is zero, so it would be right over there. What would g of zero be? Well, that would be the same thing as f of zero, because the negative of zero is zero, and f of zero is right over there. It looks like negative two. And so you can already see where this is going."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "What would g of zero be? Well, that would be the same thing as f of zero, because the negative of zero is zero, and f of zero is right over there. It looks like negative two. And so you can already see where this is going. And we've already talked about it in previous videos, that if you replace your x with a negative x, you're essentially reflecting over the y-axis. So g is going to look something like this. It is going to look something like this."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And so you can already see where this is going. And we've already talked about it in previous videos, that if you replace your x with a negative x, you're essentially reflecting over the y-axis. So g is going to look something like this. It is going to look something like this. Once again, g of negative six would be the same thing as f of six. And so that would be the graph of g. And if you're doing this on Khan Academy, you'd pick the choice that looks like this, that would give a reflection over the y-axis. Let's do another example."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "It is going to look something like this. Once again, g of negative six would be the same thing as f of six. And so that would be the graph of g. And if you're doing this on Khan Academy, you'd pick the choice that looks like this, that would give a reflection over the y-axis. Let's do another example. So here, once again, this is the graph of the function f. And then they say, what is the graph of g? And so pause this video, and at least try to sketch it on your mind what g should look like. All right, so in this situation, they didn't replace the x with negative x in f of x."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Let's do another example. So here, once again, this is the graph of the function f. And then they say, what is the graph of g? And so pause this video, and at least try to sketch it on your mind what g should look like. All right, so in this situation, they didn't replace the x with negative x in f of x. Instead, g of x is equal to the negative of all of f of x. In fact, we could rewrite g of x like this. We could say that g of x is equal to, notice, all of this right over here, that was our definition of f of x."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "All right, so in this situation, they didn't replace the x with negative x in f of x. Instead, g of x is equal to the negative of all of f of x. In fact, we could rewrite g of x like this. We could say that g of x is equal to, notice, all of this right over here, that was our definition of f of x. So g of x is equal to the negative of f of x. So instead of it being f of negative x, it's equal to the negative of f of x. So one way to think about it is we can see that f of zero is two, but g of zero is going to be the negative of that."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "We could say that g of x is equal to, notice, all of this right over here, that was our definition of f of x. So g of x is equal to the negative of f of x. So instead of it being f of negative x, it's equal to the negative of f of x. So one way to think about it is we can see that f of zero is two, but g of zero is going to be the negative of that. So it's going to be equal to negative two. And so you could keep going with that. You could see that whatever f of a certain value is, g of that value would be the negative of that."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So one way to think about it is we can see that f of zero is two, but g of zero is going to be the negative of that. So it's going to be equal to negative two. And so you could keep going with that. You could see that whatever f of a certain value is, g of that value would be the negative of that. So it would be down here. And so g of x would be a reflection of f of x about the x-axis. So g of x is going to look something, something like, something like that, a reflection about the x-axis."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "You could see that whatever f of a certain value is, g of that value would be the negative of that. So it would be down here. And so g of x would be a reflection of f of x about the x-axis. So g of x is going to look something, something like, something like that, a reflection about the x-axis. And so once again, I'm kind of getting you to pick the choice that would actually look like that. Let's do another example. This is strangely fun."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So g of x is going to look something, something like, something like that, a reflection about the x-axis. And so once again, I'm kind of getting you to pick the choice that would actually look like that. Let's do another example. This is strangely fun. All right. So here we're told functions f, so that's in solid in this blue color, and g dashed, so that's right over there, are graphed. What is the equation of g in terms of f?"}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "This is strangely fun. All right. So here we're told functions f, so that's in solid in this blue color, and g dashed, so that's right over there, are graphed. What is the equation of g in terms of f? So pause this video and try to think about it. So the key is to realize how do we transform f of x, actually they labeled it over here, this is f of x right over here, in order to get g. So f of negative x would be a reflection of f above, about the y-axis. And so it would intersect there."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "What is the equation of g in terms of f? So pause this video and try to think about it. So the key is to realize how do we transform f of x, actually they labeled it over here, this is f of x right over here, in order to get g. So f of negative x would be a reflection of f above, about the y-axis. And so it would intersect there. It would have this straight portion like this. And I'm just experimenting right now. It'd have the straight portion like this."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And so it would intersect there. It would have this straight portion like this. And I'm just experimenting right now. It'd have the straight portion like this. And then it would go up, and let's see, f of negative x, so when you input six into it, that would be f of negative six, which is six, so it would go up there. So f of negative x would look something, something like this. Something like that."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "It'd have the straight portion like this. And then it would go up, and let's see, f of negative x, so when you input six into it, that would be f of negative six, which is six, so it would go up there. So f of negative x would look something, something like this. Something like that. So the purple is f of negative x. Now that doesn't quite get us to g, but it gets us a little bit closer, because it looks like if I were to take the reflection of f of negative x, f of negative x about the x-axis, it looks like I'm going to get to g. And so how do you reflect something about the x-axis? Well we saw it in the example just now."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Something like that. So the purple is f of negative x. Now that doesn't quite get us to g, but it gets us a little bit closer, because it looks like if I were to take the reflection of f of negative x, f of negative x about the x-axis, it looks like I'm going to get to g. And so how do you reflect something about the x-axis? Well we saw it in the example just now. You multiply the entire function by a negative. So we could say that g is equal to the negative of f of negative x. It's equal to the negative of this."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Well we saw it in the example just now. You multiply the entire function by a negative. So we could say that g is equal to the negative of f of negative x. It's equal to the negative of this. So we're doing both reflections. We're flipping over the y-axis, and we're flipping over the x-axis to get to g. Let's do one more example. So once again, they've graphed f, they've graphed g, and they've said f is defined as this right over here."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "It's equal to the negative of this. So we're doing both reflections. We're flipping over the y-axis, and we're flipping over the x-axis to get to g. Let's do one more example. So once again, they've graphed f, they've graphed g, and they've said f is defined as this right over here. What is the equation of g? So they're not just asking it in terms of f. They just wanna know what is the equation of g. Pause this video and try to think about it. Well, you can see pretty clearly that this is a reflection across the y-axis."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So once again, they've graphed f, they've graphed g, and they've said f is defined as this right over here. What is the equation of g? So they're not just asking it in terms of f. They just wanna know what is the equation of g. Pause this video and try to think about it. Well, you can see pretty clearly that this is a reflection across the y-axis. And a reflection across the y-axis, you can see pretty clearly that g of x is equal to f of negative x. F of negative x. How do we know that? Well, for whenever we take f of x, and we get that value, g at the negative of that value takes on the same function value, I guess I could say."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Well, you can see pretty clearly that this is a reflection across the y-axis. And a reflection across the y-axis, you can see pretty clearly that g of x is equal to f of negative x. F of negative x. How do we know that? Well, for whenever we take f of x, and we get that value, g at the negative of that value takes on the same function value, I guess I could say. Or another way to think about it is we could just pick this point, negative eight. F of negative eight is equal to a little over four. But g of eight is equal to a little over four, is equal to that same value."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Well, for whenever we take f of x, and we get that value, g at the negative of that value takes on the same function value, I guess I could say. Or another way to think about it is we could just pick this point, negative eight. F of negative eight is equal to a little over four. But g of eight is equal to a little over four, is equal to that same value. And so what is the equation of g? Well, we just have to rewrite this so that we can write it out as an equation. And so we could write out g of x is equal to, if I were to replace all of the x's here with a negative x, what would I get?"}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "But g of eight is equal to a little over four, is equal to that same value. And so what is the equation of g? Well, we just have to rewrite this so that we can write it out as an equation. And so we could write out g of x is equal to, if I were to replace all of the x's here with a negative x, what would I get? I would get four times the square root of two minus, instead of an x, I will have a negative x. And then the minus eight is outside of the radical. And so we would have g of x is equal to four times the square root of two plus x minus eight."}, {"video_title": "Proof of quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And the quadratic formula was x, the solutions, would be equal to negative b plus or minus the square root of b squared minus 4ac, all of that over 2a. And we learned how to use it. You literally just substitute the numbers a for a, b for b, c for c, and then it gives you two answers because you have a plus or a minus right there. What I want to do in this video is actually prove it to you. Prove that using, essentially completing the square, I can get from that to that right over there. So the first thing I want to do so that I can start completing the square from this point right here is, let me rewrite the equation right here. So we have ax, let me do it in a different color."}, {"video_title": "Proof of quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "What I want to do in this video is actually prove it to you. Prove that using, essentially completing the square, I can get from that to that right over there. So the first thing I want to do so that I can start completing the square from this point right here is, let me rewrite the equation right here. So we have ax, let me do it in a different color. I have ax squared plus bx plus c is equal to 0. So the first thing I want to do is divide everything by a. So I just have a 1 out here as a coefficient."}, {"video_title": "Proof of quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So we have ax, let me do it in a different color. I have ax squared plus bx plus c is equal to 0. So the first thing I want to do is divide everything by a. So I just have a 1 out here as a coefficient. So you divide everything by a, you get x squared plus b over ax plus c over a is equal to 0 over a, which is still just 0. Now we want to, let me get the c over a term onto the right hand side. So let's subtract c over a from both sides and we get x squared plus b over ax plus, well I'll just leave a blank there because this is gone now, we've subtracted it from both sides, is equal to negative c over a."}, {"video_title": "Proof of quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So I just have a 1 out here as a coefficient. So you divide everything by a, you get x squared plus b over ax plus c over a is equal to 0 over a, which is still just 0. Now we want to, let me get the c over a term onto the right hand side. So let's subtract c over a from both sides and we get x squared plus b over ax plus, well I'll just leave a blank there because this is gone now, we've subtracted it from both sides, is equal to negative c over a. And I left a space there so that we can complete the square. And you saw in the completing the square video, you literally just take half of this coefficient right here and you square it. So what is b over a divided by 2?"}, {"video_title": "Proof of quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So let's subtract c over a from both sides and we get x squared plus b over ax plus, well I'll just leave a blank there because this is gone now, we've subtracted it from both sides, is equal to negative c over a. And I left a space there so that we can complete the square. And you saw in the completing the square video, you literally just take half of this coefficient right here and you square it. So what is b over a divided by 2? Or what is 1 half times b over a? Well that is just b over 2a and of course we are going to square it. You take half of this and you square it."}, {"video_title": "Proof of quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So what is b over a divided by 2? Or what is 1 half times b over a? Well that is just b over 2a and of course we are going to square it. You take half of this and you square it. That's what we do in completing a square so that we can turn this into the perfect square, into a perfect square of a binomial. Now, of course, we cannot just add the b over 2a squared to the left hand side. We have to add it to both sides."}, {"video_title": "Proof of quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "You take half of this and you square it. That's what we do in completing a square so that we can turn this into the perfect square, into a perfect square of a binomial. Now, of course, we cannot just add the b over 2a squared to the left hand side. We have to add it to both sides. So you have a plus b over 2a squared there as well. Now what happens? Well this over here, this expression right over here, this is the exact same thing as x plus b over 2a squared."}, {"video_title": "Proof of quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "We have to add it to both sides. So you have a plus b over 2a squared there as well. Now what happens? Well this over here, this expression right over here, this is the exact same thing as x plus b over 2a squared. And if you don't believe me, I'm going to multiply it out. That x plus b over 2a squared is x plus b over 2a times x plus b over 2a. x times x is x squared."}, {"video_title": "Proof of quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Well this over here, this expression right over here, this is the exact same thing as x plus b over 2a squared. And if you don't believe me, I'm going to multiply it out. That x plus b over 2a squared is x plus b over 2a times x plus b over 2a. x times x is x squared. x times b over 2a is plus b over 2a x. You have b over 2a times x which is another b over 2a x. And then you have b over 2a times b over 2a."}, {"video_title": "Proof of quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "x times x is x squared. x times b over 2a is plus b over 2a x. You have b over 2a times x which is another b over 2a x. And then you have b over 2a times b over 2a. That is plus b over 2a squared. That and this are the same thing because these two middle terms, b over 2a plus b over 2a, that's the same thing as 2b over 2a x, which is the same thing as b over a x. So this simplifies to x squared plus b over a x plus b over 2a squared, which is exactly what we have written right there."}, {"video_title": "Proof of quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And then you have b over 2a times b over 2a. That is plus b over 2a squared. That and this are the same thing because these two middle terms, b over 2a plus b over 2a, that's the same thing as 2b over 2a x, which is the same thing as b over a x. So this simplifies to x squared plus b over a x plus b over 2a squared, which is exactly what we have written right there. That was the whole point of adding this term to both sides. So it becomes a perfect square. So the left hand side simplifies to this."}, {"video_title": "Proof of quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So this simplifies to x squared plus b over a x plus b over 2a squared, which is exactly what we have written right there. That was the whole point of adding this term to both sides. So it becomes a perfect square. So the left hand side simplifies to this. The right hand side maybe not quite as simple. Maybe we'll leave it the way it is right now. Actually, let's simplify it a little bit."}, {"video_title": "Proof of quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So the left hand side simplifies to this. The right hand side maybe not quite as simple. Maybe we'll leave it the way it is right now. Actually, let's simplify it a little bit. So the right hand side, we can rewrite it. This is going to be equal to, well this is going to be b squared. I'll write that term first."}, {"video_title": "Proof of quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Actually, let's simplify it a little bit. So the right hand side, we can rewrite it. This is going to be equal to, well this is going to be b squared. I'll write that term first. This is b, let me do it in green so we can follow along. So that term right there can be written as b squared over 4a squared. And what would this term, what would that become?"}, {"video_title": "Proof of quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "I'll write that term first. This is b, let me do it in green so we can follow along. So that term right there can be written as b squared over 4a squared. And what would this term, what would that become? This would become, in order to have 4a squared as the denominator, we have to multiply the numerator and the denominator by 4a. So this term right here will become minus 4ac over 4a squared. And you can verify for yourself that that is the same thing as that."}, {"video_title": "Proof of quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And what would this term, what would that become? This would become, in order to have 4a squared as the denominator, we have to multiply the numerator and the denominator by 4a. So this term right here will become minus 4ac over 4a squared. And you can verify for yourself that that is the same thing as that. I just multiplied the numerator and the denominator by 4a. In fact, the 4s cancel out and then this a cancels out and you just have a c over a. So this and that are equivalent."}, {"video_title": "Proof of quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And you can verify for yourself that that is the same thing as that. I just multiplied the numerator and the denominator by 4a. In fact, the 4s cancel out and then this a cancels out and you just have a c over a. So this and that are equivalent. I just switched which I write first. And you might already be seeing the beginnings of the quadratic formula here. So this I can rewrite."}, {"video_title": "Proof of quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So this and that are equivalent. I just switched which I write first. And you might already be seeing the beginnings of the quadratic formula here. So this I can rewrite. The right hand side right here I can rewrite as b squared minus 4ac. All of that over 4a squared. This is looking very close."}, {"video_title": "Proof of quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So this I can rewrite. The right hand side right here I can rewrite as b squared minus 4ac. All of that over 4a squared. This is looking very close. Notice b squared minus 4ac is already appearing. We don't have a square root yet, but we haven't taken the square root of both sides of this equation. So let's do that."}, {"video_title": "Proof of quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "This is looking very close. Notice b squared minus 4ac is already appearing. We don't have a square root yet, but we haven't taken the square root of both sides of this equation. So let's do that. So if you take the square root of both sides, the left hand side will just become x plus b over 2a is going to be equal to the plus or minus square root of this thing. And the square root of this is the square root of the numerator over the square root of the denominator. So it's going to be the plus or minus the square root of b squared minus 4ac over the square root of 4a squared."}, {"video_title": "Proof of quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So let's do that. So if you take the square root of both sides, the left hand side will just become x plus b over 2a is going to be equal to the plus or minus square root of this thing. And the square root of this is the square root of the numerator over the square root of the denominator. So it's going to be the plus or minus the square root of b squared minus 4ac over the square root of 4a squared. Now what is the square root of 4a squared? It is 2a. 2a squared is 4a squared."}, {"video_title": "Proof of quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So it's going to be the plus or minus the square root of b squared minus 4ac over the square root of 4a squared. Now what is the square root of 4a squared? It is 2a. 2a squared is 4a squared. So the square root of this is that right here. So to go from here to here, I just took the square root of both sides of this equation. Now this is looking very close to the quadratic."}, {"video_title": "Proof of quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "2a squared is 4a squared. So the square root of this is that right here. So to go from here to here, I just took the square root of both sides of this equation. Now this is looking very close to the quadratic. We have a b squared minus 4ac over 2a. Now we just essentially have to subtract this b over 2a from both sides of the equation and we're done. So let's do that."}, {"video_title": "Proof of quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Now this is looking very close to the quadratic. We have a b squared minus 4ac over 2a. Now we just essentially have to subtract this b over 2a from both sides of the equation and we're done. So let's do that. So if you subtract the b over 2a from both sides of this equation, what do you get? You get x is equal to negative b over 2a plus or minus the square root of b squared minus 4ac over 2a, common denominator. So this is equal to negative b."}, {"video_title": "Proof of quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So let's do that. So if you subtract the b over 2a from both sides of this equation, what do you get? You get x is equal to negative b over 2a plus or minus the square root of b squared minus 4ac over 2a, common denominator. So this is equal to negative b. Let me do this in a new color. Let's do it in orange. Negative b plus or minus the square root of b squared minus 4ac."}, {"video_title": "Proof of quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So this is equal to negative b. Let me do this in a new color. Let's do it in orange. Negative b plus or minus the square root of b squared minus 4ac. All of that over 2a. And we are done. By completing the square with just general coefficients in front of a, b, and c, we were able to derive the quadratic formula."}, {"video_title": "Solving square-root equations two solutions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Let's say that we have the equation six plus three w is equal to the square root of two w plus 12 plus two w. See if you could pause the video and solve for w. And it might have more than one solution, so keep that in mind. All right, now let's work through this together. So the first thing I'd like to do whenever I see one of these radical equations is just isolate the radical on one side of the equation. So let's subtract two w from both sides. I want to get rid of that two w from the right-hand side. I just want the radical sign. And if I subtract two w from both sides, what am I left with?"}, {"video_title": "Solving square-root equations two solutions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So let's subtract two w from both sides. I want to get rid of that two w from the right-hand side. I just want the radical sign. And if I subtract two w from both sides, what am I left with? Well, on the left-hand side, I am left with six plus three w minus two w. Well, three of something, take away two of them, you're gonna be left with w. Six plus w is equal to, these cancel out, we're left with the square root of two w plus 12. Now to get rid of the radical, we're going to square both sides. And we've seen before that this process right over here, it's a little bit tricky, because when you're squaring a radical in a radical equation like this, and then you solve, you might find an extraneous solution."}, {"video_title": "Solving square-root equations two solutions Mathematics III High School Math Khan Academy.mp3", "Sentence": "And if I subtract two w from both sides, what am I left with? Well, on the left-hand side, I am left with six plus three w minus two w. Well, three of something, take away two of them, you're gonna be left with w. Six plus w is equal to, these cancel out, we're left with the square root of two w plus 12. Now to get rid of the radical, we're going to square both sides. And we've seen before that this process right over here, it's a little bit tricky, because when you're squaring a radical in a radical equation like this, and then you solve, you might find an extraneous solution. What do I mean by that? Well, we're gonna get the same result whether we square this or whether we square that, because when you square a negative, it becomes a positive. But those are fundamentally two different equations."}, {"video_title": "Solving square-root equations two solutions Mathematics III High School Math Khan Academy.mp3", "Sentence": "And we've seen before that this process right over here, it's a little bit tricky, because when you're squaring a radical in a radical equation like this, and then you solve, you might find an extraneous solution. What do I mean by that? Well, we're gonna get the same result whether we square this or whether we square that, because when you square a negative, it becomes a positive. But those are fundamentally two different equations. We only want the solutions that satisfy the one that doesn't have the negative there. So that's why we're gonna test our solutions to make sure that they're valid for our original equation. So if we square both sides, on the left-hand side, we're going to have, well, it's gonna be w squared plus two times their product, so two times six times w, so that's 12w plus six squared, 36, is equal to, now if you take the square root and square it, you're gonna be left with two w plus 12."}, {"video_title": "Solving square-root equations two solutions Mathematics III High School Math Khan Academy.mp3", "Sentence": "But those are fundamentally two different equations. We only want the solutions that satisfy the one that doesn't have the negative there. So that's why we're gonna test our solutions to make sure that they're valid for our original equation. So if we square both sides, on the left-hand side, we're going to have, well, it's gonna be w squared plus two times their product, so two times six times w, so that's 12w plus six squared, 36, is equal to, now if you take the square root and square it, you're gonna be left with two w plus 12. Now we can subtract two w and 12 from both sides, so let's do that. So then we can get into kind of a standard quadratic form. So let's subtract two w from both sides, and let's subtract 12 from both sides."}, {"video_title": "Solving square-root equations two solutions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So if we square both sides, on the left-hand side, we're going to have, well, it's gonna be w squared plus two times their product, so two times six times w, so that's 12w plus six squared, 36, is equal to, now if you take the square root and square it, you're gonna be left with two w plus 12. Now we can subtract two w and 12 from both sides, so let's do that. So then we can get into kind of a standard quadratic form. So let's subtract two w from both sides, and let's subtract 12 from both sides. So subtract 12 from the right, subtract 12 here. And once again, I just wanna get rid of this on the right-hand side. And I am going to be left with, I am going to be left with, on the left-hand side, it's gonna be w squared, see, 12w minus two w is plus 10w, and then 36 minus 12 is plus 24, is equal to zero."}, {"video_title": "Solving square-root equations two solutions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So let's subtract two w from both sides, and let's subtract 12 from both sides. So subtract 12 from the right, subtract 12 here. And once again, I just wanna get rid of this on the right-hand side. And I am going to be left with, I am going to be left with, on the left-hand side, it's gonna be w squared, see, 12w minus two w is plus 10w, and then 36 minus 12 is plus 24, is equal to zero. And let's see, to solve this, let's use this factor bar, there are two numbers that add up to 10, and whose product is 24. Well, what jumps out at me is six and four. So we can rewrite this as w plus four times w plus six is equal to zero."}, {"video_title": "Solving square-root equations two solutions Mathematics III High School Math Khan Academy.mp3", "Sentence": "And I am going to be left with, I am going to be left with, on the left-hand side, it's gonna be w squared, see, 12w minus two w is plus 10w, and then 36 minus 12 is plus 24, is equal to zero. And let's see, to solve this, let's use this factor bar, there are two numbers that add up to 10, and whose product is 24. Well, what jumps out at me is six and four. So we can rewrite this as w plus four times w plus six is equal to zero. And so, if I have the product of two things equaling zero, well, to solve this, either one or both of them could be equal to zero. Zero times anything is going to be zero. So w plus four is equal to zero, or w plus six is equal to zero."}, {"video_title": "Solving square-root equations two solutions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So we can rewrite this as w plus four times w plus six is equal to zero. And so, if I have the product of two things equaling zero, well, to solve this, either one or both of them could be equal to zero. Zero times anything is going to be zero. So w plus four is equal to zero, or w plus six is equal to zero. And over here, if you subtract four from both sides, you get w is equal to negative four, or subtract six from both sides here, w is equal to negative six. Now let's verify that these actually are solutions to our original equation. Remember, our original equation was six, I'll rewrite it here, our original equation was six plus three w is equal to the square root of two w plus 12 plus two w. So let's see, if w is equal to negative four, if w is equal to negative four right over, let me do this in a different, is equal to negative four, so that's going to be six plus three times negative four is equal to the square root of two times negative four plus 12 plus two times negative four."}, {"video_title": "Solving square-root equations two solutions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So w plus four is equal to zero, or w plus six is equal to zero. And over here, if you subtract four from both sides, you get w is equal to negative four, or subtract six from both sides here, w is equal to negative six. Now let's verify that these actually are solutions to our original equation. Remember, our original equation was six, I'll rewrite it here, our original equation was six plus three w is equal to the square root of two w plus 12 plus two w. So let's see, if w is equal to negative four, if w is equal to negative four right over, let me do this in a different, is equal to negative four, so that's going to be six plus three times negative four is equal to the square root of two times negative four plus 12 plus two times negative four. So this would be, this is negative 12 here, this is negative eight here, this is negative eight here. So you have six plus negative 12, which is negative six, is equal to the square root of negative eight plus 12 is four plus negative eight, so that would be negative six is equal to two plus negative eight, which is absolutely true. So this is definitely a solution."}, {"video_title": "Solving square-root equations two solutions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Remember, our original equation was six, I'll rewrite it here, our original equation was six plus three w is equal to the square root of two w plus 12 plus two w. So let's see, if w is equal to negative four, if w is equal to negative four right over, let me do this in a different, is equal to negative four, so that's going to be six plus three times negative four is equal to the square root of two times negative four plus 12 plus two times negative four. So this would be, this is negative 12 here, this is negative eight here, this is negative eight here. So you have six plus negative 12, which is negative six, is equal to the square root of negative eight plus 12 is four plus negative eight, so that would be negative six is equal to two plus negative eight, which is absolutely true. So this is definitely a solution. And let's try w is equal to negative six. So w is equal to negative six, so we're going to get, if we look up here, we're going to have six plus three times negative six is equal to the square root of two times negative six plus 12 plus two w. So this is going to be negative 18, this is going to be negative 12, this is negative 12, negative 12 plus 12 is zero, square root of zero, this is all zero, and then two times, and actually let me, I shouldn't have written a w there, I should have written a two times negative six. So back to what I was doing, this right over here is negative 18, this is two times negative six plus 12, this is all zero, square root of zero is zero, and then this is negative 12."}, {"video_title": "Example 1 Factoring a difference of squares with two variables Algebra II Khan Academy.mp3", "Sentence": "Factor x squared minus 49y squared. So what's interesting here is that, well, x squared is clearly a perfect square. It's the square of x. And 49y squared is also a perfect square. It's the square of 7y. So it looks like we might have a special form here. And to remind ourselves, just think about what happens if we take a plus b times a minus b. I'm just doing it in the general case so we can see a pattern here."}, {"video_title": "Example 1 Factoring a difference of squares with two variables Algebra II Khan Academy.mp3", "Sentence": "And 49y squared is also a perfect square. It's the square of 7y. So it looks like we might have a special form here. And to remind ourselves, just think about what happens if we take a plus b times a minus b. I'm just doing it in the general case so we can see a pattern here. So over here, this would be a times a, which would be a squared, plus a times negative b, which would be negative ab, plus b times a, or a times b again, which would be ab. And then you have b times negative b, so it would be minus b squared. Now these middle two terms cancel out."}, {"video_title": "Example 1 Factoring a difference of squares with two variables Algebra II Khan Academy.mp3", "Sentence": "And to remind ourselves, just think about what happens if we take a plus b times a minus b. I'm just doing it in the general case so we can see a pattern here. So over here, this would be a times a, which would be a squared, plus a times negative b, which would be negative ab, plus b times a, or a times b again, which would be ab. And then you have b times negative b, so it would be minus b squared. Now these middle two terms cancel out. Negative ab plus ab, they cancel out. And you're left with just a squared minus b squared. And that's the exact pattern we have here."}, {"video_title": "Example 1 Factoring a difference of squares with two variables Algebra II Khan Academy.mp3", "Sentence": "Now these middle two terms cancel out. Negative ab plus ab, they cancel out. And you're left with just a squared minus b squared. And that's the exact pattern we have here. We have an a squared minus a b squared. So in this case, a is equal to x, and b is equal to 7y. So we have x squared minus 7y, the whole thing squared."}, {"video_title": "Example 1 Factoring a difference of squares with two variables Algebra II Khan Academy.mp3", "Sentence": "And that's the exact pattern we have here. We have an a squared minus a b squared. So in this case, a is equal to x, and b is equal to 7y. So we have x squared minus 7y, the whole thing squared. So we can expand this as the difference of squares. Or actually, this thing right over here is the difference of squares. So we can expand this like this."}, {"video_title": "Example 1 Factoring a difference of squares with two variables Algebra II Khan Academy.mp3", "Sentence": "So we have x squared minus 7y, the whole thing squared. So we can expand this as the difference of squares. Or actually, this thing right over here is the difference of squares. So we can expand this like this. So this will be equal to x plus 7y times x minus 7y. And once again, we're just pattern matching based on this realization right here. If I take a plus b times a minus b, I get a difference of squares."}, {"video_title": "Logarithm of a power Logarithms Algebra II Khan Academy.mp3", "Sentence": "And once again, we're just going to rewrite this in a different way. You could argue whether it's going to be more simple or not. And the logarithm property that I'm guessing that we should use for this example right here is the property, if I take log base x of, let me pick some more letters here, of log base x of y to the z power, that this is the same thing as z times log base x of y. So this is a logarithm property. If I'm taking the logarithm of a given base of something to a power, I could take that power out front and multiply that times the log of the base of just the y in this case. So we apply this property over here, and in a second, once I do this problem, we'll talk about why this actually makes a lot of sense and comes straight out of exponent properties. But if we just apply that over here, we get log base 5 of x to the third."}, {"video_title": "Logarithm of a power Logarithms Algebra II Khan Academy.mp3", "Sentence": "So this is a logarithm property. If I'm taking the logarithm of a given base of something to a power, I could take that power out front and multiply that times the log of the base of just the y in this case. So we apply this property over here, and in a second, once I do this problem, we'll talk about why this actually makes a lot of sense and comes straight out of exponent properties. But if we just apply that over here, we get log base 5 of x to the third. Well, this is the exponent right over here. That's the same thing as z. So that's going to be the same thing as, let me do this in a different color, that 3 is the same thing."}, {"video_title": "Logarithm of a power Logarithms Algebra II Khan Academy.mp3", "Sentence": "But if we just apply that over here, we get log base 5 of x to the third. Well, this is the exponent right over here. That's the same thing as z. So that's going to be the same thing as, let me do this in a different color, that 3 is the same thing. I'll put it out front, that's the same thing as 3 times the logarithm base 5 of x. And we're done. This is just another way of writing it using this property."}, {"video_title": "Logarithm of a power Logarithms Algebra II Khan Academy.mp3", "Sentence": "So that's going to be the same thing as, let me do this in a different color, that 3 is the same thing. I'll put it out front, that's the same thing as 3 times the logarithm base 5 of x. And we're done. This is just another way of writing it using this property. So you could argue that this is a, maybe this is a simplification because you took the exponent outside of the logarithm and you're multiplying the logarithm by that number now. Now with that out of the way, let's think about why that actually makes sense. So let's say that we know that, and I'll just pick some arbitrary letters here, let's say that we know that a to the b power is equal to c. And so if we know that, that's written as an exponential equation."}, {"video_title": "Logarithm of a power Logarithms Algebra II Khan Academy.mp3", "Sentence": "This is just another way of writing it using this property. So you could argue that this is a, maybe this is a simplification because you took the exponent outside of the logarithm and you're multiplying the logarithm by that number now. Now with that out of the way, let's think about why that actually makes sense. So let's say that we know that, and I'll just pick some arbitrary letters here, let's say that we know that a to the b power is equal to c. And so if we know that, that's written as an exponential equation. If we wanted to write the same truth as a logarithmic equation, we would say logarithm base a of c is equal to b. To what power do I have to raise a to get c? I raise it to the bth power."}, {"video_title": "Logarithm of a power Logarithms Algebra II Khan Academy.mp3", "Sentence": "So let's say that we know that, and I'll just pick some arbitrary letters here, let's say that we know that a to the b power is equal to c. And so if we know that, that's written as an exponential equation. If we wanted to write the same truth as a logarithmic equation, we would say logarithm base a of c is equal to b. To what power do I have to raise a to get c? I raise it to the bth power. a to the b power is equal to c. Fair enough. Now let's take both sides of this equation right over here and raise it to the dth power. So let's take both sides of this equation and raise it to the dth power."}, {"video_title": "Logarithm of a power Logarithms Algebra II Khan Academy.mp3", "Sentence": "I raise it to the bth power. a to the b power is equal to c. Fair enough. Now let's take both sides of this equation right over here and raise it to the dth power. So let's take both sides of this equation and raise it to the dth power. Instead of doing it in place, I'm just going to rewrite it over here. So I wrote the original equation, a to the b is equal to c, which is just rewriting the statement. But let me take both sides of this to the dth power."}, {"video_title": "Logarithm of a power Logarithms Algebra II Khan Academy.mp3", "Sentence": "So let's take both sides of this equation and raise it to the dth power. Instead of doing it in place, I'm just going to rewrite it over here. So I wrote the original equation, a to the b is equal to c, which is just rewriting the statement. But let me take both sides of this to the dth power. And I should be consistent. I'll use all capital letters. So this should be a b."}, {"video_title": "Logarithm of a power Logarithms Algebra II Khan Academy.mp3", "Sentence": "But let me take both sides of this to the dth power. And I should be consistent. I'll use all capital letters. So this should be a b. Actually, let's say I'm using all lowercase letters. So this is a lowercase c. So let me write it this way. a to the, so I'm going to raise this to the dth power."}, {"video_title": "Logarithm of a power Logarithms Algebra II Khan Academy.mp3", "Sentence": "So this should be a b. Actually, let's say I'm using all lowercase letters. So this is a lowercase c. So let me write it this way. a to the, so I'm going to raise this to the dth power. And I'm going to raise this to the dth power. Obviously, if these two things are equal to each other, if I raise both sides to the same power, the equality is still going to hold. Now what's interesting over here, is we can now say, what we can do is we can use our, what we know about exponent properties, say, look, if I have a to the b power, and then I raise that to the d power, our exponent properties say that this is the same thing."}, {"video_title": "Logarithm of a power Logarithms Algebra II Khan Academy.mp3", "Sentence": "a to the, so I'm going to raise this to the dth power. And I'm going to raise this to the dth power. Obviously, if these two things are equal to each other, if I raise both sides to the same power, the equality is still going to hold. Now what's interesting over here, is we can now say, what we can do is we can use our, what we know about exponent properties, say, look, if I have a to the b power, and then I raise that to the d power, our exponent properties say that this is the same thing. This is equal to a to the bd power. This is equal to a to the bd. Let me write it here."}, {"video_title": "Logarithm of a power Logarithms Algebra II Khan Academy.mp3", "Sentence": "Now what's interesting over here, is we can now say, what we can do is we can use our, what we know about exponent properties, say, look, if I have a to the b power, and then I raise that to the d power, our exponent properties say that this is the same thing. This is equal to a to the bd power. This is equal to a to the bd. Let me write it here. Actually, let me do that in a different color. I've already used that green. Let me do this right over here, using what we know about exponent properties."}, {"video_title": "Logarithm of a power Logarithms Algebra II Khan Academy.mp3", "Sentence": "Let me write it here. Actually, let me do that in a different color. I've already used that green. Let me do this right over here, using what we know about exponent properties. This is the same thing as a to the bd power. So we have a to the bd power is equal to c to the dth power. And now this exponential equation, if we were to write it as a logarithmic equation, we would say log base a of c to the dth power is equal to bd."}, {"video_title": "Logarithm of a power Logarithms Algebra II Khan Academy.mp3", "Sentence": "Let me do this right over here, using what we know about exponent properties. This is the same thing as a to the bd power. So we have a to the bd power is equal to c to the dth power. And now this exponential equation, if we were to write it as a logarithmic equation, we would say log base a of c to the dth power is equal to bd. What power do I have to raise a to to get to c to the dth power? To get to this, I have to raise it to the bd power. But what do we know that b is?"}, {"video_title": "Logarithm of a power Logarithms Algebra II Khan Academy.mp3", "Sentence": "And now this exponential equation, if we were to write it as a logarithmic equation, we would say log base a of c to the dth power is equal to bd. What power do I have to raise a to to get to c to the dth power? To get to this, I have to raise it to the bd power. But what do we know that b is? We already know that b is this thing right over here. So if we substitute this in for b, and we can rewrite this as db, we get logarithm base a of c to the dth power is equal to bd, or you could also call that db, if you switch the order. So that's equal to d times b. b is just log base a of c. So there you have it."}, {"video_title": "Example 1 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Use completing the square to find the value of c that makes x squared minus 44x plus c, so we need to figure out a c, that makes it a perfect square trinomial. Trinomial is just a polynomial with three terms here. Then write the expression as the square of a binomial. So we have x squared minus 44x plus c. So how do we make this into a perfect square? Well if you just look at the traditional pattern for a perfect square, let's just think of it in terms of x plus a squared. That's the same thing as x plus a times x plus a, and we've seen this before. And if you were to multiply this out, that's x times x, which is x squared, plus x times a, which is ax, plus a times x, which is ax, plus a times a, which is a squared."}, {"video_title": "Example 1 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So we have x squared minus 44x plus c. So how do we make this into a perfect square? Well if you just look at the traditional pattern for a perfect square, let's just think of it in terms of x plus a squared. That's the same thing as x plus a times x plus a, and we've seen this before. And if you were to multiply this out, that's x times x, which is x squared, plus x times a, which is ax, plus a times x, which is ax, plus a times a, which is a squared. So it's x squared plus 2ax, these two, you have an ax plus an ax, gives you 2ax, plus a squared. So if we can get this into this pattern, where I have whatever value is here, if I take half of it, this is going to be 2a here, if I take half of it and square it over here, then this will be a perfect square. So if we look over here, this thing right here is 2a, if we want to pattern match, if we want to make this look like a perfect square, that has to be 2a."}, {"video_title": "Example 1 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "And if you were to multiply this out, that's x times x, which is x squared, plus x times a, which is ax, plus a times x, which is ax, plus a times a, which is a squared. So it's x squared plus 2ax, these two, you have an ax plus an ax, gives you 2ax, plus a squared. So if we can get this into this pattern, where I have whatever value is here, if I take half of it, this is going to be 2a here, if I take half of it and square it over here, then this will be a perfect square. So if we look over here, this thing right here is 2a, if we want to pattern match, if we want to make this look like a perfect square, that has to be 2a. So negative 44 is equal to 2a. And this right here, this c, if we pattern match, that has to be equal to, c has to be equal to a squared. So what's a?"}, {"video_title": "Example 1 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So if we look over here, this thing right here is 2a, if we want to pattern match, if we want to make this look like a perfect square, that has to be 2a. So negative 44 is equal to 2a. And this right here, this c, if we pattern match, that has to be equal to, c has to be equal to a squared. So what's a? Well if we know negative 44 is 2a, we can divide both sides of that by 2, and we know that a, or we know that negative 22 has got to be equal to a. a has got to be equal to negative 22. a is half of the coefficient right here, it's half of negative 44. And whenever you complete the square, it's always going to be half of the coefficient right here. Now, if that's a, what does c need to be?"}, {"video_title": "Example 1 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So what's a? Well if we know negative 44 is 2a, we can divide both sides of that by 2, and we know that a, or we know that negative 22 has got to be equal to a. a has got to be equal to negative 22. a is half of the coefficient right here, it's half of negative 44. And whenever you complete the square, it's always going to be half of the coefficient right here. Now, if that's a, what does c need to be? Well c needs to be a squared in order for this to be a perfect square. So c needs to equal negative 22 squared. And we can figure out what that is."}, {"video_title": "Example 1 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Now, if that's a, what does c need to be? Well c needs to be a squared in order for this to be a perfect square. So c needs to equal negative 22 squared. And we can figure out what that is. 22 times 22, we can put the negative later, actually it's going to be the same thing because the negative times the negative is a positive. 2 times 22 is 44, put a 0, 2 times 22 is 44, get a 4, get an 8, get a 4. So it's 484, so if we were to rewrite this as, if we were to rewrite this as x squared minus 44x plus 484, then this is a perfect square trinomial."}, {"video_title": "Example 1 Completing the square Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "And we can figure out what that is. 22 times 22, we can put the negative later, actually it's going to be the same thing because the negative times the negative is a positive. 2 times 22 is 44, put a 0, 2 times 22 is 44, get a 4, get an 8, get a 4. So it's 484, so if we were to rewrite this as, if we were to rewrite this as x squared minus 44x plus 484, then this is a perfect square trinomial. Or we could write it like this, this is x squared minus 2 times, minus 2 times, or maybe I should write it this way, plus 2 times negative 22x plus negative 22 squared. And when you view it that way, it's pretty clear that this is a perfect square and if you were to factor it, it's the same thing as x minus 22 times x minus 22 or x minus 22 squared. These are all equivalent statements."}, {"video_title": "Slope from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "And there's a couple of ways that you can approach it. What my brain wants to do is, well, I know a few forms where it's easy to pick out the slope. For example, if I can manipulate that equation to get me in the form y is equal to mx plus b, well then I know that this m here, the coefficient on the x term, well that's going to be my slope. And b is going to be my y intercept. We cover that in many other videos. Another option is to get into point-slope form. So the general framework or the general template for point-slope form is, if I have an equation of the form y minus y one is equal to m times x minus x one, well then I immediately know that the line that this equation describes is going to have a slope of m, once again."}, {"video_title": "Slope from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "And b is going to be my y intercept. We cover that in many other videos. Another option is to get into point-slope form. So the general framework or the general template for point-slope form is, if I have an equation of the form y minus y one is equal to m times x minus x one, well then I immediately know that the line that this equation describes is going to have a slope of m, once again. And here the y intercept doesn't jump out at you. Let me make sure you can read this over here. The y intercept doesn't jump out at you, but you know a point that is on this line."}, {"video_title": "Slope from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "So the general framework or the general template for point-slope form is, if I have an equation of the form y minus y one is equal to m times x minus x one, well then I immediately know that the line that this equation describes is going to have a slope of m, once again. And here the y intercept doesn't jump out at you. Let me make sure you can read this over here. The y intercept doesn't jump out at you, but you know a point that is on this line. In particular, you know that the point x one, y one is going to be on this line, x one comma y one. So let's look at our original example. So it might immediately jump out at you that this is actually in point-slope form."}, {"video_title": "Slope from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "The y intercept doesn't jump out at you, but you know a point that is on this line. In particular, you know that the point x one, y one is going to be on this line, x one comma y one. So let's look at our original example. So it might immediately jump out at you that this is actually in point-slope form. You might say, well, okay, I see I have a minus x one, so x one would be three. I have my slope here, and that answers our question. Our slope would be negative two."}, {"video_title": "Slope from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "So it might immediately jump out at you that this is actually in point-slope form. You might say, well, okay, I see I have a minus x one, so x one would be three. I have my slope here, and that answers our question. Our slope would be negative two. But here it says plus two. I have to subtract a y one. Well, you could just rewrite this."}, {"video_title": "Slope from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "Our slope would be negative two. But here it says plus two. I have to subtract a y one. Well, you could just rewrite this. So it says, so you have y minus negative two is equal to negative two times x minus three. And then you see it's exactly this point-slope form right over here. So our slope right over there is negative two."}, {"video_title": "Slope from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "Well, you could just rewrite this. So it says, so you have y minus negative two is equal to negative two times x minus three. And then you see it's exactly this point-slope form right over here. So our slope right over there is negative two. And then if I were to ask you, well, give me a point that sits on this line, you could say, all right, an x one would be three, and a y one would be negative two. This point sits on the line. It's not the y intercept, but it's a point on the line, and we know the slope is negative two."}, {"video_title": "Slope from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "So our slope right over there is negative two. And then if I were to ask you, well, give me a point that sits on this line, you could say, all right, an x one would be three, and a y one would be negative two. This point sits on the line. It's not the y intercept, but it's a point on the line, and we know the slope is negative two. Now, another way to approach this is to just manipulate it so that we get into slope-intercept form. So let's do that. Let's manipulate it so we get into slope-intercept form."}, {"video_title": "Slope from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "It's not the y intercept, but it's a point on the line, and we know the slope is negative two. Now, another way to approach this is to just manipulate it so that we get into slope-intercept form. So let's do that. Let's manipulate it so we get into slope-intercept form. So the first thing my brain wants to do is distribute this negative two. And if I do that, I get y plus two is equal to negative two x, negative two times negative three plus six. And then I can subtract two from both sides."}, {"video_title": "Slope from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "Let's manipulate it so we get into slope-intercept form. So the first thing my brain wants to do is distribute this negative two. And if I do that, I get y plus two is equal to negative two x, negative two times negative three plus six. And then I can subtract two from both sides. And then I get y is equal to negative two x plus four. And so here I am in slope-y-intercept form. And once again, I could say, all right, my m here, the coefficient on the x term, is my slope."}, {"video_title": "Slope from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "And then I can subtract two from both sides. And then I get y is equal to negative two x plus four. And so here I am in slope-y-intercept form. And once again, I could say, all right, my m here, the coefficient on the x term, is my slope. So my slope is negative two. Let's do another example. So here, this equation doesn't immediately go into either one of these forms."}, {"video_title": "Slope from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "And once again, I could say, all right, my m here, the coefficient on the x term, is my slope. So my slope is negative two. Let's do another example. So here, this equation doesn't immediately go into either one of these forms. So let's manipulate it. And if it's in either one of them, I like to get into slope-y-intercept form. It's a little bit easier for my brain to understand."}, {"video_title": "Slope from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "So here, this equation doesn't immediately go into either one of these forms. So let's manipulate it. And if it's in either one of them, I like to get into slope-y-intercept form. It's a little bit easier for my brain to understand. So let's do that. So let us collect, well, let's get the x's, let's just isolate the y on the right-hand side since the two y is already there. So let's add three to both sides."}, {"video_title": "Slope from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "It's a little bit easier for my brain to understand. So let's do that. So let us collect, well, let's get the x's, let's just isolate the y on the right-hand side since the two y is already there. So let's add three to both sides. I'm just trying to get rid of this negative three. So if we add three to both sides, on the left-hand side, we have negative four x plus 10 is equal to two y. These cancel out, that was the whole point."}, {"video_title": "Slope from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "So let's add three to both sides. I'm just trying to get rid of this negative three. So if we add three to both sides, on the left-hand side, we have negative four x plus 10 is equal to two y. These cancel out, that was the whole point. And now to solve for y, we just have to divide both sides by two. So if we divide everything by two, we get negative two x plus five is equal to y. So this is in slope-intercept form."}, {"video_title": "Slope from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "These cancel out, that was the whole point. And now to solve for y, we just have to divide both sides by two. So if we divide everything by two, we get negative two x plus five is equal to y. So this is in slope-intercept form. I just have the y on the right-hand side instead of the left-hand side. We have y is equal to mx plus b. And so our m is the coefficient on the x term right over here."}, {"video_title": "Slope from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "So this is in slope-intercept form. I just have the y on the right-hand side instead of the left-hand side. We have y is equal to mx plus b. And so our m is the coefficient on the x term right over here. So our slope is once again is negative two. And here our y-intercept is five in case we wanted to know it. Let's do one more example."}, {"video_title": "Slope from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "And so our m is the coefficient on the x term right over here. So our slope is once again is negative two. And here our y-intercept is five in case we wanted to know it. Let's do one more example. One more example. All right, so once again, this is in neither slope-intercept or point-slope form to begin with. So let's just try to get it to slope-intercept form."}, {"video_title": "Slope from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "Let's do one more example. One more example. All right, so once again, this is in neither slope-intercept or point-slope form to begin with. So let's just try to get it to slope-intercept form. And like always, pause the video and see if you can figure it out yourself. All right, so let's get all the y's on the left-hand side isolated and the x's on the right-hand side. So let me get rid of this negative three x."}, {"video_title": "Slope from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "So let's just try to get it to slope-intercept form. And like always, pause the video and see if you can figure it out yourself. All right, so let's get all the y's on the left-hand side isolated and the x's on the right-hand side. So let me get rid of this negative three x. So I'm gonna add three x to both sides. And let's get rid of this three y right over here. So let's subtract three y from both sides."}, {"video_title": "Slope from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "So let me get rid of this negative three x. So I'm gonna add three x to both sides. And let's get rid of this three y right over here. So let's subtract three y from both sides. You could do this undoing two steps at once, but once again, I'm trying to get rid of this three x. So I'm trying to get rid of this negative three x. So I add three x to the left, but I have to do it to the right if I wanna maintain the equality."}, {"video_title": "Slope from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "So let's subtract three y from both sides. You could do this undoing two steps at once, but once again, I'm trying to get rid of this three x. So I'm trying to get rid of this negative three x. So I add three x to the left, but I have to do it to the right if I wanna maintain the equality. And if I wanna get rid of this three y, well, I subtract three y from here, but I have to do it on the left-hand side if I wanna maintain the equality. So what do I get? That cancels out."}, {"video_title": "Slope from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "So I add three x to the left, but I have to do it to the right if I wanna maintain the equality. And if I wanna get rid of this three y, well, I subtract three y from here, but I have to do it on the left-hand side if I wanna maintain the equality. So what do I get? That cancels out. Five y minus three y is two y is equal to two x plus three x is five x. And then these two characters cancel out. And so if I wanna solve for y, I just divide both sides by two, and I get y is equal to 5 1\u20442 x, and I'm done."}, {"video_title": "Slope from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "That cancels out. Five y minus three y is two y is equal to two x plus three x is five x. And then these two characters cancel out. And so if I wanna solve for y, I just divide both sides by two, and I get y is equal to 5 1\u20442 x, and I'm done. And you might say, wait, this doesn't look exactly like slope-intercept form. Where's my b? Well, your b, if you wanted to see it, you could just write plus zero."}, {"video_title": "Slope from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "And so if I wanna solve for y, I just divide both sides by two, and I get y is equal to 5 1\u20442 x, and I'm done. And you might say, wait, this doesn't look exactly like slope-intercept form. Where's my b? Well, your b, if you wanted to see it, you could just write plus zero. B is implicitly zero right over here. So your slope, your slope is going to be the coefficient on the x term. It's going to be 5 1\u20442."}, {"video_title": "Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3", "Sentence": "We're asked to graph the circle x plus five squared plus y minus five squared equals four. I know what you're thinking, what's all of the silliness on the right-hand side? This is actually just the view we use when we're trying to debug things on Khan Academy, but we can still do the exercise. So it says drag the center point and perimeter of the circle to graph the equation. So the first thing we want to think about is, well, what's the center of this equation? Well, a standard form of a circle is x minus the x-coordinate of the center squared plus y minus the y-coordinate of the center squared is equal to the radius squared. So x minus the x-coordinate of the center."}, {"video_title": "Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3", "Sentence": "So it says drag the center point and perimeter of the circle to graph the equation. So the first thing we want to think about is, well, what's the center of this equation? Well, a standard form of a circle is x minus the x-coordinate of the center squared plus y minus the y-coordinate of the center squared is equal to the radius squared. So x minus the x-coordinate of the center. So the x-coordinate of the center must be negative five because the only way we can get a positive five is by subtracting a negative five. So the x-coordinate must be negative five and the y-coordinate must be positive five because it's y minus the y coordinate of the center. So y-coordinate is positive five."}, {"video_title": "Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3", "Sentence": "So x minus the x-coordinate of the center. So the x-coordinate of the center must be negative five because the only way we can get a positive five is by subtracting a negative five. So the x-coordinate must be negative five and the y-coordinate must be positive five because it's y minus the y coordinate of the center. So y-coordinate is positive five. And then the radius squared is going to be equal to four. So that means that the radius is equal to two. And the way it's drawn right now, I mean we could drag this out like this, but this, the way it's drawn, the radius is indeed equal to two."}, {"video_title": "Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3", "Sentence": "So y-coordinate is positive five. And then the radius squared is going to be equal to four. So that means that the radius is equal to two. And the way it's drawn right now, I mean we could drag this out like this, but this, the way it's drawn, the radius is indeed equal to two. And so we are done. And I really want to hit the point home of what I just did. So let me get my little, get my scratch pad out."}, {"video_title": "Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3", "Sentence": "And the way it's drawn right now, I mean we could drag this out like this, but this, the way it's drawn, the radius is indeed equal to two. And so we are done. And I really want to hit the point home of what I just did. So let me get my little, get my scratch pad out. So this, sorry for knocking the microphone just now. So that equation was x plus five squared plus y minus five squared is equal to four squared. And so I want to rewrite this as, this is x minus negative five, x minus negative five squared plus y minus positive five, positive five squared is equal to, and instead of writing it as four, I'll write it as two squared."}, {"video_title": "Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3", "Sentence": "So let me get my little, get my scratch pad out. So this, sorry for knocking the microphone just now. So that equation was x plus five squared plus y minus five squared is equal to four squared. And so I want to rewrite this as, this is x minus negative five, x minus negative five squared plus y minus positive five, positive five squared is equal to, and instead of writing it as four, I'll write it as two squared. And so this right over here tells us that the center of the circle is going to be, x equals negative five, y equals five, and the radius is going to be equal to two. And once again, this is no magic here. This is not just, I don't want you to just memorize this formula."}, {"video_title": "Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3", "Sentence": "And so I want to rewrite this as, this is x minus negative five, x minus negative five squared plus y minus positive five, positive five squared is equal to, and instead of writing it as four, I'll write it as two squared. And so this right over here tells us that the center of the circle is going to be, x equals negative five, y equals five, and the radius is going to be equal to two. And once again, this is no magic here. This is not just, I don't want you to just memorize this formula. I want you to appreciate that this formula comes straight out of the Pythagorean theorem, straight out of the distance formula, which comes out of the Pythagorean theorem. Remember, if you have some center, in this case it's the point negative five comma five. So negative five comma five."}, {"video_title": "Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3", "Sentence": "This is not just, I don't want you to just memorize this formula. I want you to appreciate that this formula comes straight out of the Pythagorean theorem, straight out of the distance formula, which comes out of the Pythagorean theorem. Remember, if you have some center, in this case it's the point negative five comma five. So negative five comma five. And you want to find all of the x's and y's that are two away from it. So you want to find all the x's and y's that are two away from it. So that would be one of them."}, {"video_title": "Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3", "Sentence": "So negative five comma five. And you want to find all of the x's and y's that are two away from it. So you want to find all the x's and y's that are two away from it. So that would be one of them. X comma y. This distance is two. And there's going to be a bunch of them."}, {"video_title": "Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3", "Sentence": "So that would be one of them. X comma y. This distance is two. And there's going to be a bunch of them. And when you plot all of them together, you're going to get a circle with a radius two around that center. But let's think about how we got that actual formula. Well, the distance between that coordinate, between any of these x's and y's, it could be an x and y here, it could be an x and y here, and this is going to be two."}, {"video_title": "Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3", "Sentence": "And there's going to be a bunch of them. And when you plot all of them together, you're going to get a circle with a radius two around that center. But let's think about how we got that actual formula. Well, the distance between that coordinate, between any of these x's and y's, it could be an x and y here, it could be an x and y here, and this is going to be two. So we could have our change in x. So we have x minus negative five. So that's our change in x between any point x comma y and negative five comma five."}, {"video_title": "Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3", "Sentence": "Well, the distance between that coordinate, between any of these x's and y's, it could be an x and y here, it could be an x and y here, and this is going to be two. So we could have our change in x. So we have x minus negative five. So that's our change in x between any point x comma y and negative five comma five. So our change in x squared plus our change in y squared, so that's going to be y minus the y coordinate over here, squared, is going to be equal to the radius squared. So the change in y, it's going to be from this y to that y. If this is the end point, it'd be the end minus the beginning, y minus five."}, {"video_title": "Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3", "Sentence": "So that's our change in x between any point x comma y and negative five comma five. So our change in x squared plus our change in y squared, so that's going to be y minus the y coordinate over here, squared, is going to be equal to the radius squared. So the change in y, it's going to be from this y to that y. If this is the end point, it'd be the end minus the beginning, y minus five. Y minus five squared. And so this shows for any xy that is two away from the center, this equation will hold. And it becomes, I'll just write this in neutral color, x plus five squared plus y minus five squared is equal to the radius squared, is equal to two, or let me just write that, is equal to four."}, {"video_title": "Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3", "Sentence": "If this is the end point, it'd be the end minus the beginning, y minus five. Y minus five squared. And so this shows for any xy that is two away from the center, this equation will hold. And it becomes, I'll just write this in neutral color, x plus five squared plus y minus five squared is equal to the radius squared, is equal to two, or let me just write that, is equal to four. And let me make it, I really want to, you know, I dislike it when formulas are just memorized and you don't see the connection to other things. Notice, we can construct, we can construct a nice little right triangle here. So our change in x is that right over there."}, {"video_title": "Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3", "Sentence": "And it becomes, I'll just write this in neutral color, x plus five squared plus y minus five squared is equal to the radius squared, is equal to two, or let me just write that, is equal to four. And let me make it, I really want to, you know, I dislike it when formulas are just memorized and you don't see the connection to other things. Notice, we can construct, we can construct a nice little right triangle here. So our change in x is that right over there. So that is our change in x, change in x. And our change in y, our change in y, not the change in y squared, but our change in y is that right over there. Change in y, our change in y, you could view that as y minus, so this is change in y is going to be y minus five."}, {"video_title": "Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3", "Sentence": "So our change in x is that right over there. So that is our change in x, change in x. And our change in y, our change in y, not the change in y squared, but our change in y is that right over there. Change in y, our change in y, you could view that as y minus, so this is change in y is going to be y minus five. And our change in x is x minus negative five. X minus negative five. So this is just change in x squared plus change in y squared is equal to the hypotenuse squared, which is the length of, which is this radius."}, {"video_title": "Example 2 Factoring quadratics by grouping Algebra I Khan Academy.mp3", "Sentence": "We're asked to factor 4y squared plus 4y minus 15. And whenever you have an expression like this, where you have a non-1 coefficient on the y squared, or on the second degree term, it could have been an x squared, the best way to do this is by grouping. And to factor by grouping, we need to look for 2 numbers whose product is equal to 4 times negative 15. So we're looking for 2 numbers whose product, let's call those a and b, is going to be equal to 4 times negative 15. 4 times negative 15, or negative 60. And the sum of those 2 numbers, a plus b, needs to be equal to this 4 right there. Needs to be equal to 4."}, {"video_title": "Example 2 Factoring quadratics by grouping Algebra I Khan Academy.mp3", "Sentence": "So we're looking for 2 numbers whose product, let's call those a and b, is going to be equal to 4 times negative 15. 4 times negative 15, or negative 60. And the sum of those 2 numbers, a plus b, needs to be equal to this 4 right there. Needs to be equal to 4. So let's think about all the factors of negative 60 or 60. And we're looking for ones that are essentially 4 apart, because the numbers are going to be of different signs, because their product is negative. So when you take 2 numbers of different signs and you sum them, you kind of view it as the difference of their absolute values."}, {"video_title": "Example 2 Factoring quadratics by grouping Algebra I Khan Academy.mp3", "Sentence": "Needs to be equal to 4. So let's think about all the factors of negative 60 or 60. And we're looking for ones that are essentially 4 apart, because the numbers are going to be of different signs, because their product is negative. So when you take 2 numbers of different signs and you sum them, you kind of view it as the difference of their absolute values. If that confuses you, don't worry about it. But this tells you that the numbers, since they're going to be of different size, their absolute values are going to be roughly 4 apart. So we could try out things like 5 and 12."}, {"video_title": "Example 2 Factoring quadratics by grouping Algebra I Khan Academy.mp3", "Sentence": "So when you take 2 numbers of different signs and you sum them, you kind of view it as the difference of their absolute values. If that confuses you, don't worry about it. But this tells you that the numbers, since they're going to be of different size, their absolute values are going to be roughly 4 apart. So we could try out things like 5 and 12. 5 and negative 12, because 1 has to be negative. If you add these 2, you get negative 7. If you did negative 5 and 12, you'd get positive 7."}, {"video_title": "Example 2 Factoring quadratics by grouping Algebra I Khan Academy.mp3", "Sentence": "So we could try out things like 5 and 12. 5 and negative 12, because 1 has to be negative. If you add these 2, you get negative 7. If you did negative 5 and 12, you'd get positive 7. They're just still too far apart. What if we tried 6 and negative 10? Then you'd get a negative 4 if you added these 2."}, {"video_title": "Example 2 Factoring quadratics by grouping Algebra I Khan Academy.mp3", "Sentence": "If you did negative 5 and 12, you'd get positive 7. They're just still too far apart. What if we tried 6 and negative 10? Then you'd get a negative 4 if you added these 2. We want a positive 4, so let's do negative 6 and 10. Negative 6 plus 10 is positive 4. So those will be our 2 numbers, negative 6 and positive 10."}, {"video_title": "Example 2 Factoring quadratics by grouping Algebra I Khan Academy.mp3", "Sentence": "Then you'd get a negative 4 if you added these 2. We want a positive 4, so let's do negative 6 and 10. Negative 6 plus 10 is positive 4. So those will be our 2 numbers, negative 6 and positive 10. Now, what we want to do is we want to break up this middle term here. The whole point of figuring out the negative 6 and the 10 is to break up the 4y into a negative 6y and a 10y. So let's do that."}, {"video_title": "Example 2 Factoring quadratics by grouping Algebra I Khan Academy.mp3", "Sentence": "So those will be our 2 numbers, negative 6 and positive 10. Now, what we want to do is we want to break up this middle term here. The whole point of figuring out the negative 6 and the 10 is to break up the 4y into a negative 6y and a 10y. So let's do that. So this 4y can be rewritten as negative 6y plus 10y. Because if you add those, you get 4y. And on the other sides of it, you have your 4y squared, and then you have your minus 15."}, {"video_title": "Example 2 Factoring quadratics by grouping Algebra I Khan Academy.mp3", "Sentence": "So let's do that. So this 4y can be rewritten as negative 6y plus 10y. Because if you add those, you get 4y. And on the other sides of it, you have your 4y squared, and then you have your minus 15. All I did is expand this into these 2 numbers as being the coefficients on the y. If you add these, you get the 4y again. Now, this is where the grouping comes in."}, {"video_title": "Example 2 Factoring quadratics by grouping Algebra I Khan Academy.mp3", "Sentence": "And on the other sides of it, you have your 4y squared, and then you have your minus 15. All I did is expand this into these 2 numbers as being the coefficients on the y. If you add these, you get the 4y again. Now, this is where the grouping comes in. You group the terms. So let's see. Let me do it in a different color."}, {"video_title": "Example 2 Factoring quadratics by grouping Algebra I Khan Academy.mp3", "Sentence": "Now, this is where the grouping comes in. You group the terms. So let's see. Let me do it in a different color. So if I take these 2 guys, what can I factor out of those 2 guys? Well, there's a common factor. It looks like there's a common factor of 2y."}, {"video_title": "Example 2 Factoring quadratics by grouping Algebra I Khan Academy.mp3", "Sentence": "Let me do it in a different color. So if I take these 2 guys, what can I factor out of those 2 guys? Well, there's a common factor. It looks like there's a common factor of 2y. So if we factor out 2y, we get 2y times 4y squared divided by 2y is 2y. And then negative 6y divided by 2y is negative 3. Negative 3."}, {"video_title": "Example 2 Factoring quadratics by grouping Algebra I Khan Academy.mp3", "Sentence": "It looks like there's a common factor of 2y. So if we factor out 2y, we get 2y times 4y squared divided by 2y is 2y. And then negative 6y divided by 2y is negative 3. Negative 3. So this group gets factored into 2y times 2y minus 3. Now, let's look at this other group right here. This was the whole point about breaking it up like this."}, {"video_title": "Example 2 Factoring quadratics by grouping Algebra I Khan Academy.mp3", "Sentence": "Negative 3. So this group gets factored into 2y times 2y minus 3. Now, let's look at this other group right here. This was the whole point about breaking it up like this. And in other videos, I've explained why this works. Now, here, the greatest common factor is a 5. So we can factor out a 5."}, {"video_title": "Example 2 Factoring quadratics by grouping Algebra I Khan Academy.mp3", "Sentence": "This was the whole point about breaking it up like this. And in other videos, I've explained why this works. Now, here, the greatest common factor is a 5. So we can factor out a 5. So this is equal to plus 5 times 10y divided by 5 is 2y. Negative 15 divided by 5 is 3. And so we have 2y times 2y minus 3 plus 5 times 2y minus 3."}, {"video_title": "Example 2 Factoring quadratics by grouping Algebra I Khan Academy.mp3", "Sentence": "So we can factor out a 5. So this is equal to plus 5 times 10y divided by 5 is 2y. Negative 15 divided by 5 is 3. And so we have 2y times 2y minus 3 plus 5 times 2y minus 3. So now you have 2 terms, and 2y minus 3 is a common factor to both. So let's factor out a 2y minus 3. So this is equal to 2y minus 3 times 2y, times that 2y, plus that 5."}, {"video_title": "Example 2 Factoring quadratics by grouping Algebra I Khan Academy.mp3", "Sentence": "And so we have 2y times 2y minus 3 plus 5 times 2y minus 3. So now you have 2 terms, and 2y minus 3 is a common factor to both. So let's factor out a 2y minus 3. So this is equal to 2y minus 3 times 2y, times that 2y, plus that 5. There's no magic happening here. All I did is undistribute the 2y minus 3. I factored it out of both of these guys and took it out of the parentheses."}, {"video_title": "Example 2 Factoring quadratics by grouping Algebra I Khan Academy.mp3", "Sentence": "So this is equal to 2y minus 3 times 2y, times that 2y, plus that 5. There's no magic happening here. All I did is undistribute the 2y minus 3. I factored it out of both of these guys and took it out of the parentheses. If I distributed it in, you'd get back to this expression. But we're done. We factored it."}, {"video_title": "Determining whether values are in domain of function.mp3", "Sentence": "We're asked to determine for each x value whether it is in the domain of f or not, and they have our definition of f of x up here. So pause this video and see if you can work through this before we do it together. All right, so just as a bit of a review, if x is in the domain of our function, that means that if we input our x into our function, then we are going to get a legitimate output f of x. But if for whatever reason f isn't defined at x, or if it gets some kind of undefined state, well, then it is, then x would not be in the domain. So let's try these different values. Is x equal negative five in the domain of f? Let's see what happens if we try to evaluate f of negative five."}, {"video_title": "Determining whether values are in domain of function.mp3", "Sentence": "But if for whatever reason f isn't defined at x, or if it gets some kind of undefined state, well, then it is, then x would not be in the domain. So let's try these different values. Is x equal negative five in the domain of f? Let's see what happens if we try to evaluate f of negative five. Well, then we, in the numerator, we get negative five plus five. Every place where we see an x, we replace it with a negative five. So it's negative five plus five over negative five minus three which is equal to, in our numerator we get zero, in our denominator we get negative eight."}, {"video_title": "Determining whether values are in domain of function.mp3", "Sentence": "Let's see what happens if we try to evaluate f of negative five. Well, then we, in the numerator, we get negative five plus five. Every place where we see an x, we replace it with a negative five. So it's negative five plus five over negative five minus three which is equal to, in our numerator we get zero, in our denominator we get negative eight. Now, at first you see the zero, you might get a little bit worried, but it's just a zero in the numerator, so this whole thing just evaluates to a zero, which is a completely legitimate output. So x equals negative five is in the domain. What about x equals zero?"}, {"video_title": "Determining whether values are in domain of function.mp3", "Sentence": "So it's negative five plus five over negative five minus three which is equal to, in our numerator we get zero, in our denominator we get negative eight. Now, at first you see the zero, you might get a little bit worried, but it's just a zero in the numerator, so this whole thing just evaluates to a zero, which is a completely legitimate output. So x equals negative five is in the domain. What about x equals zero? Is that in the domain? Pause the video, see if you can figure that out. Well, f of zero is going to be equal to, in our numerator we have zero plus five, in our denominator we have zero minus three."}, {"video_title": "Determining whether values are in domain of function.mp3", "Sentence": "What about x equals zero? Is that in the domain? Pause the video, see if you can figure that out. Well, f of zero is going to be equal to, in our numerator we have zero plus five, in our denominator we have zero minus three. Well, that's just going to get us five in the numerator and negative three in the denominator, this would just be negative 5 3rds, but this is a completely legitimate output. So the function is defined at x equals zero, so it's in the domain for sure. Now, what about x equals three?"}, {"video_title": "Determining whether values are in domain of function.mp3", "Sentence": "Well, f of zero is going to be equal to, in our numerator we have zero plus five, in our denominator we have zero minus three. Well, that's just going to get us five in the numerator and negative three in the denominator, this would just be negative 5 3rds, but this is a completely legitimate output. So the function is defined at x equals zero, so it's in the domain for sure. Now, what about x equals three? Pause the video, try to figure that out. Well, I'll do that up here. F of three is going to be equal to what?"}, {"video_title": "Determining whether values are in domain of function.mp3", "Sentence": "Now, what about x equals three? Pause the video, try to figure that out. Well, I'll do that up here. F of three is going to be equal to what? And you might already see some warning signs as to what's going to happen here in the denominator, but I'll just evaluate the whole thing. In the numerator we get three plus five, in the denominator we get three minus three. So this is going to be equal to eight over zero."}, {"video_title": "Determining whether values are in domain of function.mp3", "Sentence": "F of three is going to be equal to what? And you might already see some warning signs as to what's going to happen here in the denominator, but I'll just evaluate the whole thing. In the numerator we get three plus five, in the denominator we get three minus three. So this is going to be equal to eight over zero. Now, what is eight divided by zero? Well, we don't know. This is one of those fascinating things in mathematics."}, {"video_title": "Determining whether values are in domain of function.mp3", "Sentence": "So this is going to be equal to eight over zero. Now, what is eight divided by zero? Well, we don't know. This is one of those fascinating things in mathematics. We haven't defined what happens when something is divided by zero. So three is not in the domain. The function is not defined there, not in domain."}, {"video_title": "Determining whether values are in domain of function.mp3", "Sentence": "This is one of those fascinating things in mathematics. We haven't defined what happens when something is divided by zero. So three is not in the domain. The function is not defined there, not in domain. Let's do another example. Determine for each x value whether it is in the domain of g or not. So pause this video and try to work through all three of these."}, {"video_title": "Determining whether values are in domain of function.mp3", "Sentence": "The function is not defined there, not in domain. Let's do another example. Determine for each x value whether it is in the domain of g or not. So pause this video and try to work through all three of these. All right, so first of all, when x equals negative three, do we get a legitimate g of x? So let's see, g of negative three, if we try to evaluate, that's going to be the square root of three times negative three, which is equal to the square root of negative nine. Well, with just a principal square root like this, we don't know how to evaluate this."}, {"video_title": "Determining whether values are in domain of function.mp3", "Sentence": "So pause this video and try to work through all three of these. All right, so first of all, when x equals negative three, do we get a legitimate g of x? So let's see, g of negative three, if we try to evaluate, that's going to be the square root of three times negative three, which is equal to the square root of negative nine. Well, with just a principal square root like this, we don't know how to evaluate this. So this is not in the domain. What about when x equals zero? Well, g of zero is going to be equal to the square root of three times zero, which is equal to the square root of zero, which is equal to zero, so that gave us a legitimate result."}, {"video_title": "Determining whether values are in domain of function.mp3", "Sentence": "Well, with just a principal square root like this, we don't know how to evaluate this. So this is not in the domain. What about when x equals zero? Well, g of zero is going to be equal to the square root of three times zero, which is equal to the square root of zero, which is equal to zero, so that gave us a legitimate result. So that is in the domain. And what about g of two, or x equals two? Does that give us a legitimate g of two?"}, {"video_title": "Determining whether values are in domain of function.mp3", "Sentence": "Well, g of zero is going to be equal to the square root of three times zero, which is equal to the square root of zero, which is equal to zero, so that gave us a legitimate result. So that is in the domain. And what about g of two, or x equals two? Does that give us a legitimate g of two? Well, g of two is going to be equal to the square root of three times two, which is equal to the square root of six, which is a legitimate output. So x equals two is in the domain. Let's do one last example."}, {"video_title": "Determining whether values are in domain of function.mp3", "Sentence": "Does that give us a legitimate g of two? Well, g of two is going to be equal to the square root of three times two, which is equal to the square root of six, which is a legitimate output. So x equals two is in the domain. Let's do one last example. So we're told this is h of x right over here, and once again, we have to figure out whether these x values are in the domain or not. Pause this video and see if you can work through that. All right, well, let's just first think about h of negative one."}, {"video_title": "Determining whether values are in domain of function.mp3", "Sentence": "Let's do one last example. So we're told this is h of x right over here, and once again, we have to figure out whether these x values are in the domain or not. Pause this video and see if you can work through that. All right, well, let's just first think about h of negative one. What's that going to be equal to? Negative one, every place we see an x, we're going to replace it with a negative one, minus five squared. Well, this is going to be equal to negative six squared, negative six squared, which is equal to positive 36, which is a very legitimate output, and so this is definitely in the domain."}, {"video_title": "Determining whether values are in domain of function.mp3", "Sentence": "All right, well, let's just first think about h of negative one. What's that going to be equal to? Negative one, every place we see an x, we're going to replace it with a negative one, minus five squared. Well, this is going to be equal to negative six squared, negative six squared, which is equal to positive 36, which is a very legitimate output, and so this is definitely in the domain. What about five? So h of five is going to be equal to five minus five squared. Now, you might be getting worried because you're seeing a zero here, but it's not like we're trying to divide by zero."}, {"video_title": "Determining whether values are in domain of function.mp3", "Sentence": "Well, this is going to be equal to negative six squared, negative six squared, which is equal to positive 36, which is a very legitimate output, and so this is definitely in the domain. What about five? So h of five is going to be equal to five minus five squared. Now, you might be getting worried because you're seeing a zero here, but it's not like we're trying to divide by zero. We're just squaring zero, which is completely legitimate. So zero squared is just a zero, and so h of five is very much defined, so this is in the domain. Now, what about h of 10?"}, {"video_title": "Testing if a relationship is a function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "We're asked, do the points on the graph below represent a function? So in order for the points to represent a function, for every input into our function, we can only get one value. So if we look here, they've graphed the point, looks like negative 1, 3. So that's the point negative 1, 3. So if we assume that this is our x-axis, and that that is our f of x axis, and I'm just assuming it's a function, I don't know whether it really is just now, this point is telling us that if you put negative 1 into our function, or the thing that might be a function, or maybe our relation, you'll get a 3. So it's telling us that f of negative 1 is equal to 3. So, so far, it could be a reasonable function."}, {"video_title": "Testing if a relationship is a function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So that's the point negative 1, 3. So if we assume that this is our x-axis, and that that is our f of x axis, and I'm just assuming it's a function, I don't know whether it really is just now, this point is telling us that if you put negative 1 into our function, or the thing that might be a function, or maybe our relation, you'll get a 3. So it's telling us that f of negative 1 is equal to 3. So, so far, it could be a reasonable function. You give me negative 1, and I will map it to 3. Then they have, if x is 2, then our value is negative 2. This is the point 2, negative 2."}, {"video_title": "Testing if a relationship is a function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So, so far, it could be a reasonable function. You give me negative 1, and I will map it to 3. Then they have, if x is 2, then our value is negative 2. This is the point 2, negative 2. So that still seems consistent with being a function. If you pass me 2, I will map you, or I will point you to negative 2. Seems fair enough."}, {"video_title": "Testing if a relationship is a function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "This is the point 2, negative 2. So that still seems consistent with being a function. If you pass me 2, I will map you, or I will point you to negative 2. Seems fair enough. Let's see this next value here. This is the point 3, 2 right there. So once again, that says that, look, if you give me 3 into my function, into my black box, I will output a 2."}, {"video_title": "Testing if a relationship is a function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Seems fair enough. Let's see this next value here. This is the point 3, 2 right there. So once again, that says that, look, if you give me 3 into my function, into my black box, I will output a 2. Pretty reasonable. No reason why these points can't represent a function so far. Now, what about when we input 4 into the function?"}, {"video_title": "Testing if a relationship is a function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So once again, that says that, look, if you give me 3 into my function, into my black box, I will output a 2. Pretty reasonable. No reason why these points can't represent a function so far. Now, what about when we input 4 into the function? Let me do this in a new special, let me do this in magenta. So what happens if I input 4 into my function? So this is 4 right here."}, {"video_title": "Testing if a relationship is a function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Now, what about when we input 4 into the function? Let me do this in a new special, let me do this in magenta. So what happens if I input 4 into my function? So this is 4 right here. Well, according to these points, there's two points that relate to 4, that 4 can be mapped to. I could map it to the point 4, that's 5 up here, 4, 5. So that says if you give me a 4, I'll give you a 5."}, {"video_title": "Testing if a relationship is a function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So this is 4 right here. Well, according to these points, there's two points that relate to 4, that 4 can be mapped to. I could map it to the point 4, that's 5 up here, 4, 5. So that says if you give me a 4, I'll give you a 5. It also says if you give me a 4, I could also give you a negative 1 because that's the point 4, negative 1. So this is not a function. It cannot be a function if for some input into the function, you could give me two different values, and you can see that right here."}, {"video_title": "Testing if a relationship is a function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So that says if you give me a 4, I'll give you a 5. It also says if you give me a 4, I could also give you a negative 1 because that's the point 4, negative 1. So this is not a function. It cannot be a function if for some input into the function, you could give me two different values, and you can see that right here. And an easy test is to just see, look, for one value, I have two points for this relation. So this cannot be a function. So this is not a function."}, {"video_title": "Subtracting complex numbers Imaginary and complex numbers Algebra II Khan Academy.mp3", "Sentence": "So the first thing I'd like to do here is to just kind of get rid of these parentheses. So we just have a bunch of real parts and imaginary parts that we can then add up together. So we have 2 minus 3i, and then we're subtracting this entire quantity. And to get rid of the parentheses, we can just distribute the negative sign, or another way to think about it, we can say that this is negative 1 times all of this. So we can just distribute the negative sign, and negative 1 times 6 is negative 6. Let me do these in magenta. So this is negative 6, and then negative 1 times negative 18i, well, that's just going to be positive."}, {"video_title": "Subtracting complex numbers Imaginary and complex numbers Algebra II Khan Academy.mp3", "Sentence": "And to get rid of the parentheses, we can just distribute the negative sign, or another way to think about it, we can say that this is negative 1 times all of this. So we can just distribute the negative sign, and negative 1 times 6 is negative 6. Let me do these in magenta. So this is negative 6, and then negative 1 times negative 18i, well, that's just going to be positive. Positive 18i. Negative times a negative is a positive. And now we want to add the real parts, and we want to add the imaginary parts."}, {"video_title": "Subtracting complex numbers Imaginary and complex numbers Algebra II Khan Academy.mp3", "Sentence": "So this is negative 6, and then negative 1 times negative 18i, well, that's just going to be positive. Positive 18i. Negative times a negative is a positive. And now we want to add the real parts, and we want to add the imaginary parts. So here's a real part here, 2, and then we have a minus 6, so we have 2 minus 6, and we want to add the imaginary parts. We have a negative, let me do that in a different color, we have a negative 3i right over here, so negative 3i, or minus 3i right over there, and then we have a plus 18i, or positive 18i. So then we have a positive 18i."}, {"video_title": "Subtracting complex numbers Imaginary and complex numbers Algebra II Khan Academy.mp3", "Sentence": "And now we want to add the real parts, and we want to add the imaginary parts. So here's a real part here, 2, and then we have a minus 6, so we have 2 minus 6, and we want to add the imaginary parts. We have a negative, let me do that in a different color, we have a negative 3i right over here, so negative 3i, or minus 3i right over there, and then we have a plus 18i, or positive 18i. So then we have a positive 18i. If you add the real parts, 2 minus 6 is negative 4, and you add the imaginary parts, if I have negative 3 of something, and to that I add 18 of something, well, that's just going to leave me with 15 of that something. Or another way you could think about it, if I have 18 of something, and I subtract 3 of that something, I'll have 15 of that something. And in this case, the something is i, is the imaginary unit."}, {"video_title": "Solving quadratics using structure Mathematics II High School Math Khan Academy.mp3", "Sentence": "So let's try to find the solutions to this equation right over here. We have the quantity two x minus three squared, and that is equal to four x minus six. I encourage you to pause the video and give it a shot, and I'll give you a little bit of a hint. You could do this in the traditional way of expanding this out, but there might, and then turning it into kind of a classic quadratic form, but there might be a faster or a simpler way to do this if you really pay attention to the structure of both sides of this equation. Well, let's look at this. We have two x minus three squared on the left-hand side. On the right-hand side, we have four x minus six."}, {"video_title": "Solving quadratics using structure Mathematics II High School Math Khan Academy.mp3", "Sentence": "You could do this in the traditional way of expanding this out, but there might, and then turning it into kind of a classic quadratic form, but there might be a faster or a simpler way to do this if you really pay attention to the structure of both sides of this equation. Well, let's look at this. We have two x minus three squared on the left-hand side. On the right-hand side, we have four x minus six. Well, four x minus six, that's just two times two x minus three. Let me be clear there. So this is the same thing as two x minus three squared is equal to four x minus six."}, {"video_title": "Solving quadratics using structure Mathematics II High School Math Khan Academy.mp3", "Sentence": "On the right-hand side, we have four x minus six. Well, four x minus six, that's just two times two x minus three. Let me be clear there. So this is the same thing as two x minus three squared is equal to four x minus six. If I factor out a two, that's two times two x minus three. And so this is really interesting. We have something squared is equal to two times that something."}, {"video_title": "Solving quadratics using structure Mathematics II High School Math Khan Academy.mp3", "Sentence": "So this is the same thing as two x minus three squared is equal to four x minus six. If I factor out a two, that's two times two x minus three. And so this is really interesting. We have something squared is equal to two times that something. So if we can solve for the something, let me be very clear here. So the stuff in blue squared is equal to two times the stuff in blue. So if we can solve for what the stuff in blue could be equal to, then we could solve for x, and I'll show you that right now."}, {"video_title": "Solving quadratics using structure Mathematics II High School Math Khan Academy.mp3", "Sentence": "We have something squared is equal to two times that something. So if we can solve for the something, let me be very clear here. So the stuff in blue squared is equal to two times the stuff in blue. So if we can solve for what the stuff in blue could be equal to, then we could solve for x, and I'll show you that right now. So let's say, let's just replace two x minus three. We'll do a little bit of a substitution. Let's replace that with, let's replace it with p. So let's say that p is equal to two x minus three."}, {"video_title": "Solving quadratics using structure Mathematics II High School Math Khan Academy.mp3", "Sentence": "So if we can solve for what the stuff in blue could be equal to, then we could solve for x, and I'll show you that right now. So let's say, let's just replace two x minus three. We'll do a little bit of a substitution. Let's replace that with, let's replace it with p. So let's say that p is equal to two x minus three. Well, then this equation simplifies quite nicely. The left-hand side becomes p squared is equal to two times p. Because once again, two x minus three is p. Two times p. And now we just have to solve for p. And I'll switch to just one color now. So we can write this as, if we subtract two p from both sides, we can get p squared minus two p is equal to zero."}, {"video_title": "Solving quadratics using structure Mathematics II High School Math Khan Academy.mp3", "Sentence": "Let's replace that with, let's replace it with p. So let's say that p is equal to two x minus three. Well, then this equation simplifies quite nicely. The left-hand side becomes p squared is equal to two times p. Because once again, two x minus three is p. Two times p. And now we just have to solve for p. And I'll switch to just one color now. So we can write this as, if we subtract two p from both sides, we can get p squared minus two p is equal to zero. And we can factor out a p. So we get p times p minus two is equal to zero. And we've seen this show multiple times. If I have the product of two things and they equal to zero, at least one of them needs to be equal to zero."}, {"video_title": "Solving quadratics using structure Mathematics II High School Math Khan Academy.mp3", "Sentence": "So we can write this as, if we subtract two p from both sides, we can get p squared minus two p is equal to zero. And we can factor out a p. So we get p times p minus two is equal to zero. And we've seen this show multiple times. If I have the product of two things and they equal to zero, at least one of them needs to be equal to zero. So either p is equal to zero or p minus two is equal to zero. Well, if p minus two is equal to zero, that means p is equal to two. So either p equals zero or p equals two."}, {"video_title": "Solving quadratics using structure Mathematics II High School Math Khan Academy.mp3", "Sentence": "If I have the product of two things and they equal to zero, at least one of them needs to be equal to zero. So either p is equal to zero or p minus two is equal to zero. Well, if p minus two is equal to zero, that means p is equal to two. So either p equals zero or p equals two. Well, we're not quite done yet because we wanted to solve for x, not for p. But luckily, we know that two x minus three is equal to p. So now we could say either two x minus three is going to be equal to this p value, is going to be equal to zero, or two x minus three is going to be equal to this p value, is going to be equal to two. And so this is pretty straightforward to solve. Add three to both sides."}, {"video_title": "Solving quadratics using structure Mathematics II High School Math Khan Academy.mp3", "Sentence": "So either p equals zero or p equals two. Well, we're not quite done yet because we wanted to solve for x, not for p. But luckily, we know that two x minus three is equal to p. So now we could say either two x minus three is going to be equal to this p value, is going to be equal to zero, or two x minus three is going to be equal to this p value, is going to be equal to two. And so this is pretty straightforward to solve. Add three to both sides. You get two x is equal to three. Divide both sides by two. And we get x is equal to 3 halves."}, {"video_title": "Solving quadratics using structure Mathematics II High School Math Khan Academy.mp3", "Sentence": "Add three to both sides. You get two x is equal to three. Divide both sides by two. And we get x is equal to 3 halves. Or over here, if we add three to both sides, we get two x is equal to five, divide both sides by two, and you get x is equal to 5 halves. So these are the possible solutions. And this is pretty neat."}, {"video_title": "Solving quadratics using structure Mathematics II High School Math Khan Academy.mp3", "Sentence": "And we get x is equal to 3 halves. Or over here, if we add three to both sides, we get two x is equal to five, divide both sides by two, and you get x is equal to 5 halves. So these are the possible solutions. And this is pretty neat. This one right over here, you could almost do this in your head. It was nice and simple. Well, if you were to expand this out and then subtract this, it would have been a much more complex set of operations that you would have done."}, {"video_title": "Solving quadratics using structure Mathematics II High School Math Khan Academy.mp3", "Sentence": "And this is pretty neat. This one right over here, you could almost do this in your head. It was nice and simple. Well, if you were to expand this out and then subtract this, it would have been a much more complex set of operations that you would have done. You still would have hopefully gotten to the right answer. But it would have just taken a lot more steps. But here we could appreciate some patterns that we saw in our equations."}, {"video_title": "Fractional exponents with numerators other than 1 Algebra I Khan Academy.mp3", "Sentence": "We saw that this is the exact same thing as taking the cube root of 64. And because we know that 4 times 4 times 4, or 4 to the 3rd power, is equal to 64, if we're looking for the cube root of 64, we're looking for a number that that number times that number times that same number is going to be equal to 64. Well, we know that number is 4, so this thing right over here is going to be 4. Now we're going to think of slightly more complex fractional exponents. The one we see here has a 1 in the numerator. Now we're going to see something different. So what I want to do is think about what 64 to the 2 3rd power is."}, {"video_title": "Fractional exponents with numerators other than 1 Algebra I Khan Academy.mp3", "Sentence": "Now we're going to think of slightly more complex fractional exponents. The one we see here has a 1 in the numerator. Now we're going to see something different. So what I want to do is think about what 64 to the 2 3rd power is. And here I'm going to use a property of exponents that we'll study more later on. But this property of exponents is the idea that, let's say with a simple number, if I raise something to the 3rd power, and then I were to raise that to, say, the 4th power, this is going to be the same thing as raising it to the 2 to the 3 times 4 power, or 2 to the 12th power, which you could also write as raising it to the 4th power, and then the 3rd power. All this is saying is if I raise something to a power, and then raise that whole thing to a power, it's the same thing as multiplying the two exponents."}, {"video_title": "Fractional exponents with numerators other than 1 Algebra I Khan Academy.mp3", "Sentence": "So what I want to do is think about what 64 to the 2 3rd power is. And here I'm going to use a property of exponents that we'll study more later on. But this property of exponents is the idea that, let's say with a simple number, if I raise something to the 3rd power, and then I were to raise that to, say, the 4th power, this is going to be the same thing as raising it to the 2 to the 3 times 4 power, or 2 to the 12th power, which you could also write as raising it to the 4th power, and then the 3rd power. All this is saying is if I raise something to a power, and then raise that whole thing to a power, it's the same thing as multiplying the two exponents. This is the same thing as 2 to the 12th. So we could use that property here to say, well, 2 3rds is the same thing as 1 3rd times 2. So we could go in the other direction."}, {"video_title": "Fractional exponents with numerators other than 1 Algebra I Khan Academy.mp3", "Sentence": "All this is saying is if I raise something to a power, and then raise that whole thing to a power, it's the same thing as multiplying the two exponents. This is the same thing as 2 to the 12th. So we could use that property here to say, well, 2 3rds is the same thing as 1 3rd times 2. So we could go in the other direction. We could say, hey, look, well, this is going to be the same thing as 64 to the 1 3rd power, and then that thing squared. Notice, I'm raising something to a power, and then raising that to a power. If I were to multiply these two things, I would get 64 to the 2 3rds power."}, {"video_title": "Fractional exponents with numerators other than 1 Algebra I Khan Academy.mp3", "Sentence": "So we could go in the other direction. We could say, hey, look, well, this is going to be the same thing as 64 to the 1 3rd power, and then that thing squared. Notice, I'm raising something to a power, and then raising that to a power. If I were to multiply these two things, I would get 64 to the 2 3rds power. Now, why did I do this? Well, we already know what 64 to the 1 3rd power is. We just calculated it."}, {"video_title": "Fractional exponents with numerators other than 1 Algebra I Khan Academy.mp3", "Sentence": "If I were to multiply these two things, I would get 64 to the 2 3rds power. Now, why did I do this? Well, we already know what 64 to the 1 3rd power is. We just calculated it. That's equal to 4. So we could say that this is equal to, and I'll write in that same yellow color, this is equal to 4 squared. This is equal to 4 squared, which is equal to 16."}, {"video_title": "Fractional exponents with numerators other than 1 Algebra I Khan Academy.mp3", "Sentence": "We just calculated it. That's equal to 4. So we could say that this is equal to, and I'll write in that same yellow color, this is equal to 4 squared. This is equal to 4 squared, which is equal to 16. Which is equal to 16. So 64 to 2 3rds power is equal to 16. The way I think of it, let me find the cube root of 64, which is 4, and then let me square it."}, {"video_title": "Fractional exponents with numerators other than 1 Algebra I Khan Academy.mp3", "Sentence": "This is equal to 4 squared, which is equal to 16. Which is equal to 16. So 64 to 2 3rds power is equal to 16. The way I think of it, let me find the cube root of 64, which is 4, and then let me square it. And that is gonna get me to 16. Now I'll give you an even hairier problem, and I encourage you to try this one on your own before I work through it. So we're gonna work with 8 over 27."}, {"video_title": "Fractional exponents with numerators other than 1 Algebra I Khan Academy.mp3", "Sentence": "The way I think of it, let me find the cube root of 64, which is 4, and then let me square it. And that is gonna get me to 16. Now I'll give you an even hairier problem, and I encourage you to try this one on your own before I work through it. So we're gonna work with 8 over 27. And we're gonna raise this thing to the negative, to the negative, and I'll try to color code it, negative 2 over 3 power, to the negative 2 3rds power. I encourage you to pause this and try this on your own. Well, the first thing I do whenever I see a negative exponent is to say, well, how can I get rid of that negative exponent?"}, {"video_title": "Fractional exponents with numerators other than 1 Algebra I Khan Academy.mp3", "Sentence": "So we're gonna work with 8 over 27. And we're gonna raise this thing to the negative, to the negative, and I'll try to color code it, negative 2 over 3 power, to the negative 2 3rds power. I encourage you to pause this and try this on your own. Well, the first thing I do whenever I see a negative exponent is to say, well, how can I get rid of that negative exponent? And I just remind myself, well, the negative exponent really just says take the reciprocal of this to the positive exponent. So this is going to be equal to, the reciprocal of this is 27, I'm using a different color. Let me use that light mauve color."}, {"video_title": "Fractional exponents with numerators other than 1 Algebra I Khan Academy.mp3", "Sentence": "Well, the first thing I do whenever I see a negative exponent is to say, well, how can I get rid of that negative exponent? And I just remind myself, well, the negative exponent really just says take the reciprocal of this to the positive exponent. So this is going to be equal to, the reciprocal of this is 27, I'm using a different color. Let me use that light mauve color. So this is going to be equal to 27 over 8, over 8, I just took the reciprocal of this right over here. It's equal to 27 over 8 to the positive 2 3rds power, to the positive 2 3rds power. So notice, all I did, got rid of the exponent and took the reciprocal of the base right over here."}, {"video_title": "Fractional exponents with numerators other than 1 Algebra I Khan Academy.mp3", "Sentence": "Let me use that light mauve color. So this is going to be equal to 27 over 8, over 8, I just took the reciprocal of this right over here. It's equal to 27 over 8 to the positive 2 3rds power, to the positive 2 3rds power. So notice, all I did, got rid of the exponent and took the reciprocal of the base right over here. 8 over 27 is the base, negative 2 3rds is the exponent. Now, how can we handle this? Well, we've already seen that if I have a numerator to some power over denominator some power, and this is another very powerful exponent property, this is going to be the exact same thing, this is going to be the exact same thing as raising 27 to the 2 3rds power, to the 2 over 3 power, over, over 8 to the 2 3rds power, 8 to the 2 3rds power."}, {"video_title": "Fractional exponents with numerators other than 1 Algebra I Khan Academy.mp3", "Sentence": "So notice, all I did, got rid of the exponent and took the reciprocal of the base right over here. 8 over 27 is the base, negative 2 3rds is the exponent. Now, how can we handle this? Well, we've already seen that if I have a numerator to some power over denominator some power, and this is another very powerful exponent property, this is going to be the exact same thing, this is going to be the exact same thing as raising 27 to the 2 3rds power, to the 2 over 3 power, over, over 8 to the 2 3rds power, 8 to the 2 3rds power. This is another very powerful exponent property. Notice, if I have something divided by something and I'm raising the whole thing to a power, I can essentially raise the numerator of that power over the denominator raised to that power. Now, let's think about what this is."}, {"video_title": "Fractional exponents with numerators other than 1 Algebra I Khan Academy.mp3", "Sentence": "Well, we've already seen that if I have a numerator to some power over denominator some power, and this is another very powerful exponent property, this is going to be the exact same thing, this is going to be the exact same thing as raising 27 to the 2 3rds power, to the 2 over 3 power, over, over 8 to the 2 3rds power, 8 to the 2 3rds power. This is another very powerful exponent property. Notice, if I have something divided by something and I'm raising the whole thing to a power, I can essentially raise the numerator of that power over the denominator raised to that power. Now, let's think about what this is. Well, just like we saw before, this is going to be the same thing, this is going to be the same thing as 27 to the 1 3rd power, 27 to the 1 3rd power, and then that squared, and then that squared, because 1 3rd times 2 is 2 3rds, so I'm gonna raise 27 to the 1 3rd power and then square whatever that is, and that is going to be over, oh, this color coding is making this, just switch a lot of colors, this is going to be over 8 to the 1 3rd power, 8 to the 1 3rd power, to the, and then that's gonna be raised to the 2nd power, same thing we're doing in the denominator, we raise 8 to the 1 3rd and then square that. So what's this going to be? Well, 27 to the 1 3rd power, 27 to the 1 3rd power, is the cube root of 27, it's some number, that number times that same number is gonna be equal to 27."}, {"video_title": "Fractional exponents with numerators other than 1 Algebra I Khan Academy.mp3", "Sentence": "Now, let's think about what this is. Well, just like we saw before, this is going to be the same thing, this is going to be the same thing as 27 to the 1 3rd power, 27 to the 1 3rd power, and then that squared, and then that squared, because 1 3rd times 2 is 2 3rds, so I'm gonna raise 27 to the 1 3rd power and then square whatever that is, and that is going to be over, oh, this color coding is making this, just switch a lot of colors, this is going to be over 8 to the 1 3rd power, 8 to the 1 3rd power, to the, and then that's gonna be raised to the 2nd power, same thing we're doing in the denominator, we raise 8 to the 1 3rd and then square that. So what's this going to be? Well, 27 to the 1 3rd power, 27 to the 1 3rd power, is the cube root of 27, it's some number, that number times that same number is gonna be equal to 27. Well, it might jump out at you already that 3 to the 3rd is equal to 27, or that 27 to the 1 3rd is equal to 3. So the numerator, we're gonna end up with 3 squared, and then in the denominator, we are going to end up with, well, what's 8 to the 1 3rd power? Well, 2 times 2 times 2 is 8, so this is 8 to the 1 3rd is 2, and then we are going to, and then let me do that same orange color, 8 to the 1 3rd is 2, and then we're going to square that."}, {"video_title": "Simplifying quotient of powers (rational exponents) Algebra I High School Math Khan Academy.mp3", "Sentence": "So we have an interesting equation here. Let's see if we can solve for k. And we're going to assume that m is greater than zero. Like always, pause the video, try it out on your own, and then I will do it with you. All right, let's work on this a little bit. So you can imagine that the key to this is to simplify it using our knowledge of exponent properties and there's a couple of ways to think about it. First, we can look at this rational expression here, m to the 7 9ths power divided by m to the 1 3rd power. And the key realization here is that if I have x to the a over x to the b, that this is going to be equal to x to the a minus b power."}, {"video_title": "Simplifying quotient of powers (rational exponents) Algebra I High School Math Khan Academy.mp3", "Sentence": "All right, let's work on this a little bit. So you can imagine that the key to this is to simplify it using our knowledge of exponent properties and there's a couple of ways to think about it. First, we can look at this rational expression here, m to the 7 9ths power divided by m to the 1 3rd power. And the key realization here is that if I have x to the a over x to the b, that this is going to be equal to x to the a minus b power. And it actually comes straight out of the notion that x to the a over x to the b, x to the a over x to the b, is the same thing as x to the a times 1 over x to the b, which is the same thing as x to the a times 1 over x to the b, that's the same thing as x to the negative b, which is going to be the same thing as if I have a base to one exponent times the same base to another exponent, that's the same thing as that base to the sum of the exponents, A plus negative b, which is just going to be a minus b. So we got to the same place, So we can rewrite this as, so we can rewrite this part as being equal to m to the 7 9ths power minus 1 3rd power is equal to, is equal to m to the k over 9. And I think you see where this is going."}, {"video_title": "Simplifying quotient of powers (rational exponents) Algebra I High School Math Khan Academy.mp3", "Sentence": "And the key realization here is that if I have x to the a over x to the b, that this is going to be equal to x to the a minus b power. And it actually comes straight out of the notion that x to the a over x to the b, x to the a over x to the b, is the same thing as x to the a times 1 over x to the b, which is the same thing as x to the a times 1 over x to the b, that's the same thing as x to the negative b, which is going to be the same thing as if I have a base to one exponent times the same base to another exponent, that's the same thing as that base to the sum of the exponents, A plus negative b, which is just going to be a minus b. So we got to the same place, So we can rewrite this as, so we can rewrite this part as being equal to m to the 7 9ths power minus 1 3rd power is equal to, is equal to m to the k over 9. And I think you see where this is going. What is 7 9ths minus 1 3rd? Well, 1 3rd is the same thing we want to have a common denominator. 1 3rd is the same thing as 3 9ths."}, {"video_title": "Simplifying quotient of powers (rational exponents) Algebra I High School Math Khan Academy.mp3", "Sentence": "And I think you see where this is going. What is 7 9ths minus 1 3rd? Well, 1 3rd is the same thing we want to have a common denominator. 1 3rd is the same thing as 3 9ths. So I can rewrite this as 3 9ths. So 7 9ths minus 3 9ths is going to be 4 9ths. So this is the same thing as m to the, m to the 4 9ths power is going to be equal to m to the k 9ths power."}, {"video_title": "Simplifying quotient of powers (rational exponents) Algebra I High School Math Khan Academy.mp3", "Sentence": "1 3rd is the same thing as 3 9ths. So I can rewrite this as 3 9ths. So 7 9ths minus 3 9ths is going to be 4 9ths. So this is the same thing as m to the, m to the 4 9ths power is going to be equal to m to the k 9ths power. So 4 9ths must be the same thing as k 9ths. So we can say 4 9ths is equal to k 9ths. 4 over 9 is equal to k over 9, which tells us that k must be equal to 4."}, {"video_title": "Negative fractional exponent examples Algebra I Khan Academy.mp3", "Sentence": "So we already know that if I were to take nine to the 1 1\u20442 power, this is going to be equal to three. And we know that because three times three is equal to nine. This is equivalent to saying, what is the principal root of nine? Well, that is equal to three. But what would happen if I took nine to the negative 1 1\u20442 power? Now we have a negative fractional exponent. And the key to this is to just not to get too worried or intimidated by this, but just think about it step by step."}, {"video_title": "Negative fractional exponent examples Algebra I Khan Academy.mp3", "Sentence": "Well, that is equal to three. But what would happen if I took nine to the negative 1 1\u20442 power? Now we have a negative fractional exponent. And the key to this is to just not to get too worried or intimidated by this, but just think about it step by step. Just ignore for the second that this is a fraction. And just look at this negative first. Just breathe slowly and realize, OK, I got a negative."}, {"video_title": "Negative fractional exponent examples Algebra I Khan Academy.mp3", "Sentence": "And the key to this is to just not to get too worried or intimidated by this, but just think about it step by step. Just ignore for the second that this is a fraction. And just look at this negative first. Just breathe slowly and realize, OK, I got a negative. Negative exponent. That means that this is just going to be 1 over 9 to the 1 1\u20442. That's what that negative is the cue for."}, {"video_title": "Negative fractional exponent examples Algebra I Khan Academy.mp3", "Sentence": "Just breathe slowly and realize, OK, I got a negative. Negative exponent. That means that this is just going to be 1 over 9 to the 1 1\u20442. That's what that negative is the cue for. This is 1 over 9 to the 1 1\u20442. And we know that 9 to the 1 1\u20442 is equal to 3. So this is just going to be equal to 1."}, {"video_title": "Negative fractional exponent examples Algebra I Khan Academy.mp3", "Sentence": "That's what that negative is the cue for. This is 1 over 9 to the 1 1\u20442. And we know that 9 to the 1 1\u20442 is equal to 3. So this is just going to be equal to 1. This is just going to be equal to 1 3rd. Let's take things a little bit further. What would this evaluate to?"}, {"video_title": "Negative fractional exponent examples Algebra I Khan Academy.mp3", "Sentence": "So this is just going to be equal to 1. This is just going to be equal to 1 3rd. Let's take things a little bit further. What would this evaluate to? And I encourage you to pause the video after trying it. Or pause the video to try it. Negative 27 to the negative 1 3rd power."}, {"video_title": "Negative fractional exponent examples Algebra I Khan Academy.mp3", "Sentence": "What would this evaluate to? And I encourage you to pause the video after trying it. Or pause the video to try it. Negative 27 to the negative 1 3rd power. So I encourage you to pause the video and think about what this would evaluate to. So remember, just take a deep breath. You can always get rid of this negative in the exponent by taking the reciprocal and raising it to the positive."}, {"video_title": "Negative fractional exponent examples Algebra I Khan Academy.mp3", "Sentence": "Negative 27 to the negative 1 3rd power. So I encourage you to pause the video and think about what this would evaluate to. So remember, just take a deep breath. You can always get rid of this negative in the exponent by taking the reciprocal and raising it to the positive. So this is going to be equal to 1 over negative 27 to the positive 1 3rd power. And I know what you're saying. Hey, I still can't breathe easily."}, {"video_title": "Negative fractional exponent examples Algebra I Khan Academy.mp3", "Sentence": "You can always get rid of this negative in the exponent by taking the reciprocal and raising it to the positive. So this is going to be equal to 1 over negative 27 to the positive 1 3rd power. And I know what you're saying. Hey, I still can't breathe easily. I have this negative number to this fractional exponent. But this is just saying, what number, if I were to multiply it three times, so if I have that number, so whatever the number this is, I were to multiply it, if I took three of them and I multiply them together, if I multiplied 1 by that number three times, what number would I have to use here to get negative 27? Well, we already know that 3 to the 3rd, which is equal to 3 times 3 times 3 is equal to positive 27."}, {"video_title": "Negative fractional exponent examples Algebra I Khan Academy.mp3", "Sentence": "Hey, I still can't breathe easily. I have this negative number to this fractional exponent. But this is just saying, what number, if I were to multiply it three times, so if I have that number, so whatever the number this is, I were to multiply it, if I took three of them and I multiply them together, if I multiplied 1 by that number three times, what number would I have to use here to get negative 27? Well, we already know that 3 to the 3rd, which is equal to 3 times 3 times 3 is equal to positive 27. So that's a pretty good clue. What would negative 3 to the 3rd power be? Well, that's negative 3 times negative 3 times negative 3, which is negative 3 times negative 3 is positive 9 times negative 3 is negative 27."}, {"video_title": "Negative fractional exponent examples Algebra I Khan Academy.mp3", "Sentence": "Well, we already know that 3 to the 3rd, which is equal to 3 times 3 times 3 is equal to positive 27. So that's a pretty good clue. What would negative 3 to the 3rd power be? Well, that's negative 3 times negative 3 times negative 3, which is negative 3 times negative 3 is positive 9 times negative 3 is negative 27. So we just found this number, this question mark. Negative 3 times negative 3 times negative 3 is equal to negative 27. So negative 27 to the 1 3rd, this part right over here, is equal to negative 3."}, {"video_title": "Factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "So let's start with a little bit of a warmup. Let's say that we wanted to factor six x squared plus nine x times x squared minus four x plus four. Pause this video and see if you can factor this into the product of even more expressions. All right, now let's do this together. And the way that this might be a little bit different than what you've seen before is this is already partially factored. This polynomial, this higher degree polynomial, is already expressed as the product of two quadratic expressions. But as you might be able to tell, we can factor this further."}, {"video_title": "Factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "All right, now let's do this together. And the way that this might be a little bit different than what you've seen before is this is already partially factored. This polynomial, this higher degree polynomial, is already expressed as the product of two quadratic expressions. But as you might be able to tell, we can factor this further. For example, six x squared plus nine x, both six x squared and nine x are divisible by three x. So let's factor out a three x here. So this is the same thing as three x times."}, {"video_title": "Factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "But as you might be able to tell, we can factor this further. For example, six x squared plus nine x, both six x squared and nine x are divisible by three x. So let's factor out a three x here. So this is the same thing as three x times. Three x times what is six x squared? Well, three times two is six, and x times x is x squared. And then three x times what is nine x?"}, {"video_title": "Factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "So this is the same thing as three x times. Three x times what is six x squared? Well, three times two is six, and x times x is x squared. And then three x times what is nine x? Well, three x times three is nine x. And you can verify that if we were to distribute this three x you would get six x squared plus nine x. And then what about this second expression right over here?"}, {"video_title": "Factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "And then three x times what is nine x? Well, three x times three is nine x. And you can verify that if we were to distribute this three x you would get six x squared plus nine x. And then what about this second expression right over here? Can we factor this? Well, you might recognize this as a perfect square. Some of you might have said, hey, I need to come up with two numbers whose product is four and whose sum is negative four."}, {"video_title": "Factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "And then what about this second expression right over here? Can we factor this? Well, you might recognize this as a perfect square. Some of you might have said, hey, I need to come up with two numbers whose product is four and whose sum is negative four. And you might say, hey, that's negative two and negative two. And so this would be x minus two. We could write x minus two squared or we could write it as x minus two times x minus two."}, {"video_title": "Factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "Some of you might have said, hey, I need to come up with two numbers whose product is four and whose sum is negative four. And you might say, hey, that's negative two and negative two. And so this would be x minus two. We could write x minus two squared or we could write it as x minus two times x minus two. If what I just did is unfamiliar, I encourage you to go back and watch videos on factoring perfect square quadratics and things like that. But there you have it. I think we have factored this as far as we could go."}, {"video_title": "Factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "We could write x minus two squared or we could write it as x minus two times x minus two. If what I just did is unfamiliar, I encourage you to go back and watch videos on factoring perfect square quadratics and things like that. But there you have it. I think we have factored this as far as we could go. So now let's do a slightly trickier higher degree polynomial. So let's say we wanted to factor x to the third minus four x squared plus six x minus 24. And just like always, pause this video and see if you can have a go at it."}, {"video_title": "Factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "I think we have factored this as far as we could go. So now let's do a slightly trickier higher degree polynomial. So let's say we wanted to factor x to the third minus four x squared plus six x minus 24. And just like always, pause this video and see if you can have a go at it. And I'll give you a little bit of a hint. You can factor in this case by grouping. And in some ways, it's a little bit easier than what we've done in the past."}, {"video_title": "Factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "And just like always, pause this video and see if you can have a go at it. And I'll give you a little bit of a hint. You can factor in this case by grouping. And in some ways, it's a little bit easier than what we've done in the past. Historically, when we've learned factoring by grouping, we've looked at a quadratic and then we looked at the middle term, the x term of the quadratic, and we broke it up so that we had four terms. Here, we already have four terms. And see if you can have a go at that."}, {"video_title": "Factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "And in some ways, it's a little bit easier than what we've done in the past. Historically, when we've learned factoring by grouping, we've looked at a quadratic and then we looked at the middle term, the x term of the quadratic, and we broke it up so that we had four terms. Here, we already have four terms. And see if you can have a go at that. All right, now let's do it together. So you can't always factor a third degree polynomial by grouping, but sometimes you can, so it's good to look for it. So when we see it written like this, we say, okay, x to the third minus four x squared, is there a common factor here?"}, {"video_title": "Factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "And see if you can have a go at that. All right, now let's do it together. So you can't always factor a third degree polynomial by grouping, but sometimes you can, so it's good to look for it. So when we see it written like this, we say, okay, x to the third minus four x squared, is there a common factor here? Well, yeah, both x to the third and negative four x squared are divisible by x squared. So what happens if we factor out an x squared? So that's x squared times x minus four."}, {"video_title": "Factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "So when we see it written like this, we say, okay, x to the third minus four x squared, is there a common factor here? Well, yeah, both x to the third and negative four x squared are divisible by x squared. So what happens if we factor out an x squared? So that's x squared times x minus four. And what about these second two terms? Is there a common factor between six x and negative 24? Yeah, they're both divisible by six."}, {"video_title": "Factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "So that's x squared times x minus four. And what about these second two terms? Is there a common factor between six x and negative 24? Yeah, they're both divisible by six. So let's factor out a six here. So plus six times x minus four. And now you are probably seeing the home stretch where you have something times x minus four and then something else times x minus four."}, {"video_title": "Multiplying & dividing powers (integer exponents) Mathematics I High School Math Khan Academy.mp3", "Sentence": "So let's think about what four to the negative three times four to the fifth power is going to be equal to. And I encourage you to pause the video and think about it on your own. Well, there's a couple of ways to do this. One, you say, oh look, I'm multiplying two things that have the same base, so this is going to be that base, four, and then I add the exponents. Four to the negative three plus five power, which is equal to four to the second power. And that's just a straightforward exponent property, but you can also think about why does that actually make sense? Four to the negative three power, that is one over four to the third power."}, {"video_title": "Multiplying & dividing powers (integer exponents) Mathematics I High School Math Khan Academy.mp3", "Sentence": "One, you say, oh look, I'm multiplying two things that have the same base, so this is going to be that base, four, and then I add the exponents. Four to the negative three plus five power, which is equal to four to the second power. And that's just a straightforward exponent property, but you can also think about why does that actually make sense? Four to the negative three power, that is one over four to the third power. Or you could view that as one over four times four times four and then four to the fifth, that's five fours being multiplied together, so it's times four times four times four times four times four. And so notice, when you multiply this out, you're gonna have five fours in the numerator and three fours in the denominator. And so three of these in the denominator are gonna cancel out with three of these in the numerator."}, {"video_title": "Multiplying & dividing powers (integer exponents) Mathematics I High School Math Khan Academy.mp3", "Sentence": "Four to the negative three power, that is one over four to the third power. Or you could view that as one over four times four times four and then four to the fifth, that's five fours being multiplied together, so it's times four times four times four times four times four. And so notice, when you multiply this out, you're gonna have five fours in the numerator and three fours in the denominator. And so three of these in the denominator are gonna cancel out with three of these in the numerator. And so you're gonna be left with five minus three, or negative three plus five fours. So this four times four is the same thing as four squared. Now let's do one with variables."}, {"video_title": "Multiplying & dividing powers (integer exponents) Mathematics I High School Math Khan Academy.mp3", "Sentence": "And so three of these in the denominator are gonna cancel out with three of these in the numerator. And so you're gonna be left with five minus three, or negative three plus five fours. So this four times four is the same thing as four squared. Now let's do one with variables. So let's say that you have a to the negative four power times a to the, let's say a squared. What is that going to be? Well, once again, you have the same base."}, {"video_title": "Multiplying & dividing powers (integer exponents) Mathematics I High School Math Khan Academy.mp3", "Sentence": "Now let's do one with variables. So let's say that you have a to the negative four power times a to the, let's say a squared. What is that going to be? Well, once again, you have the same base. In this case, it's a. And so, and since I'm multiplying them, you can just add the exponents. So it's gonna be a to the negative four plus two power, which is equal to a to the negative two power."}, {"video_title": "Multiplying & dividing powers (integer exponents) Mathematics I High School Math Khan Academy.mp3", "Sentence": "Well, once again, you have the same base. In this case, it's a. And so, and since I'm multiplying them, you can just add the exponents. So it's gonna be a to the negative four plus two power, which is equal to a to the negative two power. And once again, it should make sense. This right over here, that is one over a times a times a times a. And then this is times a times a."}, {"video_title": "Multiplying & dividing powers (integer exponents) Mathematics I High School Math Khan Academy.mp3", "Sentence": "So it's gonna be a to the negative four plus two power, which is equal to a to the negative two power. And once again, it should make sense. This right over here, that is one over a times a times a times a. And then this is times a times a. So that cancels with that, that cancels with that, and you're still left with one over a times a, which is the same thing as a to the negative two power. Now let's do it with some quotients. So what if I were to ask you, what is 12 to the negative seven divided by 12 to the negative five power?"}, {"video_title": "Multiplying & dividing powers (integer exponents) Mathematics I High School Math Khan Academy.mp3", "Sentence": "And then this is times a times a. So that cancels with that, that cancels with that, and you're still left with one over a times a, which is the same thing as a to the negative two power. Now let's do it with some quotients. So what if I were to ask you, what is 12 to the negative seven divided by 12 to the negative five power? Well, when you're dividing, you subtract exponents if you have the same base. So this is going to be equal to 12 to the negative seven minus negative five power. You're subtracting the bottom exponent."}, {"video_title": "Multiplying & dividing powers (integer exponents) Mathematics I High School Math Khan Academy.mp3", "Sentence": "So what if I were to ask you, what is 12 to the negative seven divided by 12 to the negative five power? Well, when you're dividing, you subtract exponents if you have the same base. So this is going to be equal to 12 to the negative seven minus negative five power. You're subtracting the bottom exponent. And so this is going to be equal to 12 to the, well, subtracting a negative is the same thing as adding the positive, a 12 to the negative two power. And once again, we just have to think about why does this actually make sense? Well, you can actually rewrite this."}, {"video_title": "Multiplying & dividing powers (integer exponents) Mathematics I High School Math Khan Academy.mp3", "Sentence": "You're subtracting the bottom exponent. And so this is going to be equal to 12 to the, well, subtracting a negative is the same thing as adding the positive, a 12 to the negative two power. And once again, we just have to think about why does this actually make sense? Well, you can actually rewrite this. 12 to the negative seven divided by 12 to the negative five, that's the same thing as 12 to the negative seven times 12 to the fifth power. If we take the reciprocal of, if we take the reciprocal of this right over here, you would make the exponent positive. And then you get exactly what we were doing in those previous examples with products."}, {"video_title": "Multiplying & dividing powers (integer exponents) Mathematics I High School Math Khan Academy.mp3", "Sentence": "Well, you can actually rewrite this. 12 to the negative seven divided by 12 to the negative five, that's the same thing as 12 to the negative seven times 12 to the fifth power. If we take the reciprocal of, if we take the reciprocal of this right over here, you would make the exponent positive. And then you get exactly what we were doing in those previous examples with products. And so let's just do one more with variables for good measure. Let's say I have x to the negative 20th power divided by x to the fifth power. Well, once again, we have the same base and we're taking a quotient."}, {"video_title": "Multiplying & dividing powers (integer exponents) Mathematics I High School Math Khan Academy.mp3", "Sentence": "And then you get exactly what we were doing in those previous examples with products. And so let's just do one more with variables for good measure. Let's say I have x to the negative 20th power divided by x to the fifth power. Well, once again, we have the same base and we're taking a quotient. So this is going to be x to the negative 20 minus five because we have this one right over here in the denominator. So this is going to be equal to x to the negative 25th power. And once again, you could view our original expression as x to the negative 20th."}, {"video_title": "Using multiple logarithm properties to simplify Logarithms Algebra II Khan Academy.mp3", "Sentence": "We're asked to simplify log base 5 of 25 to the x power over y. So we can use some logarithm properties, and I do agree that this does require some simplification over here, that this, having this right over here inside of the logarithm is not a pleasant thing to look at. So the first thing that we realize, and this is the logarithm, one of our logarithm properties, is logarithm for a given base, so let's say that the base is x, of a over b, that is equal to log base x of a minus log base x of b. And here we have 25 to the x over y, so we can simplify, so let me write this down, I'll do this in blue, log base 5 of 25 to the x over y, using this property means that it's the same thing as log base 5 of 25 to the x power minus log base 5 of y. Now, this looks like we can do a little bit of simplifying, and it seems like the relevant logarithm property here is if I have log base x of a to the b power, that's the same thing as b times log base x of a, that this exponent over here can be moved out front, which is what we did right over there. This part right over here can be rewritten as x times the logarithm base 5 of 25, and then of course we have minus log base 5 of y. And this is useful because log base 5 of 25 is actually fairly easy to think about."}, {"video_title": "Using multiple logarithm properties to simplify Logarithms Algebra II Khan Academy.mp3", "Sentence": "And here we have 25 to the x over y, so we can simplify, so let me write this down, I'll do this in blue, log base 5 of 25 to the x over y, using this property means that it's the same thing as log base 5 of 25 to the x power minus log base 5 of y. Now, this looks like we can do a little bit of simplifying, and it seems like the relevant logarithm property here is if I have log base x of a to the b power, that's the same thing as b times log base x of a, that this exponent over here can be moved out front, which is what we did right over there. This part right over here can be rewritten as x times the logarithm base 5 of 25, and then of course we have minus log base 5 of y. And this is useful because log base 5 of 25 is actually fairly easy to think about. This part right here is asking us, what power do I have to raise 5 to to get to 25? So we have to raise 5 to the second power to get to 25, so this simplifies to 2. So then we are left with, this is equal to 2, and I'll write it in front of the x now, 2 times x minus log base 5 of y."}, {"video_title": "Change of base formula proof Logarithms Algebra II Khan Academy.mp3", "Sentence": "Change of base formula, which tells us that if I want to figure out the logarithm base a of x, that I can figure this out by taking logarithms with a different base. That this would be equal to the logarithm base b, so some other base, base b of x divided by the logarithm base b of a. This is a really useful result if your calculator only has natural logarithm or log base 10. You can now use this to figure out the logarithm using any base. If you want to figure out the log base 2, let me make it clear. If you want to figure out the logarithm base 3 of 25, you can use your calculator either using log base 10 or log base 2. You can say that this is going to be equal to log base 10 of 25, and most calculators have a button for that, divided by log base 10 of 3."}, {"video_title": "Change of base formula proof Logarithms Algebra II Khan Academy.mp3", "Sentence": "You can now use this to figure out the logarithm using any base. If you want to figure out the log base 2, let me make it clear. If you want to figure out the logarithm base 3 of 25, you can use your calculator either using log base 10 or log base 2. You can say that this is going to be equal to log base 10 of 25, and most calculators have a button for that, divided by log base 10 of 3. This is an application of the change of base formula, but let's actually prove it. Let's say that we want to set logarithm base a of x to be equal to some new variable. Let's call that variable equal to y."}, {"video_title": "Change of base formula proof Logarithms Algebra II Khan Academy.mp3", "Sentence": "You can say that this is going to be equal to log base 10 of 25, and most calculators have a button for that, divided by log base 10 of 3. This is an application of the change of base formula, but let's actually prove it. Let's say that we want to set logarithm base a of x to be equal to some new variable. Let's call that variable equal to y. This right over here, we are just setting that equal to y. This is just another way of saying that a to the y power is equal to x. We can rewrite this as a to the y power is equal to x. I'll write the x out here."}, {"video_title": "Change of base formula proof Logarithms Algebra II Khan Academy.mp3", "Sentence": "Let's call that variable equal to y. This right over here, we are just setting that equal to y. This is just another way of saying that a to the y power is equal to x. We can rewrite this as a to the y power is equal to x. I'll write the x out here. These two things are equal. This is just another way of restating what we just wrote up here. Let's introduce the logarithm base b."}, {"video_title": "Change of base formula proof Logarithms Algebra II Khan Academy.mp3", "Sentence": "We can rewrite this as a to the y power is equal to x. I'll write the x out here. These two things are equal. This is just another way of restating what we just wrote up here. Let's introduce the logarithm base b. To introduce it, I'm just going to take log base b of both sides of this equation. Let's take logarithm base b of the left-hand side and logarithm base b of the right-hand side. We know from our logarithm properties that the logarithm of something to a power is the exact same thing as the power times the logarithm of that something."}, {"video_title": "Change of base formula proof Logarithms Algebra II Khan Academy.mp3", "Sentence": "Let's introduce the logarithm base b. To introduce it, I'm just going to take log base b of both sides of this equation. Let's take logarithm base b of the left-hand side and logarithm base b of the right-hand side. We know from our logarithm properties that the logarithm of something to a power is the exact same thing as the power times the logarithm of that something. Logarithm base b of a to the y is the same thing as y times the logarithm of base b of a. This is just a traditional logarithm property. We prove it elsewhere."}, {"video_title": "Change of base formula proof Logarithms Algebra II Khan Academy.mp3", "Sentence": "We know from our logarithm properties that the logarithm of something to a power is the exact same thing as the power times the logarithm of that something. Logarithm base b of a to the y is the same thing as y times the logarithm of base b of a. This is just a traditional logarithm property. We prove it elsewhere. We already know it's going to be equal to the right-hand side. It's going to be equal to log base b of x. Now, let's just solve for y."}, {"video_title": "Change of base formula proof Logarithms Algebra II Khan Academy.mp3", "Sentence": "We prove it elsewhere. We already know it's going to be equal to the right-hand side. It's going to be equal to log base b of x. Now, let's just solve for y. This is exciting because y was this thing right over here. Now, if we solve for y, we're going to be solving for y in terms of logarithm base b. To solve for y, we just have to divide both sides of this equation by log base b of a."}, {"video_title": "Change of base formula proof Logarithms Algebra II Khan Academy.mp3", "Sentence": "Now, let's just solve for y. This is exciting because y was this thing right over here. Now, if we solve for y, we're going to be solving for y in terms of logarithm base b. To solve for y, we just have to divide both sides of this equation by log base b of a. We divide by log base b of a on the left-hand side. We divide by log base b of a on the right-hand side. On the left-hand side, these two characters are going to cancel out."}, {"video_title": "Change of base formula proof Logarithms Algebra II Khan Academy.mp3", "Sentence": "To solve for y, we just have to divide both sides of this equation by log base b of a. We divide by log base b of a on the left-hand side. We divide by log base b of a on the right-hand side. On the left-hand side, these two characters are going to cancel out. We are left with, and we deserve a drum roll now, that y is equal to log base b of x divided by log base b of a. Let me write it. Let's copy and paste this so I don't have to keep switching colors."}, {"video_title": "Change of base formula proof Logarithms Algebra II Khan Academy.mp3", "Sentence": "On the left-hand side, these two characters are going to cancel out. We are left with, and we deserve a drum roll now, that y is equal to log base b of x divided by log base b of a. Let me write it. Let's copy and paste this so I don't have to keep switching colors. Let me paste this. There you have it. You have your change of base formula."}, {"video_title": "Change of base formula proof Logarithms Algebra II Khan Academy.mp3", "Sentence": "Let's copy and paste this so I don't have to keep switching colors. Let me paste this. There you have it. You have your change of base formula. Remember, y is the same thing as this thing right over here. y is log of a. Actually, let me make it clear."}, {"video_title": "Change of base formula proof Logarithms Algebra II Khan Academy.mp3", "Sentence": "You have your change of base formula. Remember, y is the same thing as this thing right over here. y is log of a. Actually, let me make it clear. y, which is equal to log of a, which is equal to log base a of x, so copy and paste, y, which is equal to this thing, which is how we defined it right over here, y is equal to log base a of x, is we've just shown is also equal to this if we write it in terms of base. If we write it in terms of base b. And we have our change of base formula."}, {"video_title": "Compound inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "Negative 16 is less than or equal to 3x plus 5, which is less than or equal to 20. And really, there's two ways to approach it, which are really the same way. And I'll do both of them. And I'll actually do both of them simultaneously. So one is to just solve this compound inequality all at once. And I'll just rewrite it. Negative 16 is less than or equal to 3x plus 5, which is less than or equal to 20."}, {"video_title": "Compound inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "And I'll actually do both of them simultaneously. So one is to just solve this compound inequality all at once. And I'll just rewrite it. Negative 16 is less than or equal to 3x plus 5, which is less than or equal to 20. And the other way is to think of it as two separate inequalities, but both of them need to be true. So you could also view it as negative 16 has to be less than or equal to 3x plus 5, and 3x plus 5 needs to be less than or equal to 20. This statement and this statement are equivalent."}, {"video_title": "Compound inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "Negative 16 is less than or equal to 3x plus 5, which is less than or equal to 20. And the other way is to think of it as two separate inequalities, but both of them need to be true. So you could also view it as negative 16 has to be less than or equal to 3x plus 5, and 3x plus 5 needs to be less than or equal to 20. This statement and this statement are equivalent. This one might seem a little bit more familiar because we can independently solve each of these inequalities and just remember the and. This one might seem a little less traditional, because now we have three sides to the statement. We have three parts of this compound inequality."}, {"video_title": "Compound inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "This statement and this statement are equivalent. This one might seem a little bit more familiar because we can independently solve each of these inequalities and just remember the and. This one might seem a little less traditional, because now we have three sides to the statement. We have three parts of this compound inequality. But what we can see is that we're actually going to solve it the exact same way. In any situation, we really just want to isolate the x on one side of the inequality, or in this case, one part of the compound inequality. Well, the best way to isolate this x right here is to first get rid of this positive 5 that's sitting in the middle."}, {"video_title": "Compound inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "We have three parts of this compound inequality. But what we can see is that we're actually going to solve it the exact same way. In any situation, we really just want to isolate the x on one side of the inequality, or in this case, one part of the compound inequality. Well, the best way to isolate this x right here is to first get rid of this positive 5 that's sitting in the middle. So let's subtract 5 from every part of this compound inequality. So I'm going to subtract 5 there, subtract 5 there, and subtract 5 over there. And so we get negative 16 minus 5 is negative 21, is less than or equal to 3x plus 5 minus 5 is 3x, which is less than or equal to 20 minus 5, which is 15."}, {"video_title": "Compound inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "Well, the best way to isolate this x right here is to first get rid of this positive 5 that's sitting in the middle. So let's subtract 5 from every part of this compound inequality. So I'm going to subtract 5 there, subtract 5 there, and subtract 5 over there. And so we get negative 16 minus 5 is negative 21, is less than or equal to 3x plus 5 minus 5 is 3x, which is less than or equal to 20 minus 5, which is 15. And we could essentially do the same thing here. If we want to isolate the 3x, we can subtract 5 from both sides. We get negative 21 is less than or equal to 3x."}, {"video_title": "Compound inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "And so we get negative 16 minus 5 is negative 21, is less than or equal to 3x plus 5 minus 5 is 3x, which is less than or equal to 20 minus 5, which is 15. And we could essentially do the same thing here. If we want to isolate the 3x, we can subtract 5 from both sides. We get negative 21 is less than or equal to 3x. And we get, subtracting 5 from both sides, and notice, we're just subtracting 5 from every part of this compound inequality. We get 3x is less than or equal to 15. So this statement and this statement, once again, are the exact same thing."}, {"video_title": "Compound inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "We get negative 21 is less than or equal to 3x. And we get, subtracting 5 from both sides, and notice, we're just subtracting 5 from every part of this compound inequality. We get 3x is less than or equal to 15. So this statement and this statement, once again, are the exact same thing. Now going back here, if we want to isolate the x, we can divide by 3. And we have to do it to every part of the inequality. And since 3 is positive, we don't have to change the sign."}, {"video_title": "Compound inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "So this statement and this statement, once again, are the exact same thing. Now going back here, if we want to isolate the x, we can divide by 3. And we have to do it to every part of the inequality. And since 3 is positive, we don't have to change the sign. So let's divide every part of this compound inequality by 3. You divide every part by 3. This is equivalent to dividing every part of each of these inequalities."}, {"video_title": "Compound inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "And since 3 is positive, we don't have to change the sign. So let's divide every part of this compound inequality by 3. You divide every part by 3. This is equivalent to dividing every part of each of these inequalities. Every part of these inequalities by 3. And then we get negative 21 divided by 3 is negative 7, is less than or equal to x, which is less than or equal to 15 divided by 3 is 5. You do it here, you get negative 7 is less than or equal to x."}, {"video_title": "Compound inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "This is equivalent to dividing every part of each of these inequalities. Every part of these inequalities by 3. And then we get negative 21 divided by 3 is negative 7, is less than or equal to x, which is less than or equal to 15 divided by 3 is 5. You do it here, you get negative 7 is less than or equal to x. And x is less than or equal to 15 over 3, which is 5. This statement and this statement are completely equal. And we've solved for x."}, {"video_title": "Compound inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "You do it here, you get negative 7 is less than or equal to x. And x is less than or equal to 15 over 3, which is 5. This statement and this statement are completely equal. And we've solved for x. We've given you the solution set. And if we want to graph it on a number line, it would look like this. It would look like this is 0, this is 5, this is negative 7."}, {"video_title": "Compound inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "And we've solved for x. We've given you the solution set. And if we want to graph it on a number line, it would look like this. It would look like this is 0, this is 5, this is negative 7. Our solution set includes everything between negative 7 and 5, including negative 7 and 5. So we have to fill in the circles on negative 7 and positive 5. And it is everything in between."}, {"video_title": "Compound inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "It would look like this is 0, this is 5, this is negative 7. Our solution set includes everything between negative 7 and 5, including negative 7 and 5. So we have to fill in the circles on negative 7 and positive 5. And it is everything in between. That's our solution set. And so we can verify that these work. You could try out a number that's well inside of our solution set, like 0."}, {"video_title": "Compound inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "And it is everything in between. That's our solution set. And so we can verify that these work. You could try out a number that's well inside of our solution set, like 0. 3 times 0 is 0. So you're just left with 5 is greater than or equal to negative 16, which is true. And 5 is less than or equal to 20."}, {"video_title": "Compound inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "You could try out a number that's well inside of our solution set, like 0. 3 times 0 is 0. So you're just left with 5 is greater than or equal to negative 16, which is true. And 5 is less than or equal to 20. Or negative 16 is less than or equal to 5, which is less than or equal to 20. So that works. And that makes sense."}, {"video_title": "Compound inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "And 5 is less than or equal to 20. Or negative 16 is less than or equal to 5, which is less than or equal to 20. So that works. And that makes sense. You could try 5. If you put 5 here, you get 3 times 5 plus 5. Well, that's just 20."}, {"video_title": "Compound inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "And that makes sense. You could try 5. If you put 5 here, you get 3 times 5 plus 5. Well, that's just 20. Negative 16 is less than or equal to 20, which is less than or equal to 20. That works. Negative 7 should also work."}, {"video_title": "Compound inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "Well, that's just 20. Negative 16 is less than or equal to 20, which is less than or equal to 20. That works. Negative 7 should also work. 3 times negative 7 is negative 21, plus 5 is negative 16. So you get negative 16, which is less than or equal to negative 16, which is less than or equal to 20. And you could try other values."}, {"video_title": "Compound inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "Negative 7 should also work. 3 times negative 7 is negative 21, plus 5 is negative 16. So you get negative 16, which is less than or equal to negative 16, which is less than or equal to 20. And you could try other values. You could go outside of our solution set. Try something like 10. 10 should not work."}, {"video_title": "Compound inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "And you could try other values. You could go outside of our solution set. Try something like 10. 10 should not work. And if you see here, if you put 10 here, you get 3 times 10 plus 5 is 35. Negative 16 is less than or equal to 35, but 35 is not less than or equal to 20. And that's why 10 is not part of our solution set."}, {"video_title": "Using specific values to test for inverses Precalculus Khan Academy.mp3", "Sentence": "So for example, let's say we have f of x is equal to x squared plus three. And let's say that g of x is equal to the square root, the principal root of x minus three. Pause this video and think about whether f and g are inverses of each other. All right, now one approach is to try out some values. So for example, let me make a little table here for f. So this is x, and then this would be f of x. And then let me do the same thing for g. So we have x, and then we have g of x. Now, first, let's try a simple value."}, {"video_title": "Using specific values to test for inverses Precalculus Khan Academy.mp3", "Sentence": "All right, now one approach is to try out some values. So for example, let me make a little table here for f. So this is x, and then this would be f of x. And then let me do the same thing for g. So we have x, and then we have g of x. Now, first, let's try a simple value. If we try out the value one, what is f of one? Well, it's gonna be one squared plus three. That's one plus three, that is four."}, {"video_title": "Using specific values to test for inverses Precalculus Khan Academy.mp3", "Sentence": "Now, first, let's try a simple value. If we try out the value one, what is f of one? Well, it's gonna be one squared plus three. That's one plus three, that is four. So if g is an inverse of f, then if I input four here, I should get one. Now, that wouldn't prove that they're inverses, but if it is an inverse, we should at least be able to get that. So let's see if that's true."}, {"video_title": "Using specific values to test for inverses Precalculus Khan Academy.mp3", "Sentence": "That's one plus three, that is four. So if g is an inverse of f, then if I input four here, I should get one. Now, that wouldn't prove that they're inverses, but if it is an inverse, we should at least be able to get that. So let's see if that's true. So if we take four here, four minus three is one. The principal root of that is one. So that's looking pretty good."}, {"video_title": "Using specific values to test for inverses Precalculus Khan Academy.mp3", "Sentence": "So let's see if that's true. So if we take four here, four minus three is one. The principal root of that is one. So that's looking pretty good. Let's try one more value here. Let's try two. Two squared plus three is seven."}, {"video_title": "Using specific values to test for inverses Precalculus Khan Academy.mp3", "Sentence": "So that's looking pretty good. Let's try one more value here. Let's try two. Two squared plus three is seven. Now let's try out seven here. Seven minus three is four. The principal root of that is two."}, {"video_title": "Using specific values to test for inverses Precalculus Khan Academy.mp3", "Sentence": "Two squared plus three is seven. Now let's try out seven here. Seven minus three is four. The principal root of that is two. So so far, it is looking pretty good. But then what happens if we try a negative value? Pause the video and think about that."}, {"video_title": "Using specific values to test for inverses Precalculus Khan Academy.mp3", "Sentence": "The principal root of that is two. So so far, it is looking pretty good. But then what happens if we try a negative value? Pause the video and think about that. So let's do that. Let me put a, let me put a negative two right over here. Now, if I have negative two squared, that's positive four plus three is seven."}, {"video_title": "Using specific values to test for inverses Precalculus Khan Academy.mp3", "Sentence": "Pause the video and think about that. So let's do that. Let me put a, let me put a negative two right over here. Now, if I have negative two squared, that's positive four plus three is seven. So I have seven here. But we already know that when we input seven into g, we don't get negative two, we get two. In fact, there's no way to get negative two out of this function right over here."}, {"video_title": "Using specific values to test for inverses Precalculus Khan Academy.mp3", "Sentence": "Now, if I have negative two squared, that's positive four plus three is seven. So I have seven here. But we already know that when we input seven into g, we don't get negative two, we get two. In fact, there's no way to get negative two out of this function right over here. So we have just found a case, and frankly, any negative number that you try to use would be a case where you can show that these are not inverses of each other, not inverses. So you actually can use specific points to determine that two functions like this, especially functions that are defined over really an infinite number of values, these are continuous functions, that using specific points, you can show examples where they are not inverses. But you actually can't use specific points to prove that they're inverses because there's an infinite number of values that you could input into these functions, and there's no way that you're going to be able to try out every value."}, {"video_title": "Using specific values to test for inverses Precalculus Khan Academy.mp3", "Sentence": "In fact, there's no way to get negative two out of this function right over here. So we have just found a case, and frankly, any negative number that you try to use would be a case where you can show that these are not inverses of each other, not inverses. So you actually can use specific points to determine that two functions like this, especially functions that are defined over really an infinite number of values, these are continuous functions, that using specific points, you can show examples where they are not inverses. But you actually can't use specific points to prove that they're inverses because there's an infinite number of values that you could input into these functions, and there's no way that you're going to be able to try out every value. For example, if I were to tell you that h of x, really simple functions, h of x is equal to four x, and let's say that j of x is equal to x over four. We know that these are inverses of each other. We'll prove it in other ways in future videos."}, {"video_title": "Equations with rational expressions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So we have a nice little equation here dealing with rational expressions. And I encourage you to pause the video and see if you can figure out what values of x satisfy this equation. All right, let's work through this together. So the first thing I'd like to do is just see if I can simplify this at all, and maybe by finding some common factors between numerators and denominators, or common factors on either side of the equal sign. So let's factor all of these, all of the numerators and denominators, all the ones on the right-hand side are already done. So this thing up here, I could rewrite this as, let's see, what product is 21? What two numbers when I take their product is 21, positive 21, so they're gonna have the same sign."}, {"video_title": "Equations with rational expressions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So the first thing I'd like to do is just see if I can simplify this at all, and maybe by finding some common factors between numerators and denominators, or common factors on either side of the equal sign. So let's factor all of these, all of the numerators and denominators, all the ones on the right-hand side are already done. So this thing up here, I could rewrite this as, let's see, what product is 21? What two numbers when I take their product is 21, positive 21, so they're gonna have the same sign. And when I add them, I get negative 10. Well, negative seven and negative three, so this could be rewritten as x minus seven times x minus three. This over here, both are divisible by three."}, {"video_title": "Equations with rational expressions Mathematics III High School Math Khan Academy.mp3", "Sentence": "What two numbers when I take their product is 21, positive 21, so they're gonna have the same sign. And when I add them, I get negative 10. Well, negative seven and negative three, so this could be rewritten as x minus seven times x minus three. This over here, both are divisible by three. I could rewrite this as three times x minus four. And these are already factored. So the one thing that jumps out at me is I have x minus four in the denominator on the left-hand side and on the right-hand side."}, {"video_title": "Equations with rational expressions Mathematics III High School Math Khan Academy.mp3", "Sentence": "This over here, both are divisible by three. I could rewrite this as three times x minus four. And these are already factored. So the one thing that jumps out at me is I have x minus four in the denominator on the left-hand side and on the right-hand side. And so if I were to multiply both sides by x minus four, so actually let me just, let me formally replace this with that. And up here, it's not so obvious that it's going to be valuable for me to keep this factored form, so I'm just going to keep it in this yellow form, in the expanded out form. So let me just scratch that out for now because once I, well, let me multiply by x minus four."}, {"video_title": "Equations with rational expressions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So the one thing that jumps out at me is I have x minus four in the denominator on the left-hand side and on the right-hand side. And so if I were to multiply both sides by x minus four, so actually let me just, let me formally replace this with that. And up here, it's not so obvious that it's going to be valuable for me to keep this factored form, so I'm just going to keep it in this yellow form, in the expanded out form. So let me just scratch that out for now because once I, well, let me multiply by x minus four. So if we multiply both sides by x minus four, and once again, why am I doing this? So I get rid of the x minus fours in the denominator. X minus four, and then x minus four."}, {"video_title": "Equations with rational expressions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So let me just scratch that out for now because once I, well, let me multiply by x minus four. So if we multiply both sides by x minus four, and once again, why am I doing this? So I get rid of the x minus fours in the denominator. X minus four, and then x minus four. That and that cancels, that and that cancels. And then we're left with, in the numerator, we're left with our x squared minus 10x plus 21, and let's see, divided by three, divided by three is equal to x minus five. Let's see, now what we could do, and actually I could have done it in the last step, is I could multiply both sides by three, multiply both sides, do that in another color just so it sticks out a little bit more."}, {"video_title": "Equations with rational expressions Mathematics III High School Math Khan Academy.mp3", "Sentence": "X minus four, and then x minus four. That and that cancels, that and that cancels. And then we're left with, in the numerator, we're left with our x squared minus 10x plus 21, and let's see, divided by three, divided by three is equal to x minus five. Let's see, now what we could do, and actually I could have done it in the last step, is I could multiply both sides by three, multiply both sides, do that in another color just so it sticks out a little bit more. So I can multiply both sides by three. So multiply both sides by three. On the left-hand side, that and that cancels, and I'll just be left with x squared minus 10x plus 21, and I don't have a denominator anymore."}, {"video_title": "Equations with rational expressions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Let's see, now what we could do, and actually I could have done it in the last step, is I could multiply both sides by three, multiply both sides, do that in another color just so it sticks out a little bit more. So I can multiply both sides by three. So multiply both sides by three. On the left-hand side, that and that cancels, and I'll just be left with x squared minus 10x plus 21, and I don't have a denominator anymore. My denominator is one, so I don't need to write it. It's going to be equal to three times, let's distribute the three. Three times x is three x."}, {"video_title": "Equations with rational expressions Mathematics III High School Math Khan Academy.mp3", "Sentence": "On the left-hand side, that and that cancels, and I'll just be left with x squared minus 10x plus 21, and I don't have a denominator anymore. My denominator is one, so I don't need to write it. It's going to be equal to three times, let's distribute the three. Three times x is three x. Three times negative five is negative 15. And now I can get this in standard quadratic form by getting all of these terms onto the left-hand side. The best way to do that, let's subtract three x from the right, but I can't just do it from the right, otherwise the equality won't hold."}, {"video_title": "Equations with rational expressions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Three times x is three x. Three times negative five is negative 15. And now I can get this in standard quadratic form by getting all of these terms onto the left-hand side. The best way to do that, let's subtract three x from the right, but I can't just do it from the right, otherwise the equality won't hold. I have to do it from both sides if I want the equality to hold. And I want to get rid of this negative 15, so I can add 15 to both sides. So let's do that."}, {"video_title": "Equations with rational expressions Mathematics III High School Math Khan Academy.mp3", "Sentence": "The best way to do that, let's subtract three x from the right, but I can't just do it from the right, otherwise the equality won't hold. I have to do it from both sides if I want the equality to hold. And I want to get rid of this negative 15, so I can add 15 to both sides. So let's do that. And what we are left with, scroll down a little bit so I have a little more space. What we are going to be left with is x squared minus 13x and then plus, what is this? Plus 36, plus 36."}, {"video_title": "Equations with rational expressions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So let's do that. And what we are left with, scroll down a little bit so I have a little more space. What we are going to be left with is x squared minus 13x and then plus, what is this? Plus 36, plus 36. Did I do that right? Yeah, plus 36 is equal to zero. All right, now let's see."}, {"video_title": "Equations with rational expressions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Plus 36, plus 36. Did I do that right? Yeah, plus 36 is equal to zero. All right, now let's see. We have this quadratic in the standard form. How can we solve this? So the first thing, can we factor this?"}, {"video_title": "Equations with rational expressions Mathematics III High School Math Khan Academy.mp3", "Sentence": "All right, now let's see. We have this quadratic in the standard form. How can we solve this? So the first thing, can we factor this? Product of two numbers, 36. If I add them, I get negative 13. They're both going to be negative since they have to have the same sign to get their product to be positive."}, {"video_title": "Equations with rational expressions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So the first thing, can we factor this? Product of two numbers, 36. If I add them, I get negative 13. They're both going to be negative since they have to have the same sign to get their product to be positive. And let's see, nine and four seem to do the trick, or negative nine and negative four. So x minus four times x minus nine is equal to zero. Well, that's going to happen if either x minus four is equal to zero or x minus nine is equal to zero."}, {"video_title": "Equations with rational expressions Mathematics III High School Math Khan Academy.mp3", "Sentence": "They're both going to be negative since they have to have the same sign to get their product to be positive. And let's see, nine and four seem to do the trick, or negative nine and negative four. So x minus four times x minus nine is equal to zero. Well, that's going to happen if either x minus four is equal to zero or x minus nine is equal to zero. Well, add four to both sides of this. This happens when x is equal to four. Add nine to both sides of this."}, {"video_title": "Equations with rational expressions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Well, that's going to happen if either x minus four is equal to zero or x minus nine is equal to zero. Well, add four to both sides of this. This happens when x is equal to four. Add nine to both sides of this. This happens when x is equal to nine. So we could say that the solutions are x equals four or x equals nine. So x is equal to four or x equals nine, but we need to be careful because we have to remember in our original expression, x minus four was a factor of both denominators."}, {"video_title": "Equations with rational expressions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Add nine to both sides of this. This happens when x is equal to nine. So we could say that the solutions are x equals four or x equals nine. So x is equal to four or x equals nine, but we need to be careful because we have to remember in our original expression, x minus four was a factor of both denominators. And so if we actually tried to test x minus four in the original equation, not one of these intermediary steps, in the original equation, I would end up dividing by zero right over here, and actually I would end up dividing by zero right over there as well. So the original equations, if I tried to substitute four, they don't make sense. So this is actually an extraneous solution."}, {"video_title": "Parallel lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "We are asked, which of these lines are parallel? So parallel lines are lines that have the same slope, and they're different lines, so they never, ever intersect. So we need to look for different lines that have the exact same slope. And lucky for us, all of these lines are in y equals mx plus b, or slope intercept form. So you can really just look at these lines and figure out their slopes. The slope for line A, m is equal to 2. We see it right over there."}, {"video_title": "Parallel lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "And lucky for us, all of these lines are in y equals mx plus b, or slope intercept form. So you can really just look at these lines and figure out their slopes. The slope for line A, m is equal to 2. We see it right over there. For line B, our slope is equal to 3. So these two guys are not parallel. And I'll graph it in a second, and you'll see that."}, {"video_title": "Parallel lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "We see it right over there. For line B, our slope is equal to 3. So these two guys are not parallel. And I'll graph it in a second, and you'll see that. And then finally, for line C, I'll do it in purple, the slope is 2. So m is equal to 2. I don't know if that purple is too dark for you."}, {"video_title": "Parallel lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "And I'll graph it in a second, and you'll see that. And then finally, for line C, I'll do it in purple, the slope is 2. So m is equal to 2. I don't know if that purple is too dark for you. So line C and line A have the same slope, but they're different lines. They have different y-intercepts. So they're going to be parallel."}, {"video_title": "Parallel lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "I don't know if that purple is too dark for you. So line C and line A have the same slope, but they're different lines. They have different y-intercepts. So they're going to be parallel. And to see that, let's actually graph all of these characters. So line A, our y-intercept is negative 6, so the point 0, 1, 2, 3, 4, 5, 6. And our slope is 2."}, {"video_title": "Parallel lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "So they're going to be parallel. And to see that, let's actually graph all of these characters. So line A, our y-intercept is negative 6, so the point 0, 1, 2, 3, 4, 5, 6. And our slope is 2. So if we move 1 in the positive x direction, we go up 2 in the positive y direction. 1 in x, up 2 in y. If we go 2 in x, we're going to go up 4 in y."}, {"video_title": "Parallel lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "And our slope is 2. So if we move 1 in the positive x direction, we go up 2 in the positive y direction. 1 in x, up 2 in y. If we go 2 in x, we're going to go up 4 in y. We're going to go up 4 in y. And I could just do up 2, then we're going to go 2, 4. And you're going to see it's all on the same line."}, {"video_title": "Parallel lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "If we go 2 in x, we're going to go up 4 in y. We're going to go up 4 in y. And I could just do up 2, then we're going to go 2, 4. And you're going to see it's all on the same line. So line A is going to look something like, do my best to draw it as straight as possible. Line A, I could do a better version than that. Line A is going to look like, well, that's about just as good as what I just drew."}, {"video_title": "Parallel lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "And you're going to see it's all on the same line. So line A is going to look something like, do my best to draw it as straight as possible. Line A, I could do a better version than that. Line A is going to look like, well, that's about just as good as what I just drew. That is line A. Now let's do line B. Line B, the y-intercept is negative 6."}, {"video_title": "Parallel lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "Line A is going to look like, well, that's about just as good as what I just drew. That is line A. Now let's do line B. Line B, the y-intercept is negative 6. So it has the same y-intercept, but its slope is 3. So if x goes up by 1, y will go up by 3. So x goes up by 1, y goes up by 3."}, {"video_title": "Parallel lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "Line B, the y-intercept is negative 6. So it has the same y-intercept, but its slope is 3. So if x goes up by 1, y will go up by 3. So x goes up by 1, y goes up by 3. If x goes up by 2, y is going to go up by 6. 2, 4, 6. So 2, and then you go 2, 4, 6."}, {"video_title": "Parallel lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "So x goes up by 1, y goes up by 3. If x goes up by 2, y is going to go up by 6. 2, 4, 6. So 2, and then you go 2, 4, 6. So this line is going to look something like this. Trying my best to connect the dots. It has a steeper slope."}, {"video_title": "Parallel lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "So 2, and then you go 2, 4, 6. So this line is going to look something like this. Trying my best to connect the dots. It has a steeper slope. And you see that when x increases, this blue line increases by more in the y direction. So that is line B. And notice they do intersect."}, {"video_title": "Parallel lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "It has a steeper slope. And you see that when x increases, this blue line increases by more in the y direction. So that is line B. And notice they do intersect. There's definitely not two parallel lines. And then finally, let's look at line C. The y-intercept is 5. So 0, 1, 2, 3, 4, 5."}, {"video_title": "Parallel lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "And notice they do intersect. There's definitely not two parallel lines. And then finally, let's look at line C. The y-intercept is 5. So 0, 1, 2, 3, 4, 5. The point 0, 5, it's y-intercept. And its slope is 2. So you increase by 1 in the x direction."}, {"video_title": "Parallel lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "So 0, 1, 2, 3, 4, 5. The point 0, 5, it's y-intercept. And its slope is 2. So you increase by 1 in the x direction. You're going to go up by 2 in the y direction. If you decrease by 1, you're going to go down 2 in the y direction. If you increase by, well, you're going to go to that point."}, {"video_title": "Parallel lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "So you increase by 1 in the x direction. You're going to go up by 2 in the y direction. If you decrease by 1, you're going to go down 2 in the y direction. If you increase by, well, you're going to go to that point. You're going to have a bunch of these points. And then if I were to graph the line, let me do it one more time. If I were to decrease by 2, I'm going to have to go down by 4."}, {"video_title": "Parallel lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "If you increase by, well, you're going to go to that point. You're going to have a bunch of these points. And then if I were to graph the line, let me do it one more time. If I were to decrease by 2, I'm going to have to go down by 4. Negative 4 over negative 2 is still a slope of 2. So 1, 2, 3, 4. And I can do that one more time."}, {"video_title": "Parallel lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "If I were to decrease by 2, I'm going to have to go down by 4. Negative 4 over negative 2 is still a slope of 2. So 1, 2, 3, 4. And I can do that one more time. Get right over there. And then you'll see the line. The line will look like that."}, {"video_title": "Parallel lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "And I can do that one more time. Get right over there. And then you'll see the line. The line will look like that. It will look just like that. And notice that line C and line A never intersect. They have the exact same slope."}, {"video_title": "Parallel lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "The line will look like that. It will look just like that. And notice that line C and line A never intersect. They have the exact same slope. Different y-intercepts, same slope. So they're increasing at the exact same rate. But they're never going to intersect each other."}, {"video_title": "Factoring quadratics as (x+a)(x+b) Mathematics II High School Math Khan Academy.mp3", "Sentence": "And what I'd like to do in this video is I'd like to factor it as the product of two binomials. Or to put it another way, I wanna write it as the product x plus a, that's one binomial, times x plus b, where we need to figure out what a and b are going to be. So I encourage you to pause the video and see if you can figure out what a and b need to be. Can we rewrite this expression as a product of two binomials where we know what a and b are? So let's work through this together now. And I'll highlight a and b in different colors. So I'll put a in yellow and I'll put b in magenta."}, {"video_title": "Factoring quadratics as (x+a)(x+b) Mathematics II High School Math Khan Academy.mp3", "Sentence": "Can we rewrite this expression as a product of two binomials where we know what a and b are? So let's work through this together now. And I'll highlight a and b in different colors. So I'll put a in yellow and I'll put b in magenta. So if you, one way to think about it, so let's just multiply these two binomials using a and b. And we've done this in previous videos. You might wanna review multiplying binomials if any of this looks strange to you."}, {"video_title": "Factoring quadratics as (x+a)(x+b) Mathematics II High School Math Khan Academy.mp3", "Sentence": "So I'll put a in yellow and I'll put b in magenta. So if you, one way to think about it, so let's just multiply these two binomials using a and b. And we've done this in previous videos. You might wanna review multiplying binomials if any of this looks strange to you. But if you were to multiply what we have on the right-hand side out, it would be equal to, you're going to have the x times the x, which is going to be x squared. Then you are going to have the a times the x, which is ax. And then you're gonna have the b times the x, which is bx."}, {"video_title": "Factoring quadratics as (x+a)(x+b) Mathematics II High School Math Khan Academy.mp3", "Sentence": "You might wanna review multiplying binomials if any of this looks strange to you. But if you were to multiply what we have on the right-hand side out, it would be equal to, you're going to have the x times the x, which is going to be x squared. Then you are going to have the a times the x, which is ax. And then you're gonna have the b times the x, which is bx. Actually, I'm not gonna skip any steps here just to see it this time. But this is all review, or it should be review. So then we have, so we did x times x to get x squared."}, {"video_title": "Factoring quadratics as (x+a)(x+b) Mathematics II High School Math Khan Academy.mp3", "Sentence": "And then you're gonna have the b times the x, which is bx. Actually, I'm not gonna skip any steps here just to see it this time. But this is all review, or it should be review. So then we have, so we did x times x to get x squared. Then we have a times x to get ax, to get ax. And then we're gonna have x times b. So we're multiplying each term times every other term."}, {"video_title": "Factoring quadratics as (x+a)(x+b) Mathematics II High School Math Khan Academy.mp3", "Sentence": "So then we have, so we did x times x to get x squared. Then we have a times x to get ax, to get ax. And then we're gonna have x times b. So we're multiplying each term times every other term. So then we have x times b to get bx. So plus bx, bx. And then finally we have plus the a times the b, which is of course going to be ab."}, {"video_title": "Factoring quadratics as (x+a)(x+b) Mathematics II High School Math Khan Academy.mp3", "Sentence": "So we're multiplying each term times every other term. So then we have x times b to get bx. So plus bx, bx. And then finally we have plus the a times the b, which is of course going to be ab. And now we can simplify this. And you might have been able to go straight to this if you are familiar with multiplying binomials. This would be x squared plus, we can add these two coefficients because they're both on the first degree terms."}, {"video_title": "Factoring quadratics as (x+a)(x+b) Mathematics II High School Math Khan Academy.mp3", "Sentence": "And then finally we have plus the a times the b, which is of course going to be ab. And now we can simplify this. And you might have been able to go straight to this if you are familiar with multiplying binomials. This would be x squared plus, we can add these two coefficients because they're both on the first degree terms. They're both multiplied by x. If I have ax's and I add bx's to that, I'm gonna have a plus bx's. So let me write that down."}, {"video_title": "Factoring quadratics as (x+a)(x+b) Mathematics II High School Math Khan Academy.mp3", "Sentence": "This would be x squared plus, we can add these two coefficients because they're both on the first degree terms. They're both multiplied by x. If I have ax's and I add bx's to that, I'm gonna have a plus bx's. So let me write that down. a plus bx's. And then finally I have the plus, let me do that blue color. Finally I have it as plus ab."}, {"video_title": "Factoring quadratics as (x+a)(x+b) Mathematics II High School Math Khan Academy.mp3", "Sentence": "So let me write that down. a plus bx's. And then finally I have the plus, let me do that blue color. Finally I have it as plus ab. Plus ab. And now we can use this, and now we can use this to think about what a and b need to be. If we do a little bit of pattern matching, we see we have an x squared there, we have an x squared there."}, {"video_title": "Factoring quadratics as (x+a)(x+b) Mathematics II High School Math Khan Academy.mp3", "Sentence": "Finally I have it as plus ab. Plus ab. And now we can use this, and now we can use this to think about what a and b need to be. If we do a little bit of pattern matching, we see we have an x squared there, we have an x squared there. We have something, we have something times x. In this case it's a negative three times x. And here we have something times x."}, {"video_title": "Factoring quadratics as (x+a)(x+b) Mathematics II High School Math Khan Academy.mp3", "Sentence": "If we do a little bit of pattern matching, we see we have an x squared there, we have an x squared there. We have something, we have something times x. In this case it's a negative three times x. And here we have something times x. So one way to think about it is that a plus b needs to be equal to negative three. They need to add up to be this coefficient. So let me write that down."}, {"video_title": "Factoring quadratics as (x+a)(x+b) Mathematics II High School Math Khan Academy.mp3", "Sentence": "And here we have something times x. So one way to think about it is that a plus b needs to be equal to negative three. They need to add up to be this coefficient. So let me write that down. So we have a plus b, a plus b, needs to be equal to negative three. Needs to be equal to negative three. And we're not done yet."}, {"video_title": "Factoring quadratics as (x+a)(x+b) Mathematics II High School Math Khan Academy.mp3", "Sentence": "So let me write that down. So we have a plus b, a plus b, needs to be equal to negative three. Needs to be equal to negative three. And we're not done yet. We finally look at this last term, we have a times b. Well a times b needs to be equal to negative 10. So let's write that down."}, {"video_title": "Factoring quadratics as (x+a)(x+b) Mathematics II High School Math Khan Academy.mp3", "Sentence": "And we're not done yet. We finally look at this last term, we have a times b. Well a times b needs to be equal to negative 10. So let's write that down. So we have a times b needs to be equal to, needs to be equal to negative 10. And in general, whenever you're factoring something, a quadratic expression that has a one on the second degree term, so it has a one coefficient on x squared, or you don't even see it, but it's implicitly there, you could write this as one x squared. Way to factor it is say, well can I come up with two numbers that add up to the coefficient on the first degree term?"}, {"video_title": "Factoring quadratics as (x+a)(x+b) Mathematics II High School Math Khan Academy.mp3", "Sentence": "So let's write that down. So we have a times b needs to be equal to, needs to be equal to negative 10. And in general, whenever you're factoring something, a quadratic expression that has a one on the second degree term, so it has a one coefficient on x squared, or you don't even see it, but it's implicitly there, you could write this as one x squared. Way to factor it is say, well can I come up with two numbers that add up to the coefficient on the first degree term? So two numbers that add up to negative three. And if I multiply those same two numbers, I'm going to get negative 10. So two numbers that add up to negative three to add up to the coefficient here."}, {"video_title": "Factoring quadratics as (x+a)(x+b) Mathematics II High School Math Khan Academy.mp3", "Sentence": "Way to factor it is say, well can I come up with two numbers that add up to the coefficient on the first degree term? So two numbers that add up to negative three. And if I multiply those same two numbers, I'm going to get negative 10. So two numbers that add up to negative three to add up to the coefficient here. And now when I multiply it, I get the constant term. I get this right over here. Two numbers when I multiply, I get negative 10."}, {"video_title": "Factoring quadratics as (x+a)(x+b) Mathematics II High School Math Khan Academy.mp3", "Sentence": "So two numbers that add up to negative three to add up to the coefficient here. And now when I multiply it, I get the constant term. I get this right over here. Two numbers when I multiply, I get negative 10. Well what could those numbers be? Well, since when you multiply them, we get a negative number, we know that they're going to have different signs. And so let's see how we could think about it."}, {"video_title": "Factoring quadratics as (x+a)(x+b) Mathematics II High School Math Khan Academy.mp3", "Sentence": "Two numbers when I multiply, I get negative 10. Well what could those numbers be? Well, since when you multiply them, we get a negative number, we know that they're going to have different signs. And so let's see how we could think about it. And since when we add them, we get a negative number, we know that the negative number must be the larger one. So if I were to just factor 10, 10, I could, 10, you could view that as one times 10, one times 10, or two times five. And two and five are interesting because if one of them are negative, their difference is three."}, {"video_title": "Factoring quadratics as (x+a)(x+b) Mathematics II High School Math Khan Academy.mp3", "Sentence": "And so let's see how we could think about it. And since when we add them, we get a negative number, we know that the negative number must be the larger one. So if I were to just factor 10, 10, I could, 10, you could view that as one times 10, one times 10, or two times five. And two and five are interesting because if one of them are negative, their difference is three. So if one is negative, so let's see, if we're talking about negative 10, negative 10, you could say negative two times five. And when you multiply them, you do get negative 10. But if you add these two, you're going to get positive three."}, {"video_title": "Factoring quadratics as (x+a)(x+b) Mathematics II High School Math Khan Academy.mp3", "Sentence": "And two and five are interesting because if one of them are negative, their difference is three. So if one is negative, so let's see, if we're talking about negative 10, negative 10, you could say negative two times five. And when you multiply them, you do get negative 10. But if you add these two, you're going to get positive three. But what if you went positive two times negative five? Now this is interesting because still when you multiply them you get negative 10. And when you add them, two plus negative five is going to be negative three."}, {"video_title": "Factoring quadratics as (x+a)(x+b) Mathematics II High School Math Khan Academy.mp3", "Sentence": "But if you add these two, you're going to get positive three. But what if you went positive two times negative five? Now this is interesting because still when you multiply them you get negative 10. And when you add them, two plus negative five is going to be negative three. So we have just figured out, we have just figured out our two numbers. We could say that A, and we could say that A is two, or we could say that B is two, but I'll just say that A is two. So A is equal to two, and B is equal to negative five."}, {"video_title": "Factoring quadratics as (x+a)(x+b) Mathematics II High School Math Khan Academy.mp3", "Sentence": "And when you add them, two plus negative five is going to be negative three. So we have just figured out, we have just figured out our two numbers. We could say that A, and we could say that A is two, or we could say that B is two, but I'll just say that A is two. So A is equal to two, and B is equal to negative five. B is equal to negative five. And so our original expression, we can rewrite as, so we can rewrite x squared minus three x minus 10. We can say that that is going to be equal to x plus two, x plus two times, times x, instead of saying plus negative five, which we could say, we could just say, see let me write that down."}, {"video_title": "Factoring quadratics as (x+a)(x+b) Mathematics II High School Math Khan Academy.mp3", "Sentence": "So A is equal to two, and B is equal to negative five. B is equal to negative five. And so our original expression, we can rewrite as, so we can rewrite x squared minus three x minus 10. We can say that that is going to be equal to x plus two, x plus two times, times x, instead of saying plus negative five, which we could say, we could just say, see let me write that down. I could write just plus negative five right over there because that's our B. I could just write x minus, x minus five, and we're done. We've just factored it as a product of two binomials. Now, I did it fairly involved, mainly so you see where all this came from, but in the future, whenever you see a quadratic expression, and you have a one coefficient on the second degree term right over here, you could say, all right, well I need to figure out two numbers that add up to the coefficient on the first degree term, on the x term, and those same two numbers, when I multiply them, need to be equal to this constant term, need to be equal to negative 10."}, {"video_title": "Factoring quadratics as (x+a)(x+b) Mathematics II High School Math Khan Academy.mp3", "Sentence": "We can say that that is going to be equal to x plus two, x plus two times, times x, instead of saying plus negative five, which we could say, we could just say, see let me write that down. I could write just plus negative five right over there because that's our B. I could just write x minus, x minus five, and we're done. We've just factored it as a product of two binomials. Now, I did it fairly involved, mainly so you see where all this came from, but in the future, whenever you see a quadratic expression, and you have a one coefficient on the second degree term right over here, you could say, all right, well I need to figure out two numbers that add up to the coefficient on the first degree term, on the x term, and those same two numbers, when I multiply them, need to be equal to this constant term, need to be equal to negative 10. You say, okay, well let's see. They're gonna be different signs because when I multiply them, I get a negative number. The negative one's gonna be the larger one since when I add them, I got a negative number."}, {"video_title": "Factoring quadratics as (x+a)(x+b) Mathematics II High School Math Khan Academy.mp3", "Sentence": "Now, I did it fairly involved, mainly so you see where all this came from, but in the future, whenever you see a quadratic expression, and you have a one coefficient on the second degree term right over here, you could say, all right, well I need to figure out two numbers that add up to the coefficient on the first degree term, on the x term, and those same two numbers, when I multiply them, need to be equal to this constant term, need to be equal to negative 10. You say, okay, well let's see. They're gonna be different signs because when I multiply them, I get a negative number. The negative one's gonna be the larger one since when I add them, I got a negative number. So let's see. Five and two seem interesting. Well, negative five and positive two."}, {"video_title": "Solve by completing the square Non-integer solutions Algebra I Khan Academy.mp3", "Sentence": "Let's say we're told that zero is equal to x squared plus six x plus three. What is an x or what are x's that would satisfy this equation? Pause this video and try to figure it out. All right, now let's work through it together. So the first thing that I would try to do is see if I could factor this right-hand expression. I have some expression, it's equal to zero, so if I could factor it, that might help solve. So let's see, can I think of two numbers that when I add them, I get six, and when I take their product, I get positive three?"}, {"video_title": "Solve by completing the square Non-integer solutions Algebra I Khan Academy.mp3", "Sentence": "All right, now let's work through it together. So the first thing that I would try to do is see if I could factor this right-hand expression. I have some expression, it's equal to zero, so if I could factor it, that might help solve. So let's see, can I think of two numbers that when I add them, I get six, and when I take their product, I get positive three? Well, if I'm thinking just in terms of integers, three is a prime number, it only has two factors, one and three, and let's see, one plus three is not equal to six, so it doesn't look like factoring is going to help me much. So the next thing I'll turn to is completing the square. In fact, completing the square, if there are x values that would satisfy this equation, completing the square will help us solve it."}, {"video_title": "Solve by completing the square Non-integer solutions Algebra I Khan Academy.mp3", "Sentence": "So let's see, can I think of two numbers that when I add them, I get six, and when I take their product, I get positive three? Well, if I'm thinking just in terms of integers, three is a prime number, it only has two factors, one and three, and let's see, one plus three is not equal to six, so it doesn't look like factoring is going to help me much. So the next thing I'll turn to is completing the square. In fact, completing the square, if there are x values that would satisfy this equation, completing the square will help us solve it. And the way I do it, I'll say zero is equal to, let me rewrite the first part, x squared plus six x, and then I'm gonna write the plus three out here, and my goal is to add something to this equation or to the right-hand expression right over here, and then I'm gonna subtract that same thing, so I'm not really changing the value of the right-hand side, and I wanna add something here that I'm later going to subtract so that what I have in parentheses is a perfect square. Well, the way to make it a perfect square, and we've talked about this in other videos when we introduced ourselves to completing the square, is we'll look at this first degree coefficient right over here, this positive six, and say, okay, half of that is positive three, and if we were to square that, we would get nine. So let's add a nine there, and then we could also subtract a nine."}, {"video_title": "Solve by completing the square Non-integer solutions Algebra I Khan Academy.mp3", "Sentence": "In fact, completing the square, if there are x values that would satisfy this equation, completing the square will help us solve it. And the way I do it, I'll say zero is equal to, let me rewrite the first part, x squared plus six x, and then I'm gonna write the plus three out here, and my goal is to add something to this equation or to the right-hand expression right over here, and then I'm gonna subtract that same thing, so I'm not really changing the value of the right-hand side, and I wanna add something here that I'm later going to subtract so that what I have in parentheses is a perfect square. Well, the way to make it a perfect square, and we've talked about this in other videos when we introduced ourselves to completing the square, is we'll look at this first degree coefficient right over here, this positive six, and say, okay, half of that is positive three, and if we were to square that, we would get nine. So let's add a nine there, and then we could also subtract a nine. Notice, we haven't changed the value of the right-hand side expression. We're adding nine and we're subtracting nine, and actually the parentheses are just there to help it make a little bit more visually clear to us, but you don't even need the parentheses. You would essentially get the same result."}, {"video_title": "Solve by completing the square Non-integer solutions Algebra I Khan Academy.mp3", "Sentence": "So let's add a nine there, and then we could also subtract a nine. Notice, we haven't changed the value of the right-hand side expression. We're adding nine and we're subtracting nine, and actually the parentheses are just there to help it make a little bit more visually clear to us, but you don't even need the parentheses. You would essentially get the same result. But then what happens if we simplify this a little bit? Well, what I just showed you, and let me do it in this green-blue color, this thing can be rewritten as x plus three squared. That's why we added nine there."}, {"video_title": "Solve by completing the square Non-integer solutions Algebra I Khan Academy.mp3", "Sentence": "You would essentially get the same result. But then what happens if we simplify this a little bit? Well, what I just showed you, and let me do it in this green-blue color, this thing can be rewritten as x plus three squared. That's why we added nine there. We said, all right, we're gonna be dealing with a three because three is half of six, and if we squared three, we get a nine there, and then this second part right over here, three minus nine, that's equal to negative six. So we could write it like this. Zero is equal to x plus three squared minus six."}, {"video_title": "Solve by completing the square Non-integer solutions Algebra I Khan Academy.mp3", "Sentence": "That's why we added nine there. We said, all right, we're gonna be dealing with a three because three is half of six, and if we squared three, we get a nine there, and then this second part right over here, three minus nine, that's equal to negative six. So we could write it like this. Zero is equal to x plus three squared minus six. And now what we can do is isolate this x plus three squared by adding six to both sides. So let's do that. Let's add six there."}, {"video_title": "Solve by completing the square Non-integer solutions Algebra I Khan Academy.mp3", "Sentence": "Zero is equal to x plus three squared minus six. And now what we can do is isolate this x plus three squared by adding six to both sides. So let's do that. Let's add six there. Let's add six there. And what we get on the left-hand side, we get six is equal to, on the right-hand side, we just get x plus three squared. And now we can take the square root of both sides, and we could say that the plus or minus square root of six is equal to x plus three."}, {"video_title": "Solve by completing the square Non-integer solutions Algebra I Khan Academy.mp3", "Sentence": "Let's add six there. Let's add six there. And what we get on the left-hand side, we get six is equal to, on the right-hand side, we just get x plus three squared. And now we can take the square root of both sides, and we could say that the plus or minus square root of six is equal to x plus three. And if this doesn't make full sense, just pause the video a little bit and think about it. If I'm saying that something squared is equal to six, that means that the something is either going to be the positive square root of six or the negative square root of six. And so now we can, if we wanna solve for x, we can just subtract three from both sides."}, {"video_title": "Solve by completing the square Non-integer solutions Algebra I Khan Academy.mp3", "Sentence": "And now we can take the square root of both sides, and we could say that the plus or minus square root of six is equal to x plus three. And if this doesn't make full sense, just pause the video a little bit and think about it. If I'm saying that something squared is equal to six, that means that the something is either going to be the positive square root of six or the negative square root of six. And so now we can, if we wanna solve for x, we can just subtract three from both sides. So let's subtract three from both sides. And what do we get? We get, on the right-hand side, we just are left with an x, and that's going to be equal to negative three plus or minus the square root of six."}, {"video_title": "Example 4 Factoring quadratics as a perfect square of a difference (a-b)^2 Khan Academy.mp3", "Sentence": "Factor 25x squared minus 30x plus 9. And so we have a leading coefficient that's not a 1, and it doesn't look like there are any common factors. Both 25 and 30 are divisible by 5, but 9 isn't divisible by 5. So we could factor this by grouping, but if we look a little bit more carefully here, see something interesting. 25 is a perfect square, and so 25x squared is a perfect square. It's the square of 5x. And then 9 is also a perfect square."}, {"video_title": "Example 4 Factoring quadratics as a perfect square of a difference (a-b)^2 Khan Academy.mp3", "Sentence": "So we could factor this by grouping, but if we look a little bit more carefully here, see something interesting. 25 is a perfect square, and so 25x squared is a perfect square. It's the square of 5x. And then 9 is also a perfect square. It's the square of 3, or actually it could be the square of negative 3. This could also be the square of negative 5x. So maybe, just maybe, this could be a perfect square."}, {"video_title": "Example 4 Factoring quadratics as a perfect square of a difference (a-b)^2 Khan Academy.mp3", "Sentence": "And then 9 is also a perfect square. It's the square of 3, or actually it could be the square of negative 3. This could also be the square of negative 5x. So maybe, just maybe, this could be a perfect square. So let's just think about what happens when we take the perfect square of a binomial, especially when the coefficient on the x term is not a 1. So if we have ax plus b squared, what will this look like when we expand this into a trinomial? Well, this is the same thing as ax plus b times ax plus b, which is the same thing as ax times ax."}, {"video_title": "Example 4 Factoring quadratics as a perfect square of a difference (a-b)^2 Khan Academy.mp3", "Sentence": "So maybe, just maybe, this could be a perfect square. So let's just think about what happens when we take the perfect square of a binomial, especially when the coefficient on the x term is not a 1. So if we have ax plus b squared, what will this look like when we expand this into a trinomial? Well, this is the same thing as ax plus b times ax plus b, which is the same thing as ax times ax. ax times ax is a squared x squared plus ax times b, which is abx, plus b times ax, which is another. You could call it bax or abx, plus b times b, so plus b squared. So this is equal to a squared x squared plus, these two are the same term, plus 2abx plus b squared."}, {"video_title": "Example 4 Factoring quadratics as a perfect square of a difference (a-b)^2 Khan Academy.mp3", "Sentence": "Well, this is the same thing as ax plus b times ax plus b, which is the same thing as ax times ax. ax times ax is a squared x squared plus ax times b, which is abx, plus b times ax, which is another. You could call it bax or abx, plus b times b, so plus b squared. So this is equal to a squared x squared plus, these two are the same term, plus 2abx plus b squared. So this is what happens when you square a binomial. Now, this pattern seems to work out pretty good. Let me rewrite our problem right below it."}, {"video_title": "Example 4 Factoring quadratics as a perfect square of a difference (a-b)^2 Khan Academy.mp3", "Sentence": "So this is equal to a squared x squared plus, these two are the same term, plus 2abx plus b squared. So this is what happens when you square a binomial. Now, this pattern seems to work out pretty good. Let me rewrite our problem right below it. We have 25x squared minus 30x plus 9. So if this is a perfect square, then that means that the a squared part right over here is 25, and then that means that the b squared part, and that means that the, let me do this in a different color, the b squared part is 9. So that tells us that a could be plus or minus 5, and that b could be plus or minus 3."}, {"video_title": "Example 4 Factoring quadratics as a perfect square of a difference (a-b)^2 Khan Academy.mp3", "Sentence": "Let me rewrite our problem right below it. We have 25x squared minus 30x plus 9. So if this is a perfect square, then that means that the a squared part right over here is 25, and then that means that the b squared part, and that means that the, let me do this in a different color, the b squared part is 9. So that tells us that a could be plus or minus 5, and that b could be plus or minus 3. Now let's see if this gels with this middle term. If these, for this middle term to work out, for this middle term to work out, I'm trying to look for good colors, 2ab, this part right over here, 2ab needs to be equal to negative 30. Or another way, let me write it over here, 2ab needs to be equal to negative 30, or if we divide both sides by 2, ab needs to be equal to negative 15."}, {"video_title": "Example 4 Factoring quadratics as a perfect square of a difference (a-b)^2 Khan Academy.mp3", "Sentence": "So that tells us that a could be plus or minus 5, and that b could be plus or minus 3. Now let's see if this gels with this middle term. If these, for this middle term to work out, for this middle term to work out, I'm trying to look for good colors, 2ab, this part right over here, 2ab needs to be equal to negative 30. Or another way, let me write it over here, 2ab needs to be equal to negative 30, or if we divide both sides by 2, ab needs to be equal to negative 15. So that tells us, since the product is negative, one has to be positive and one has to be negative. Now, lucky for us, the product of 5 and 3 is 15. So if we make one of them positive and one of them negative, we'll get up to negative 15."}, {"video_title": "Example 4 Factoring quadratics as a perfect square of a difference (a-b)^2 Khan Academy.mp3", "Sentence": "Or another way, let me write it over here, 2ab needs to be equal to negative 30, or if we divide both sides by 2, ab needs to be equal to negative 15. So that tells us, since the product is negative, one has to be positive and one has to be negative. Now, lucky for us, the product of 5 and 3 is 15. So if we make one of them positive and one of them negative, we'll get up to negative 15. So it looks like things are going to work out. So we could select a is equal to positive 5, and b is equal to negative 3. Those would work out to ab being equal to negative 15."}, {"video_title": "Example 4 Factoring quadratics as a perfect square of a difference (a-b)^2 Khan Academy.mp3", "Sentence": "So if we make one of them positive and one of them negative, we'll get up to negative 15. So it looks like things are going to work out. So we could select a is equal to positive 5, and b is equal to negative 3. Those would work out to ab being equal to negative 15. Or we could make a is equal to negative 5, and b is equal to positive 3. So either of these will work. So if we factor this out, this could be either, a is, let's do this first one, it could either be a is 5, b is negative 3."}, {"video_title": "Example 4 Factoring quadratics as a perfect square of a difference (a-b)^2 Khan Academy.mp3", "Sentence": "Those would work out to ab being equal to negative 15. Or we could make a is equal to negative 5, and b is equal to positive 3. So either of these will work. So if we factor this out, this could be either, a is, let's do this first one, it could either be a is 5, b is negative 3. So this could either be 5x minus 3 squared. a is 5, b is negative 3. It could be that."}, {"video_title": "Example 4 Factoring quadratics as a perfect square of a difference (a-b)^2 Khan Academy.mp3", "Sentence": "So if we factor this out, this could be either, a is, let's do this first one, it could either be a is 5, b is negative 3. So this could either be 5x minus 3 squared. a is 5, b is negative 3. It could be that. Or we could switch the signs on the two terms. Or a could be negative 5, and b could be positive 3. Or it could be negative 5x plus 3 squared."}, {"video_title": "Example 4 Factoring quadratics as a perfect square of a difference (a-b)^2 Khan Academy.mp3", "Sentence": "It could be that. Or we could switch the signs on the two terms. Or a could be negative 5, and b could be positive 3. Or it could be negative 5x plus 3 squared. So either of these are possible ways to factor this term out here. And you're going to say, wait, how does this work out? How can both of these multiply to the same thing?"}, {"video_title": "Example 4 Factoring quadratics as a perfect square of a difference (a-b)^2 Khan Academy.mp3", "Sentence": "Or it could be negative 5x plus 3 squared. So either of these are possible ways to factor this term out here. And you're going to say, wait, how does this work out? How can both of these multiply to the same thing? Well, this term, remember, this negative 5x plus 3, we could factor out a negative 1. So this right here is the same thing as negative 1 times 5x minus 3, the whole thing squared. And that's the same thing as negative 1 squared times 5x minus 3 squared."}, {"video_title": "Example 4 Factoring quadratics as a perfect square of a difference (a-b)^2 Khan Academy.mp3", "Sentence": "How can both of these multiply to the same thing? Well, this term, remember, this negative 5x plus 3, we could factor out a negative 1. So this right here is the same thing as negative 1 times 5x minus 3, the whole thing squared. And that's the same thing as negative 1 squared times 5x minus 3 squared. And negative 1 squared is clearly equal to 1. So that's why this and this are the same thing. This comes out to the same thing as 5x minus 3 squared, which is the same thing as that over there."}, {"video_title": "Examples finding the domain of functions.mp3", "Sentence": "So let's say that we have the function f of x is equal to x plus five over x minus two. What is going to be the domain of this function? Pause this video and try to figure that out. All right, now let's do it together. Now, the domain is the set of all x values that if we input it into this function, we're going to get a legitimate output. We're going to get a legitimate f of x. And so what's a situation where we would not get a legitimate f of x?"}, {"video_title": "Examples finding the domain of functions.mp3", "Sentence": "All right, now let's do it together. Now, the domain is the set of all x values that if we input it into this function, we're going to get a legitimate output. We're going to get a legitimate f of x. And so what's a situation where we would not get a legitimate f of x? Well, if we input an x value that makes this denominator equal to zero, then we're going to divide by zero, and that's going to be undefined. And so we could say the domain, the domain here is all real values of x such that x minus two does not equal zero. Now, typically, people would not wanna just see that such that x minus two does not equal zero."}, {"video_title": "Examples finding the domain of functions.mp3", "Sentence": "And so what's a situation where we would not get a legitimate f of x? Well, if we input an x value that makes this denominator equal to zero, then we're going to divide by zero, and that's going to be undefined. And so we could say the domain, the domain here is all real values of x such that x minus two does not equal zero. Now, typically, people would not wanna just see that such that x minus two does not equal zero. And so we can simplify this a little bit so that we just have an x on the left-hand side. So if we add two to both sides of this, we would get, actually, let me just do that. Let me add two to both sides."}, {"video_title": "Examples finding the domain of functions.mp3", "Sentence": "Now, typically, people would not wanna just see that such that x minus two does not equal zero. And so we can simplify this a little bit so that we just have an x on the left-hand side. So if we add two to both sides of this, we would get, actually, let me just do that. Let me add two to both sides. So x minus two not equaling zero is the same thing as x not equaling two. And you could have done that in your head as well. If you wanted to keep x minus two from being zero, x just can't be equal to two."}, {"video_title": "Examples finding the domain of functions.mp3", "Sentence": "Let me add two to both sides. So x minus two not equaling zero is the same thing as x not equaling two. And you could have done that in your head as well. If you wanted to keep x minus two from being zero, x just can't be equal to two. And so typically, people would say that the domain here is all real values of x such that x does not equal two. Let's do another example. Let's say that we're told that g of x is equal to the principal root of x minus seven."}, {"video_title": "Examples finding the domain of functions.mp3", "Sentence": "If you wanted to keep x minus two from being zero, x just can't be equal to two. And so typically, people would say that the domain here is all real values of x such that x does not equal two. Let's do another example. Let's say that we're told that g of x is equal to the principal root of x minus seven. What's the domain in this situation? What's the domain of g of x? Pause the video and try to figure that out."}, {"video_title": "Examples finding the domain of functions.mp3", "Sentence": "Let's say that we're told that g of x is equal to the principal root of x minus seven. What's the domain in this situation? What's the domain of g of x? Pause the video and try to figure that out. Well, we could say the domain, the domain is going to be all real values of x such that, are we going to have to put any constraints on this? Well, when does a principal root function break down? Well, if we tried to find the principal root, the square root of a negative number, well, that would then break down."}, {"video_title": "Examples finding the domain of functions.mp3", "Sentence": "Pause the video and try to figure that out. Well, we could say the domain, the domain is going to be all real values of x such that, are we going to have to put any constraints on this? Well, when does a principal root function break down? Well, if we tried to find the principal root, the square root of a negative number, well, that would then break down. And so x minus seven, whatever we have under the radical here, needs to be greater than or equal to zero, so such that x minus seven needs to be greater than or equal to zero. Now, another way to say that is if we add seven to both sides of that, that would be saying that x needs to be greater than or equal to seven. So let me just write it that way."}, {"video_title": "Examples finding the domain of functions.mp3", "Sentence": "Well, if we tried to find the principal root, the square root of a negative number, well, that would then break down. And so x minus seven, whatever we have under the radical here, needs to be greater than or equal to zero, so such that x minus seven needs to be greater than or equal to zero. Now, another way to say that is if we add seven to both sides of that, that would be saying that x needs to be greater than or equal to seven. So let me just write it that way. So such that x is greater than or equal to seven. So all I did is I said, all right, where could this thing break down? Well, if I get x values where this thing is negative, we're in trouble."}, {"video_title": "Examples finding the domain of functions.mp3", "Sentence": "So let me just write it that way. So such that x is greater than or equal to seven. So all I did is I said, all right, where could this thing break down? Well, if I get x values where this thing is negative, we're in trouble. So x needs to be greater, x minus seven, whatever we have in this under the radical, needs to be greater than or equal to zero. And so if you say that x minus seven needs to be greater than or equal to zero, you add seven to both sides, you get x needs to be greater than or equal to positive seven. Let's do one last example."}, {"video_title": "Examples finding the domain of functions.mp3", "Sentence": "Well, if I get x values where this thing is negative, we're in trouble. So x needs to be greater, x minus seven, whatever we have in this under the radical, needs to be greater than or equal to zero. And so if you say that x minus seven needs to be greater than or equal to zero, you add seven to both sides, you get x needs to be greater than or equal to positive seven. Let's do one last example. Let's say we're told that h of x is equal to x minus five squared. What's the domain here? So let me write this down."}, {"video_title": "Examples finding the domain of functions.mp3", "Sentence": "Let's do one last example. Let's say we're told that h of x is equal to x minus five squared. What's the domain here? So let me write this down. The domain is all real values of x. All real values of x. Now are we going to have to constrain this a little bit?"}, {"video_title": "Examples finding the domain of functions.mp3", "Sentence": "So let me write this down. The domain is all real values of x. All real values of x. Now are we going to have to constrain this a little bit? Well, is there anything that would cause this to not evaluate to a defined value? Well, we can square any value. You have to give me any real number, and if I square it, I'm just gonna get another real number."}, {"video_title": "Examples finding the domain of functions.mp3", "Sentence": "Now are we going to have to constrain this a little bit? Well, is there anything that would cause this to not evaluate to a defined value? Well, we can square any value. You have to give me any real number, and if I square it, I'm just gonna get another real number. And so x minus five can be equal to anything. And so x can be equal to anything. So here, the domain is all real values of x."}, {"video_title": "Examples finding the domain of functions.mp3", "Sentence": "You have to give me any real number, and if I square it, I'm just gonna get another real number. And so x minus five can be equal to anything. And so x can be equal to anything. So here, the domain is all real values of x. We didn't have to constrain it in any way like we did the other two. The other two, when you deal with something in a denominator that could be equal to zero, then you gotta make sure that that doesn't happen because that would get you an undefined value. And similarly, for a radical, you can't take the square root of a negative."}, {"video_title": "Understanding inverse functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So for example, if this table is our function definition, if someone were to say, well what is f of negative nine? You'd say, okay, if we input negative nine into our function, if x is negative nine, this table tells us that f of x is going to be equal to five. And you might already have experience with doing composite functions, where you say f of, let's say actually f of, f of negative nine plus one. So this is interesting. It seems very daunting, but you say, oh well, we know what f of negative nine is. This is going to be five. So it's gonna be f of five plus one."}, {"video_title": "Understanding inverse functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So this is interesting. It seems very daunting, but you say, oh well, we know what f of negative nine is. This is going to be five. So it's gonna be f of five plus one. So this is going to be equal to f of six. And if we look at our table, f of six is equal to negative seven. So all of that is a review so far."}, {"video_title": "Understanding inverse functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So it's gonna be f of five plus one. So this is going to be equal to f of six. And if we look at our table, f of six is equal to negative seven. So all of that is a review so far. But what I wanna now do is start evaluating the inverse of functions. And this function f is invertible, because it's a one-to-one mapping between the x's and the f of x's. No two x's map to the same f of x."}, {"video_title": "Understanding inverse functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So all of that is a review so far. But what I wanna now do is start evaluating the inverse of functions. And this function f is invertible, because it's a one-to-one mapping between the x's and the f of x's. No two x's map to the same f of x. And so this is an invertible function. So with that in mind, let's see if we can evaluate something like f inverse, f inverse of eight. What is that going to be?"}, {"video_title": "Understanding inverse functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "No two x's map to the same f of x. And so this is an invertible function. So with that in mind, let's see if we can evaluate something like f inverse, f inverse of eight. What is that going to be? I encourage you to pause the video and try to think about it. All right, so f of x, just as a reminder of what functions do, f of x is going to map from this domain, from a value in its domain, to a corresponding value in the range. In the range."}, {"video_title": "Understanding inverse functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "What is that going to be? I encourage you to pause the video and try to think about it. All right, so f of x, just as a reminder of what functions do, f of x is going to map from this domain, from a value in its domain, to a corresponding value in the range. In the range. So this is what f does. So this is domain, domain, and this right over here is going to be the range. Now f inverse, if you pass it the value in the range, it'll map it back to the corresponding value in the domain."}, {"video_title": "Understanding inverse functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "In the range. So this is what f does. So this is domain, domain, and this right over here is going to be the range. Now f inverse, if you pass it the value in the range, it'll map it back to the corresponding value in the domain. But how do we think about it like this? Well f inverse of eight, this is whatever maps to eight. So if this was eight, we have to say, well what mapped to eight?"}, {"video_title": "Understanding inverse functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Now f inverse, if you pass it the value in the range, it'll map it back to the corresponding value in the domain. But how do we think about it like this? Well f inverse of eight, this is whatever maps to eight. So if this was eight, we have to say, well what mapped to eight? Well we see here f of nine is eight. f of nine is eight. So f inverse of eight is going to be, is going to be equal to, and actually we do this in that same color, is going to be equal to nine."}, {"video_title": "Understanding inverse functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So if this was eight, we have to say, well what mapped to eight? Well we see here f of nine is eight. f of nine is eight. So f inverse of eight is going to be, is going to be equal to, and actually we do this in that same color, is going to be equal to nine. And if it makes it easier, we could actually construct a table here. And this is actually what I probably would do, just to make sure I'm not doing something strange. Where I could say x and f inverse of x, and what I'd essentially do is I'd swap these two columns."}, {"video_title": "Understanding inverse functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So f inverse of eight is going to be, is going to be equal to, and actually we do this in that same color, is going to be equal to nine. And if it makes it easier, we could actually construct a table here. And this is actually what I probably would do, just to make sure I'm not doing something strange. Where I could say x and f inverse of x, and what I'd essentially do is I'd swap these two columns. So f of x goes from negative nine to five, f inverse of x is going to go from five to negative nine. All I did is I swapped these two. Now we're mapping from this to that."}, {"video_title": "Understanding inverse functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Where I could say x and f inverse of x, and what I'd essentially do is I'd swap these two columns. So f of x goes from negative nine to five, f inverse of x is going to go from five to negative nine. All I did is I swapped these two. Now we're mapping from this to that. So f inverse of x is gonna map from seven to negative seven. Notice, instead of going from this, mapping from this thing to that thing, we're now gonna map from that thing to that thing. So f inverse is gonna map from 13 to five."}, {"video_title": "Understanding inverse functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Now we're mapping from this to that. So f inverse of x is gonna map from seven to negative seven. Notice, instead of going from this, mapping from this thing to that thing, we're now gonna map from that thing to that thing. So f inverse is gonna map from 13 to five. 13 to five is gonna map from negative seven to six. Negative seven to six, it's gonna map from eight to nine. Eight to nine and it's gonna map from 12 to 11."}, {"video_title": "Understanding inverse functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So f inverse is gonna map from 13 to five. 13 to five is gonna map from negative seven to six. Negative seven to six, it's gonna map from eight to nine. Eight to nine and it's gonna map from 12 to 11. From 12 to 11. So let's see, did I do that right? Yeah, it looks like I got all of, yep."}, {"video_title": "Understanding inverse functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Eight to nine and it's gonna map from 12 to 11. From 12 to 11. So let's see, did I do that right? Yeah, it looks like I got all of, yep. So all I did is really I just swapped these columns. It's mapping, the f inverse maps from this column to that column. So I just swapped them out."}, {"video_title": "Understanding inverse functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Yeah, it looks like I got all of, yep. So all I did is really I just swapped these columns. It's mapping, the f inverse maps from this column to that column. So I just swapped them out. So now it becomes a little bit clearer that way. You see it right over here, f inverse of eight. If you input eight into f inverse, you get nine."}, {"video_title": "Understanding inverse functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So I just swapped them out. So now it becomes a little bit clearer that way. You see it right over here, f inverse of eight. If you input eight into f inverse, you get nine. So now we can use that to start doing fancier things. We can evaluate something like f inverse, f inverse of, actually let's evaluate this. Let's evaluate f of, f of f inverse of, let's see, f of f inverse of seven."}, {"video_title": "Understanding inverse functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "If you input eight into f inverse, you get nine. So now we can use that to start doing fancier things. We can evaluate something like f inverse, f inverse of, actually let's evaluate this. Let's evaluate f of, f of f inverse of, let's see, f of f inverse of seven. What is this going to be? Well, let's first evaluate f inverse of seven. F inverse of seven maps from seven to negative seven."}, {"video_title": "Understanding inverse functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Let's evaluate f of, f of f inverse of, let's see, f of f inverse of seven. What is this going to be? Well, let's first evaluate f inverse of seven. F inverse of seven maps from seven to negative seven. Maps from seven to negative seven. So this is going to be f of this stuff in here. I think we do it in yellow."}, {"video_title": "Understanding inverse functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "F inverse of seven maps from seven to negative seven. Maps from seven to negative seven. So this is going to be f of this stuff in here. I think we do it in yellow. This stuff in here, f inverse of seven, we see is negative seven. So it's going to be f of negative seven. And then to evaluate the function, well, f of negative seven, that's just going to be seven again."}, {"video_title": "Understanding inverse functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "I think we do it in yellow. This stuff in here, f inverse of seven, we see is negative seven. So it's going to be f of negative seven. And then to evaluate the function, well, f of negative seven, that's just going to be seven again. And that makes complete sense. We essentially went, we mapped from seven, f inverse of seven went from seven to negative seven, and then evaluating the function of that went back to, went back to seven. So let's do one more of these, just to really feel comfortable with mapping back and forth between these two sets, between applying the function and the inverse of the function."}, {"video_title": "Understanding inverse functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And then to evaluate the function, well, f of negative seven, that's just going to be seven again. And that makes complete sense. We essentially went, we mapped from seven, f inverse of seven went from seven to negative seven, and then evaluating the function of that went back to, went back to seven. So let's do one more of these, just to really feel comfortable with mapping back and forth between these two sets, between applying the function and the inverse of the function. So let's evaluate what, do it in purple, f, I mean, no, that wasn't purple. So it's going, let's try to evaluate f, f inverse of, f inverse of 13, of f inverse of 13. What is that going to be?"}, {"video_title": "Understanding inverse functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So let's do one more of these, just to really feel comfortable with mapping back and forth between these two sets, between applying the function and the inverse of the function. So let's evaluate what, do it in purple, f, I mean, no, that wasn't purple. So it's going, let's try to evaluate f, f inverse of, f inverse of 13, of f inverse of 13. What is that going to be? I encourage you to pause the video and try to figure it out. Well, what's f inverse of 13? Well, that's looking at this table right here."}, {"video_title": "Understanding inverse functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "What is that going to be? I encourage you to pause the video and try to figure it out. Well, what's f inverse of 13? Well, that's looking at this table right here. f inverse goes from 13 to five. And you see it over here. f went from five to 13, so f inverse is going to go from 13 to five."}, {"video_title": "Understanding inverse functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Well, that's looking at this table right here. f inverse goes from 13 to five. And you see it over here. f went from five to 13, so f inverse is going to go from 13 to five. So this right over here, f inverse of 13 is just going to be five. So this whole thing is the same thing as f inverse of five. And f inverse of five, well, f inverse goes from five to negative nine."}, {"video_title": "Understanding inverse functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "f went from five to 13, so f inverse is going to go from 13 to five. So this right over here, f inverse of 13 is just going to be five. So this whole thing is the same thing as f inverse of five. And f inverse of five, well, f inverse goes from five to negative nine. So this is going to be equal to negative nine. Once again, f inverse goes from five, f goes from negative nine to five, so f inverse is gonna go from five to negative nine. So at first, when you start doing these kind of inverse and function and inverse of a function, it looks a little confusing."}, {"video_title": "Understanding inverse functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And f inverse of five, well, f inverse goes from five to negative nine. So this is going to be equal to negative nine. Once again, f inverse goes from five, f goes from negative nine to five, so f inverse is gonna go from five to negative nine. So at first, when you start doing these kind of inverse and function and inverse of a function, it looks a little confusing. Hey, I'm going back and forth. But you just have to remember, a function maps from one set of numbers to another set of numbers, and the inverse of that function goes the other way. So if the function goes from nine to eight, the inverse is gonna go from eight to nine."}, {"video_title": "Compound inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "Solve for y. We have 3y plus 7 is less than 2y, and 4y plus 8 is greater than negative 48. So we have to find all the y's that meet both of these constraints. So let's just solve for y in each of the constraints. And just remember that this and is here. So we have 3y plus 7 is less than 2y. So let's isolate the y's on the left-hand side."}, {"video_title": "Compound inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "So let's just solve for y in each of the constraints. And just remember that this and is here. So we have 3y plus 7 is less than 2y. So let's isolate the y's on the left-hand side. So let's get rid of this 2y on the right-hand side. And we can do that by subtracting 2y from both sides. So we're going to subtract 2y from both sides."}, {"video_title": "Compound inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "So let's isolate the y's on the left-hand side. So let's get rid of this 2y on the right-hand side. And we can do that by subtracting 2y from both sides. So we're going to subtract 2y from both sides. The left-hand side, we have 3y minus 2y, which is just y, plus 7 is less than 2y minus 2y. And there's nothing else there. That's just going to be 0."}, {"video_title": "Compound inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "So we're going to subtract 2y from both sides. The left-hand side, we have 3y minus 2y, which is just y, plus 7 is less than 2y minus 2y. And there's nothing else there. That's just going to be 0. And then we can get rid of this 7 here by subtracting 7 from both sides. So let's subtract 7 from both sides. Left-hand side, y plus 7 minus 7, those cancel out."}, {"video_title": "Compound inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "That's just going to be 0. And then we can get rid of this 7 here by subtracting 7 from both sides. So let's subtract 7 from both sides. Left-hand side, y plus 7 minus 7, those cancel out. We just have a y, is less than 0 minus 7, which is negative 7. So that's one of the constraints. That's this constraint right over here."}, {"video_title": "Compound inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "Left-hand side, y plus 7 minus 7, those cancel out. We just have a y, is less than 0 minus 7, which is negative 7. So that's one of the constraints. That's this constraint right over here. Now let's work on this constraint. We have 4y plus 8 is greater than negative 48. So let's get rid of the 8 from the left-hand side."}, {"video_title": "Compound inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "That's this constraint right over here. Now let's work on this constraint. We have 4y plus 8 is greater than negative 48. So let's get rid of the 8 from the left-hand side. So we can subtract 8 from both sides. The left-hand side, we're just left with a 4y, because these guys cancel out. 4y is greater than negative 48 minus 8."}, {"video_title": "Compound inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "So let's get rid of the 8 from the left-hand side. So we can subtract 8 from both sides. The left-hand side, we're just left with a 4y, because these guys cancel out. 4y is greater than negative 48 minus 8. So we're going to go another 8 negative. So 48 plus 8 would be 56. So this is going to be negative 56."}, {"video_title": "Compound inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "4y is greater than negative 48 minus 8. So we're going to go another 8 negative. So 48 plus 8 would be 56. So this is going to be negative 56. And now to isolate the y, we can divide both sides by positive 4. And we don't have to swap the inequality, since we're dividing by a positive number. So let's divide both sides by 4 over here."}, {"video_title": "Compound inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "So this is going to be negative 56. And now to isolate the y, we can divide both sides by positive 4. And we don't have to swap the inequality, since we're dividing by a positive number. So let's divide both sides by 4 over here. So we get y is greater than, what is 56 over 4? Negative 56 over 4. Let's see, 40 is 10 times 4."}, {"video_title": "Compound inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "So let's divide both sides by 4 over here. So we get y is greater than, what is 56 over 4? Negative 56 over 4. Let's see, 40 is 10 times 4. And then we have another 16 to worry about. So it's 14 times 4. So y is greater than negative 14."}, {"video_title": "Compound inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "Let's see, 40 is 10 times 4. And then we have another 16 to worry about. So it's 14 times 4. So y is greater than negative 14. Is that right? 4 times 10 is 40. 4 times 4 is 16."}, {"video_title": "Compound inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "So y is greater than negative 14. Is that right? 4 times 10 is 40. 4 times 4 is 16. Yep, 56. So y is greater than negative 14. And let's remember, we have this and here."}, {"video_title": "Compound inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "4 times 4 is 16. Yep, 56. So y is greater than negative 14. And let's remember, we have this and here. And y is less than negative 7. So we have to meet both of these constraints over here. So let's draw them on the number line."}, {"video_title": "Compound inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "And let's remember, we have this and here. And y is less than negative 7. So we have to meet both of these constraints over here. So let's draw them on the number line. So I have my number line over here. And let's say negative 14 is over here. So you have negative 13, 12, 11, 10, 9, 8, 7."}, {"video_title": "Compound inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "So let's draw them on the number line. So I have my number line over here. And let's say negative 14 is over here. So you have negative 13, 12, 11, 10, 9, 8, 7. That's negative 7. And then negative 6, 5, 4, 3, 2, 1. This would be 0."}, {"video_title": "Compound inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "So you have negative 13, 12, 11, 10, 9, 8, 7. That's negative 7. And then negative 6, 5, 4, 3, 2, 1. This would be 0. Then you could keep going up more positive. And so we're looking for all of the y's that are less than negative 7. So let's look at this."}, {"video_title": "Compound inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "This would be 0. Then you could keep going up more positive. And so we're looking for all of the y's that are less than negative 7. So let's look at this. Less than negative 7. So not including negative 7. So we'll do an open circle around negative 7."}, {"video_title": "Compound inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "So let's look at this. Less than negative 7. So not including negative 7. So we'll do an open circle around negative 7. And less than negative 7. And if that was the only constraint, we would keep going to the left. But we have this other constraint."}, {"video_title": "Compound inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "So we'll do an open circle around negative 7. And less than negative 7. And if that was the only constraint, we would keep going to the left. But we have this other constraint. And y has to be greater than negative 14. And y is greater than negative 14. So you make a circle around negative 14 and everything that's greater than that."}, {"video_title": "Compound inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "But we have this other constraint. And y has to be greater than negative 14. And y is greater than negative 14. So you make a circle around negative 14 and everything that's greater than that. And if you didn't have this other constraint, you would keep going. But the y's that satisfy both of them are all of the y's in between. These are the y's that are both less than negative 7 and greater than negative 14."}, {"video_title": "Compound inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "So you make a circle around negative 14 and everything that's greater than that. And if you didn't have this other constraint, you would keep going. But the y's that satisfy both of them are all of the y's in between. These are the y's that are both less than negative 7 and greater than negative 14. We can verify that things here work. So let's try some values out. So a value that would work, well, let me just do negative 10 is right here."}, {"video_title": "Compound inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "These are the y's that are both less than negative 7 and greater than negative 14. We can verify that things here work. So let's try some values out. So a value that would work, well, let me just do negative 10 is right here. 8, 9. This is negative 10. That should work."}, {"video_title": "Compound inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "So a value that would work, well, let me just do negative 10 is right here. 8, 9. This is negative 10. That should work. So let's try it out. So we'd have 3 times negative 10 plus 7 should be less than 2 times negative 10. So this is negative 30 plus 7 is negative 23, which is indeed less than negative 20."}, {"video_title": "Compound inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "That should work. So let's try it out. So we'd have 3 times negative 10 plus 7 should be less than 2 times negative 10. So this is negative 30 plus 7 is negative 23, which is indeed less than negative 20. So that works. And negative 10 has to work for this one as well. So you have 4 times negative 10, which is negative 40, plus 8."}, {"video_title": "Compound inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "So this is negative 30 plus 7 is negative 23, which is indeed less than negative 20. So that works. And negative 10 has to work for this one as well. So you have 4 times negative 10, which is negative 40, plus 8. Negative 40 plus 8 should be greater than negative 48. Well, negative 40 plus 8 is negative 32. We're going 8 in the positive direction."}, {"video_title": "Compound inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "So you have 4 times negative 10, which is negative 40, plus 8. Negative 40 plus 8 should be greater than negative 48. Well, negative 40 plus 8 is negative 32. We're going 8 in the positive direction. So we're getting less negative. And negative 32 is greater than negative 48. It's less negative."}, {"video_title": "Compound inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "We're going 8 in the positive direction. So we're getting less negative. And negative 32 is greater than negative 48. It's less negative. So this works. So negative 10 works. Now let's just verify things that shouldn't work."}, {"video_title": "Compound inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "It's less negative. So this works. So negative 10 works. Now let's just verify things that shouldn't work. So 0 should not work. It's not in the solution set. So let's try it out."}, {"video_title": "Compound inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "Now let's just verify things that shouldn't work. So 0 should not work. It's not in the solution set. So let's try it out. We fit 3 times 0 plus 7. That would be 7. And 7 is not less than 0."}, {"video_title": "Compound inequalities 2 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "So let's try it out. We fit 3 times 0 plus 7. That would be 7. And 7 is not less than 0. So it would violate this condition right over here if you put a 0 over here. If you put a negative 15 over here, it should violate this condition right over here because it wasn't in this guy's solution set. Anyway, hopefully you found that useful."}, {"video_title": "Introduction to vertex form of a quadratic.mp3", "Sentence": "They are in different forms. This is the equation in sometimes called standard form for a quadratic. This is the quadratic in factored form. Notice this has been factored right over here. And this last form is what we're going to focus on in this video. This is sometimes known as vertex form. And we're not gonna focus on how do you get from one of these other forms to vertex form in this video."}, {"video_title": "Introduction to vertex form of a quadratic.mp3", "Sentence": "Notice this has been factored right over here. And this last form is what we're going to focus on in this video. This is sometimes known as vertex form. And we're not gonna focus on how do you get from one of these other forms to vertex form in this video. We'll do that in future videos. But what we're going to do is appreciate why this is called vertex, vertex form. Now to start, let's just remind ourselves what a vertex is."}, {"video_title": "Introduction to vertex form of a quadratic.mp3", "Sentence": "And we're not gonna focus on how do you get from one of these other forms to vertex form in this video. We'll do that in future videos. But what we're going to do is appreciate why this is called vertex, vertex form. Now to start, let's just remind ourselves what a vertex is. So as you might remember from other videos, if we have a quadratic, if we're graphing y is equal to some quadratic expression in terms of x, the graph of that will be a parabola. And it might be an upward opening parabola or a downward opening parabola. So this one in particular is going to be an upward opening parabola."}, {"video_title": "Introduction to vertex form of a quadratic.mp3", "Sentence": "Now to start, let's just remind ourselves what a vertex is. So as you might remember from other videos, if we have a quadratic, if we're graphing y is equal to some quadratic expression in terms of x, the graph of that will be a parabola. And it might be an upward opening parabola or a downward opening parabola. So this one in particular is going to be an upward opening parabola. And so it might look something like this. So it might look something, something like this right over here. And for an upward opening parabola like this, the vertex is this point right over here."}, {"video_title": "Introduction to vertex form of a quadratic.mp3", "Sentence": "So this one in particular is going to be an upward opening parabola. And so it might look something like this. So it might look something, something like this right over here. And for an upward opening parabola like this, the vertex is this point right over here. You could view it as this minimum point. You have your x coordinate of the vertex right over there, and you have your y coordinate of the vertex right over here. Now the reason why this is called vertex form is it's fairly straightforward to pick out the coordinates of this vertex from this form."}, {"video_title": "Introduction to vertex form of a quadratic.mp3", "Sentence": "And for an upward opening parabola like this, the vertex is this point right over here. You could view it as this minimum point. You have your x coordinate of the vertex right over there, and you have your y coordinate of the vertex right over here. Now the reason why this is called vertex form is it's fairly straightforward to pick out the coordinates of this vertex from this form. How do we do that? Well to do that, we just have to appreciate the structure that's in this expression. Let me just rewrite it again."}, {"video_title": "Introduction to vertex form of a quadratic.mp3", "Sentence": "Now the reason why this is called vertex form is it's fairly straightforward to pick out the coordinates of this vertex from this form. How do we do that? Well to do that, we just have to appreciate the structure that's in this expression. Let me just rewrite it again. We have y is equal to three times x plus two squared minus 27. The important thing to realize is that this part of the expression is never going to be negative. No matter what you have here, if you square it, you're never going to get a negative value."}, {"video_title": "Introduction to vertex form of a quadratic.mp3", "Sentence": "Let me just rewrite it again. We have y is equal to three times x plus two squared minus 27. The important thing to realize is that this part of the expression is never going to be negative. No matter what you have here, if you square it, you're never going to get a negative value. And so if this is never going to be negative, and we're multiplying it by a positive right over here, this whole thing right over here is going to be greater than or equal to zero. So another way to think about it, it's only going to be additive to negative 27. So your minimum point for this curve right over here, for your parabola, is going to happen when this expression is equal to zero, when you're not adding anything to negative 27."}, {"video_title": "Introduction to vertex form of a quadratic.mp3", "Sentence": "No matter what you have here, if you square it, you're never going to get a negative value. And so if this is never going to be negative, and we're multiplying it by a positive right over here, this whole thing right over here is going to be greater than or equal to zero. So another way to think about it, it's only going to be additive to negative 27. So your minimum point for this curve right over here, for your parabola, is going to happen when this expression is equal to zero, when you're not adding anything to negative 27. And so when will this equal zero? Well it's going to be equal to zero when x plus two is going to be equal to zero. So you could just say, if you want to find the x-coordinate of the vertex, well for what x value does x plus two equal zero?"}, {"video_title": "Introduction to vertex form of a quadratic.mp3", "Sentence": "So your minimum point for this curve right over here, for your parabola, is going to happen when this expression is equal to zero, when you're not adding anything to negative 27. And so when will this equal zero? Well it's going to be equal to zero when x plus two is going to be equal to zero. So you could just say, if you want to find the x-coordinate of the vertex, well for what x value does x plus two equal zero? And of course we can subtract two from both sides and you get x is equal to negative two. So we know that this x-coordinate right over here is negative two. And then what's the y-coordinate of the vertex?"}, {"video_title": "Introduction to vertex form of a quadratic.mp3", "Sentence": "So you could just say, if you want to find the x-coordinate of the vertex, well for what x value does x plus two equal zero? And of course we can subtract two from both sides and you get x is equal to negative two. So we know that this x-coordinate right over here is negative two. And then what's the y-coordinate of the vertex? You could say, hey, what is the minimum y that this curve takes on? Well, when x is equal to negative two, this whole thing is zero and y is equal to negative 27. Y is equal to negative 27."}, {"video_title": "Introduction to vertex form of a quadratic.mp3", "Sentence": "And then what's the y-coordinate of the vertex? You could say, hey, what is the minimum y that this curve takes on? Well, when x is equal to negative two, this whole thing is zero and y is equal to negative 27. Y is equal to negative 27. So this right over here is negative 27. And so the coordinates of the vertex here are negative two comma negative 27. And you were able to pick that out just by looking at the quadratic in vertex form."}, {"video_title": "Introduction to vertex form of a quadratic.mp3", "Sentence": "Y is equal to negative 27. So this right over here is negative 27. And so the coordinates of the vertex here are negative two comma negative 27. And you were able to pick that out just by looking at the quadratic in vertex form. Now let's get a few more examples under our belt so that we can really get good at picking out the vertex when a quadratic is written in vertex form. So let's say, let's pick a scenario where we have a downward opening parabola where y is equal to, let's just say, negative two times x plus five. Actually, let me make it x minus five."}, {"video_title": "Introduction to vertex form of a quadratic.mp3", "Sentence": "And you were able to pick that out just by looking at the quadratic in vertex form. Now let's get a few more examples under our belt so that we can really get good at picking out the vertex when a quadratic is written in vertex form. So let's say, let's pick a scenario where we have a downward opening parabola where y is equal to, let's just say, negative two times x plus five. Actually, let me make it x minus five. X minus five squared, and then let's say plus 10. Well, here, this is going to be downward opening, and let's appreciate why that is. So here, this part is still always going to be non-negative, but it's being multiplied by a negative two."}, {"video_title": "Introduction to vertex form of a quadratic.mp3", "Sentence": "Actually, let me make it x minus five. X minus five squared, and then let's say plus 10. Well, here, this is going to be downward opening, and let's appreciate why that is. So here, this part is still always going to be non-negative, but it's being multiplied by a negative two. So it's actually always going to be non-positive. So this whole thing right over here is going to be less than or equal to zero for all x's. So it can only take away from the 10."}, {"video_title": "Introduction to vertex form of a quadratic.mp3", "Sentence": "So here, this part is still always going to be non-negative, but it's being multiplied by a negative two. So it's actually always going to be non-positive. So this whole thing right over here is going to be less than or equal to zero for all x's. So it can only take away from the 10. So where do we hit a maximum point? Well, we hit a maximum point when x minus five is equal to zero, when we're not taking anything away from the 10. And so x minus five is equal to zero."}, {"video_title": "Introduction to vertex form of a quadratic.mp3", "Sentence": "So it can only take away from the 10. So where do we hit a maximum point? Well, we hit a maximum point when x minus five is equal to zero, when we're not taking anything away from the 10. And so x minus five is equal to zero. Well, that, of course, is going to happen when x is equal to five. And that, indeed, is the x-coordinate for the vertex. And what's the y-coordinate for the vertex?"}, {"video_title": "Introduction to vertex form of a quadratic.mp3", "Sentence": "And so x minus five is equal to zero. Well, that, of course, is going to happen when x is equal to five. And that, indeed, is the x-coordinate for the vertex. And what's the y-coordinate for the vertex? Well, if x is equal to five, this thing is zero. You're not going to be taking anything away from the 10, and so y is going to be equal to 10. And so the vertex here is x equals five, which, and I'm just gonna eyeball it, maybe it's right over here, x equals five, and y is equal to 10."}, {"video_title": "Introduction to vertex form of a quadratic.mp3", "Sentence": "And what's the y-coordinate for the vertex? Well, if x is equal to five, this thing is zero. You're not going to be taking anything away from the 10, and so y is going to be equal to 10. And so the vertex here is x equals five, which, and I'm just gonna eyeball it, maybe it's right over here, x equals five, and y is equal to 10. If this is negative 27, this would be positive 27, 10 would be something like this, not using the same scales for the x and y-axis, but there you have it. So it's five comma 10, and our curve is gonna look something like this. I don't know exactly where it intersects the x-axis, but it's going to be a downward-opening parabola."}, {"video_title": "Introduction to vertex form of a quadratic.mp3", "Sentence": "And so the vertex here is x equals five, which, and I'm just gonna eyeball it, maybe it's right over here, x equals five, and y is equal to 10. If this is negative 27, this would be positive 27, 10 would be something like this, not using the same scales for the x and y-axis, but there you have it. So it's five comma 10, and our curve is gonna look something like this. I don't know exactly where it intersects the x-axis, but it's going to be a downward-opening parabola. Let's do one more example, just so that we get really fluent at identifying the vertex from vertex form. So let's say, I'm just gonna make this up, we have y is equal to negative pi times x minus 2.8 squared plus, I don't know, plus 7.1. What is the vertex of the parabola here?"}, {"video_title": "Introduction to vertex form of a quadratic.mp3", "Sentence": "I don't know exactly where it intersects the x-axis, but it's going to be a downward-opening parabola. Let's do one more example, just so that we get really fluent at identifying the vertex from vertex form. So let's say, I'm just gonna make this up, we have y is equal to negative pi times x minus 2.8 squared plus, I don't know, plus 7.1. What is the vertex of the parabola here? Well, the x-coordinate is going to be the x-value that makes this equal to zero, which is 2.8, and then if this is equal to zero, then this whole thing is going to be equal to zero, and y is going to be 7.1. So now you hopefully appreciate why this is called vertex form. It's quite straightforward to pick out the vertex when you have something written in this way."}, {"video_title": "Finding average rate of change of polynomials Algebra 2 Khan Academy.mp3", "Sentence": "All right, now let's work through it together. So there's a couple of ways that we can conceptualize average rate of change of a function. One way to think about it is it's our change in the value of our function divided by our change in x, or it's our change in the value of our function per x on average. So you can view it as change in the value of function divided by your change in x. If you say that y is equal to f of x, you could also express it as change in y over change in x. On average, how much does a function change per unit change in x on average? Now we could do this with a table, or we could try to conceptualize it visually."}, {"video_title": "Finding average rate of change of polynomials Algebra 2 Khan Academy.mp3", "Sentence": "So you can view it as change in the value of function divided by your change in x. If you say that y is equal to f of x, you could also express it as change in y over change in x. On average, how much does a function change per unit change in x on average? Now we could do this with a table, or we could try to conceptualize it visually. But let's just do this one with a table, and then we'll try to connect the dots a little bit with a visual. So if we have x here, and then if we have y is equal to f of x right over here, when x is equal to negative two, what is y going to be equal to, or what is f of negative two? Well, let's see, f of, so y is equal to f of negative two, which is going to be equal to negative eight."}, {"video_title": "Finding average rate of change of polynomials Algebra 2 Khan Academy.mp3", "Sentence": "Now we could do this with a table, or we could try to conceptualize it visually. But let's just do this one with a table, and then we'll try to connect the dots a little bit with a visual. So if we have x here, and then if we have y is equal to f of x right over here, when x is equal to negative two, what is y going to be equal to, or what is f of negative two? Well, let's see, f of, so y is equal to f of negative two, which is going to be equal to negative eight. That's negative two to the third power, minus four times negative two, so that's minus negative eight, so that's plus eight, that equals zero. And then when x is equal to three, I'm going to the right end of that interval, well now y is equal to f of three, which is equal to 27, three to the third power, minus four times three, minus 12, which is equal to 15. So what is our change in y over our change in x over this interval?"}, {"video_title": "Finding average rate of change of polynomials Algebra 2 Khan Academy.mp3", "Sentence": "Well, let's see, f of, so y is equal to f of negative two, which is going to be equal to negative eight. That's negative two to the third power, minus four times negative two, so that's minus negative eight, so that's plus eight, that equals zero. And then when x is equal to three, I'm going to the right end of that interval, well now y is equal to f of three, which is equal to 27, three to the third power, minus four times three, minus 12, which is equal to 15. So what is our change in y over our change in x over this interval? Well, our y went from zero to 15, so we have a increase of 15 in y. And what was our change in x? Well, we went from negative two to positive three, so we had a plus five change in x, so our change in x is plus five."}, {"video_title": "Finding average rate of change of polynomials Algebra 2 Khan Academy.mp3", "Sentence": "So what is our change in y over our change in x over this interval? Well, our y went from zero to 15, so we have a increase of 15 in y. And what was our change in x? Well, we went from negative two to positive three, so we had a plus five change in x, so our change in x is plus five. And so our average rate of change of y with respect to x, or the rate of change of our function with respect to x over the interval is going to be equal to three. If you wanted to think about this visually, I could try to sketch this. So this is the x-axis, the y-axis, and our function does something like this."}, {"video_title": "Finding average rate of change of polynomials Algebra 2 Khan Academy.mp3", "Sentence": "Well, we went from negative two to positive three, so we had a plus five change in x, so our change in x is plus five. And so our average rate of change of y with respect to x, or the rate of change of our function with respect to x over the interval is going to be equal to three. If you wanted to think about this visually, I could try to sketch this. So this is the x-axis, the y-axis, and our function does something like this. So at x equals negative two, f of x is zero. And then it goes up, and then it comes back down, and then it does something like this. It does something like this, and it does, and it was going before this."}, {"video_title": "Finding average rate of change of polynomials Algebra 2 Khan Academy.mp3", "Sentence": "So this is the x-axis, the y-axis, and our function does something like this. So at x equals negative two, f of x is zero. And then it goes up, and then it comes back down, and then it does something like this. It does something like this, and it does, and it was going before this. And so the interval that we care about, we're going from negative two to three, which is right about there. So that's x equals negative two to x equals three. And so what we wanna do at the left end of the interval, our function is equal to zero, so we're at this point right over here."}, {"video_title": "Finding average rate of change of polynomials Algebra 2 Khan Academy.mp3", "Sentence": "It does something like this, and it does, and it was going before this. And so the interval that we care about, we're going from negative two to three, which is right about there. So that's x equals negative two to x equals three. And so what we wanna do at the left end of the interval, our function is equal to zero, so we're at this point right over here. I'll do this in a new color. We're at this point right over there. And at the right end of our function, f of three is 15."}, {"video_title": "Finding average rate of change of polynomials Algebra 2 Khan Academy.mp3", "Sentence": "And so what we wanna do at the left end of the interval, our function is equal to zero, so we're at this point right over here. I'll do this in a new color. We're at this point right over there. And at the right end of our function, f of three is 15. So we are up here someplace. Let me connect the curve a little bit. We are going to be up there."}, {"video_title": "Finding average rate of change of polynomials Algebra 2 Khan Academy.mp3", "Sentence": "And at the right end of our function, f of three is 15. So we are up here someplace. Let me connect the curve a little bit. We are going to be up there. And so when we're thinking about the average rate of change over this interval, we're really thinking about the slope of the line that connects these two points. So the line that connects these two points looks something like this. And we're just calculating what is our change in y, which is going to be this, our change in y."}, {"video_title": "Finding average rate of change of polynomials Algebra 2 Khan Academy.mp3", "Sentence": "We are going to be up there. And so when we're thinking about the average rate of change over this interval, we're really thinking about the slope of the line that connects these two points. So the line that connects these two points looks something like this. And we're just calculating what is our change in y, which is going to be this, our change in y. And we see that the value of our function increased by 15 divided by our change in x. So this right over here is our change in x, which we see we went from negative two to three. That's going to be equal to five."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And what you're going to see here, and it might be a little bit difficult to fully appreciate right from the get-go, it's a more bizarre number than some of the other wacky numbers we learn in mathematics like pi or e. And it's more bizarre because it doesn't have a tangible value in the sense that we're normally used to defining numbers. i is defined as the number whose square is equal to negative 1. This is the definition of i. And it leads to all sorts of interesting things. Now some places you will see i defined this way, i as being equal to the principal square root of negative 1. I wanted to just point out to you that this is not wrong, and it might make sense to you. If something squared is negative 1, then maybe it's the principal square root of negative 1."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And it leads to all sorts of interesting things. Now some places you will see i defined this way, i as being equal to the principal square root of negative 1. I wanted to just point out to you that this is not wrong, and it might make sense to you. If something squared is negative 1, then maybe it's the principal square root of negative 1. And so these seem to be almost the same statement. But I just want to make you a little careful. When you do this, some people will even go as far as saying this is wrong."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "If something squared is negative 1, then maybe it's the principal square root of negative 1. And so these seem to be almost the same statement. But I just want to make you a little careful. When you do this, some people will even go as far as saying this is wrong. And it actually turns out that they are wrong to say that this is wrong. But when you do this, you have to be a little bit careful about what it means to take a principal square root of a negative number and it being defined for imaginary, and as we'll learn in the future, complex numbers. But for your understanding right now, you don't have to differentiate them."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "When you do this, some people will even go as far as saying this is wrong. And it actually turns out that they are wrong to say that this is wrong. But when you do this, you have to be a little bit careful about what it means to take a principal square root of a negative number and it being defined for imaginary, and as we'll learn in the future, complex numbers. But for your understanding right now, you don't have to differentiate them. You don't have to split hairs between any of these definitions. Now, with this definition, let's just think about what the different powers of i are. Because you can imagine, if something squared is negative 1, if I take it to all sorts of powers, maybe that should give us all sorts of weird things."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "But for your understanding right now, you don't have to differentiate them. You don't have to split hairs between any of these definitions. Now, with this definition, let's just think about what the different powers of i are. Because you can imagine, if something squared is negative 1, if I take it to all sorts of powers, maybe that should give us all sorts of weird things. And what we'll see is that the powers of i are kind of neat, because they kind of cycle, or they do cycle, through a set of values. So I could start with i to the 0th power. And so you might say, hey, look, anything to the 0th power is 1."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Because you can imagine, if something squared is negative 1, if I take it to all sorts of powers, maybe that should give us all sorts of weird things. And what we'll see is that the powers of i are kind of neat, because they kind of cycle, or they do cycle, through a set of values. So I could start with i to the 0th power. And so you might say, hey, look, anything to the 0th power is 1. So i to the 0th power is 1. And that is true. And you could actually derive that even from this definition."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And so you might say, hey, look, anything to the 0th power is 1. So i to the 0th power is 1. And that is true. And you could actually derive that even from this definition. But this is pretty straightforward. Anything to the 0th power, including i, is 1. Then you say, OK, what is i to the 1st power?"}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And you could actually derive that even from this definition. But this is pretty straightforward. Anything to the 0th power, including i, is 1. Then you say, OK, what is i to the 1st power? Well, anything to the 1st power is just that number times itself once. So that's just going to be i, really by the definition of what it means to take an exponent. So that completely makes sense."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Then you say, OK, what is i to the 1st power? Well, anything to the 1st power is just that number times itself once. So that's just going to be i, really by the definition of what it means to take an exponent. So that completely makes sense. And then you have i to the 2nd power. Well, by definition, i to the 2nd power is equal to negative 1. Let's try i to the 3rd power."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So that completely makes sense. And then you have i to the 2nd power. Well, by definition, i to the 2nd power is equal to negative 1. Let's try i to the 3rd power. I'll do this in a color I haven't used. i to the 3rd power. Well, that's going to be i to the 2nd power times i."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Let's try i to the 3rd power. I'll do this in a color I haven't used. i to the 3rd power. Well, that's going to be i to the 2nd power times i. And we know that i to the 2nd power is negative 1. So it's negative 1 times i. Let me make it clear."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Well, that's going to be i to the 2nd power times i. And we know that i to the 2nd power is negative 1. So it's negative 1 times i. Let me make it clear. This is the same thing as this, which is the same thing as that. i squared is negative 1. So when you multiply it out, negative 1 times i we'll write as negative i."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Let me make it clear. This is the same thing as this, which is the same thing as that. i squared is negative 1. So when you multiply it out, negative 1 times i we'll write as negative i. Now, what happens if we take i to the 4th power? I'll do it up here. i to the 4th power."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So when you multiply it out, negative 1 times i we'll write as negative i. Now, what happens if we take i to the 4th power? I'll do it up here. i to the 4th power. Well, once again, this is going to be i times i to the 3rd power. So that's i times i to the 3rd power. But what was i to the 3rd power?"}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "i to the 4th power. Well, once again, this is going to be i times i to the 3rd power. So that's i times i to the 3rd power. But what was i to the 3rd power? i to the 3rd power was negative i. This over here is negative i. And so i times i would give you negative 1."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "But what was i to the 3rd power? i to the 3rd power was negative i. This over here is negative i. And so i times i would give you negative 1. But you have a negative out here. So it's i times i is negative 1. And then you have a negative."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And so i times i would give you negative 1. But you have a negative out here. So it's i times i is negative 1. And then you have a negative. That gives you positive 1. Let me write it down. So this is the same thing as i times negative i, which is the same thing as negative 1 times, remember, multiplication is commutative."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And then you have a negative. That gives you positive 1. Let me write it down. So this is the same thing as i times negative i, which is the same thing as negative 1 times, remember, multiplication is commutative. If we're just multiplying a bunch of numbers, we can switch the order. So this is the same thing as negative 1 times i times i. i times i, by definition, is negative 1. Negative 1 times negative 1 is equal to 1."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So this is the same thing as i times negative i, which is the same thing as negative 1 times, remember, multiplication is commutative. If we're just multiplying a bunch of numbers, we can switch the order. So this is the same thing as negative 1 times i times i. i times i, by definition, is negative 1. Negative 1 times negative 1 is equal to 1. So i to the 4th is the same thing as i to the 0th power. Now let's try i to the 5th power. Well, that's just going to be i to the 4th times i."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Negative 1 times negative 1 is equal to 1. So i to the 4th is the same thing as i to the 0th power. Now let's try i to the 5th power. Well, that's just going to be i to the 4th times i. And we know what i to the 4th is. It is 1. So it's 1 times i, or it is just i again."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Well, that's just going to be i to the 4th times i. And we know what i to the 4th is. It is 1. So it's 1 times i, or it is just i again. So once again, it is exactly the same thing as i to the 1st power. Let's try, and just to see the pattern keep going, let's try i to the 6th power. Well, that's i times i to the 5th power."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So it's 1 times i, or it is just i again. So once again, it is exactly the same thing as i to the 1st power. Let's try, and just to see the pattern keep going, let's try i to the 6th power. Well, that's i times i to the 5th power. That's i times i to the 5th. i to the 5th, we already established, is just i. So it's i times i."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Well, that's i times i to the 5th power. That's i times i to the 5th. i to the 5th, we already established, is just i. So it's i times i. It is equal to, by definition, i times i is negative 1. And then let's finish up. Well, we could keep going this way."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So it's i times i. It is equal to, by definition, i times i is negative 1. And then let's finish up. Well, we could keep going this way. We can keep putting higher and higher powers of i here. And we'll see that it keeps cycling back. And in the next video, I'll teach you how, taking an arbitrarily high power of i, how you can figure out what that's going to be."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Well, we could keep going this way. We can keep putting higher and higher powers of i here. And we'll see that it keeps cycling back. And in the next video, I'll teach you how, taking an arbitrarily high power of i, how you can figure out what that's going to be. But let's just verify this cycle keeps going. i to the 7th power is equal to i times i to the 6th power. i to the 6th power is negative 1. i times negative 1 is negative i."}, {"video_title": "Simplifying rational expressions introduction Algebra II Khan Academy.mp3", "Sentence": "When we first started learning about fractions or rational numbers, we learned about the idea of putting things in lowest terms. So if we saw something like 3, 6, we knew that 3 and 6 share a common factor. We know that the numerator, well 3 is just 3, but that 6 could be written as 2 times 3. And since they share a common factor, the 3 in this case, we could divide the numerator by 3 and the denominator by 3, or we could say that this is just 3 over 3, and they would cancel out. And in lowest terms, this fraction would be 1 half. Or, just to kind of hit the point home, if we had 8 over 24, once again, we know that this is the same thing as 8 over 3 times 8, or this is the same thing as 1 over 3 times 8 over 8. The 8's cancel out, and we get this in lowest terms as 1 third."}, {"video_title": "Simplifying rational expressions introduction Algebra II Khan Academy.mp3", "Sentence": "And since they share a common factor, the 3 in this case, we could divide the numerator by 3 and the denominator by 3, or we could say that this is just 3 over 3, and they would cancel out. And in lowest terms, this fraction would be 1 half. Or, just to kind of hit the point home, if we had 8 over 24, once again, we know that this is the same thing as 8 over 3 times 8, or this is the same thing as 1 over 3 times 8 over 8. The 8's cancel out, and we get this in lowest terms as 1 third. The same exact idea applies to rational expressions. These are rational numbers. Rational expressions are essentially the same thing, but instead of the numerator being an actual number and the denominator being an actual number, they're expressions involving variables."}, {"video_title": "Simplifying rational expressions introduction Algebra II Khan Academy.mp3", "Sentence": "The 8's cancel out, and we get this in lowest terms as 1 third. The same exact idea applies to rational expressions. These are rational numbers. Rational expressions are essentially the same thing, but instead of the numerator being an actual number and the denominator being an actual number, they're expressions involving variables. So let me show you what I'm talking about. Let's say that I had 9x plus 3 over 12x plus 4. Now this numerator up here, we can factor it."}, {"video_title": "Simplifying rational expressions introduction Algebra II Khan Academy.mp3", "Sentence": "Rational expressions are essentially the same thing, but instead of the numerator being an actual number and the denominator being an actual number, they're expressions involving variables. So let me show you what I'm talking about. Let's say that I had 9x plus 3 over 12x plus 4. Now this numerator up here, we can factor it. We can factor out a 3. This is equal to 3 times 3x plus 1. That's what our numerator is equal to."}, {"video_title": "Simplifying rational expressions introduction Algebra II Khan Academy.mp3", "Sentence": "Now this numerator up here, we can factor it. We can factor out a 3. This is equal to 3 times 3x plus 1. That's what our numerator is equal to. And our denominator, we can factor out a 4. This is the same thing as 4 times 3x. 12 divided by 4 is 3."}, {"video_title": "Simplifying rational expressions introduction Algebra II Khan Academy.mp3", "Sentence": "That's what our numerator is equal to. And our denominator, we can factor out a 4. This is the same thing as 4 times 3x. 12 divided by 4 is 3. 12x over 4 is 3x. Plus 4 divided by 4 is 1. So here, just like there, the numerator and the denominator have a common factor."}, {"video_title": "Simplifying rational expressions introduction Algebra II Khan Academy.mp3", "Sentence": "12 divided by 4 is 3. 12x over 4 is 3x. Plus 4 divided by 4 is 1. So here, just like there, the numerator and the denominator have a common factor. In this case, it's 3x plus 1. In this case, it's a variable expression. It's not an actual number, but we can do the exact same thing."}, {"video_title": "Simplifying rational expressions introduction Algebra II Khan Academy.mp3", "Sentence": "So here, just like there, the numerator and the denominator have a common factor. In this case, it's 3x plus 1. In this case, it's a variable expression. It's not an actual number, but we can do the exact same thing. They cancel out. So if we were to write this rational expression in lowest terms, we could say that this is equal to 3 over 4. That's equal to 3 over 4."}, {"video_title": "Simplifying rational expressions introduction Algebra II Khan Academy.mp3", "Sentence": "It's not an actual number, but we can do the exact same thing. They cancel out. So if we were to write this rational expression in lowest terms, we could say that this is equal to 3 over 4. That's equal to 3 over 4. Let's do another one. Let's say that we had x squared minus 9 over 5x plus 15. What is this going to be equal to?"}, {"video_title": "Simplifying rational expressions introduction Algebra II Khan Academy.mp3", "Sentence": "That's equal to 3 over 4. Let's do another one. Let's say that we had x squared minus 9 over 5x plus 15. What is this going to be equal to? The numerator, we can factor. It's the difference of squares. We have x plus 3 times x minus 3."}, {"video_title": "Simplifying rational expressions introduction Algebra II Khan Academy.mp3", "Sentence": "What is this going to be equal to? The numerator, we can factor. It's the difference of squares. We have x plus 3 times x minus 3. In the denominator, we can factor a 5 out. This is 5 times x plus 3. Once again, common factor in the numerator and in the denominator."}, {"video_title": "Simplifying rational expressions introduction Algebra II Khan Academy.mp3", "Sentence": "We have x plus 3 times x minus 3. In the denominator, we can factor a 5 out. This is 5 times x plus 3. Once again, common factor in the numerator and in the denominator. We can cancel them out. But we touched on this a couple of videos ago. We have to be very careful."}, {"video_title": "Simplifying rational expressions introduction Algebra II Khan Academy.mp3", "Sentence": "Once again, common factor in the numerator and in the denominator. We can cancel them out. But we touched on this a couple of videos ago. We have to be very careful. We can cancel them out. We can say that this is going to be equal to x minus 3 over 5. But we have to exclude the values of x that would have made this denominator equal to 0."}, {"video_title": "Simplifying rational expressions introduction Algebra II Khan Academy.mp3", "Sentence": "We have to be very careful. We can cancel them out. We can say that this is going to be equal to x minus 3 over 5. But we have to exclude the values of x that would have made this denominator equal to 0. That would have made the entire expression undefined. We could write this as being equal to x minus 3 over 5. But x cannot be equal to negative 3."}, {"video_title": "Simplifying rational expressions introduction Algebra II Khan Academy.mp3", "Sentence": "But we have to exclude the values of x that would have made this denominator equal to 0. That would have made the entire expression undefined. We could write this as being equal to x minus 3 over 5. But x cannot be equal to negative 3. Negative 3 would make this 0. Or it would make this whole thing 0. This and this whole thing are equivalent."}, {"video_title": "Simplifying rational expressions introduction Algebra II Khan Academy.mp3", "Sentence": "But x cannot be equal to negative 3. Negative 3 would make this 0. Or it would make this whole thing 0. This and this whole thing are equivalent. This is not equivalent to this right here. Because this is defined at x is equal to negative 3. While this isn't defined at x is equal to negative 3."}, {"video_title": "Simplifying rational expressions introduction Algebra II Khan Academy.mp3", "Sentence": "This and this whole thing are equivalent. This is not equivalent to this right here. Because this is defined at x is equal to negative 3. While this isn't defined at x is equal to negative 3. To make them the same, I also have to add the extra condition that x cannot equal negative 3. Likewise, over here, if this was a function. If this was y is equal to 9x plus 3 over 12x plus 4 and we wanted to graph it."}, {"video_title": "Simplifying rational expressions introduction Algebra II Khan Academy.mp3", "Sentence": "While this isn't defined at x is equal to negative 3. To make them the same, I also have to add the extra condition that x cannot equal negative 3. Likewise, over here, if this was a function. If this was y is equal to 9x plus 3 over 12x plus 4 and we wanted to graph it. When we simplify it, the temptation is that we factored out a 3x plus 1 in the numerator and denominator. They cancel out. The temptation is to say this is the same graph as y is equal to the constant 3 fourths."}, {"video_title": "Simplifying rational expressions introduction Algebra II Khan Academy.mp3", "Sentence": "If this was y is equal to 9x plus 3 over 12x plus 4 and we wanted to graph it. When we simplify it, the temptation is that we factored out a 3x plus 1 in the numerator and denominator. They cancel out. The temptation is to say this is the same graph as y is equal to the constant 3 fourths. Which is just a horizontal line at y is equal to 3 fourths. But we have to add one condition. We have to exclude the x values that would have made this thing right here equal to 0."}, {"video_title": "Simplifying rational expressions introduction Algebra II Khan Academy.mp3", "Sentence": "The temptation is to say this is the same graph as y is equal to the constant 3 fourths. Which is just a horizontal line at y is equal to 3 fourths. But we have to add one condition. We have to exclude the x values that would have made this thing right here equal to 0. That would have been 0 if x is equal to negative 1 third. If x is equal to negative 1 third, this or this denominator would be equal to 0. Even over here, we would have to say x cannot be equal to negative 1 third."}, {"video_title": "Simplifying rational expressions introduction Algebra II Khan Academy.mp3", "Sentence": "We have to exclude the x values that would have made this thing right here equal to 0. That would have been 0 if x is equal to negative 1 third. If x is equal to negative 1 third, this or this denominator would be equal to 0. Even over here, we would have to say x cannot be equal to negative 1 third. That condition is what really makes this equal to that. That x cannot be equal to negative 1 third. Let's do a couple more of these."}, {"video_title": "Simplifying rational expressions introduction Algebra II Khan Academy.mp3", "Sentence": "Even over here, we would have to say x cannot be equal to negative 1 third. That condition is what really makes this equal to that. That x cannot be equal to negative 1 third. Let's do a couple more of these. I'll do these in pink. Let's say that I had x squared plus 6x plus 8 over x squared plus 4x. Actually, even better."}, {"video_title": "Simplifying rational expressions introduction Algebra II Khan Academy.mp3", "Sentence": "Let's do a couple more of these. I'll do these in pink. Let's say that I had x squared plus 6x plus 8 over x squared plus 4x. Actually, even better. x squared plus 6x plus 5 over x squared minus x minus 2. Once again, we want to factor the numerator and denominator. Just like we did with traditional numbers when we first learned about fractions and lowest terms."}, {"video_title": "Simplifying rational expressions introduction Algebra II Khan Academy.mp3", "Sentence": "Actually, even better. x squared plus 6x plus 5 over x squared minus x minus 2. Once again, we want to factor the numerator and denominator. Just like we did with traditional numbers when we first learned about fractions and lowest terms. If we factor the numerator, what two numbers when I multiply them equal 5 and I add them equal 6? The numbers that pop into my head are 5 and 1. The numerator is x plus 5 times x minus 1."}, {"video_title": "Simplifying rational expressions introduction Algebra II Khan Academy.mp3", "Sentence": "Just like we did with traditional numbers when we first learned about fractions and lowest terms. If we factor the numerator, what two numbers when I multiply them equal 5 and I add them equal 6? The numbers that pop into my head are 5 and 1. The numerator is x plus 5 times x minus 1. Then our denominator. Two numbers. Multiply negative 2, add them negative 1."}, {"video_title": "Simplifying rational expressions introduction Algebra II Khan Academy.mp3", "Sentence": "The numerator is x plus 5 times x minus 1. Then our denominator. Two numbers. Multiply negative 2, add them negative 1. Negative 2 and positive 1 pop out of my head. This is a positive 1. x plus 5 times x plus 1. 1 times 5 is 5."}, {"video_title": "Simplifying rational expressions introduction Algebra II Khan Academy.mp3", "Sentence": "Multiply negative 2, add them negative 1. Negative 2 and positive 1 pop out of my head. This is a positive 1. x plus 5 times x plus 1. 1 times 5 is 5. 5x plus 1x is 6x. Here we have a positive 1 and a negative 2. x minus 2 times x plus 1. We have a common factor in the numerator and the denominator."}, {"video_title": "Simplifying rational expressions introduction Algebra II Khan Academy.mp3", "Sentence": "1 times 5 is 5. 5x plus 1x is 6x. Here we have a positive 1 and a negative 2. x minus 2 times x plus 1. We have a common factor in the numerator and the denominator. These cancel out. You could say that this is equal to x plus 5 over x minus 2. For them to really be equal, we have to add the condition."}, {"video_title": "Simplifying rational expressions introduction Algebra II Khan Academy.mp3", "Sentence": "We have a common factor in the numerator and the denominator. These cancel out. You could say that this is equal to x plus 5 over x minus 2. For them to really be equal, we have to add the condition. We have to add the condition that x cannot be equal to negative 1. If x is equal to negative 1, this is undefined. We have to add that condition because this by itself is defined at x is equal to negative 1."}, {"video_title": "Simplifying rational expressions introduction Algebra II Khan Academy.mp3", "Sentence": "For them to really be equal, we have to add the condition. We have to add the condition that x cannot be equal to negative 1. If x is equal to negative 1, this is undefined. We have to add that condition because this by itself is defined at x is equal to negative 1. You could put negative 1 here and you're going to get a number. But this is not defined at x is equal to negative 1. We have to add this condition for this to truly be equal to that."}, {"video_title": "Fancier logarithm expressions Logarithms Algebra II Khan Academy.mp3", "Sentence": "What does this evaluate to? Well, it's asking us, or it will evaluate to the power that I have to, or the exponent that I have to raise our base to, that I have to raise two to, to get to eight. So two to the first power is two, two to the second power is four, two to the third power is eight. So this right over here is going to be equal to three. Fair enough, we did examples like that in the last video. Let's do something a little bit more interesting. What is, and I'll color code it, what is log base eight?"}, {"video_title": "Fancier logarithm expressions Logarithms Algebra II Khan Academy.mp3", "Sentence": "So this right over here is going to be equal to three. Fair enough, we did examples like that in the last video. Let's do something a little bit more interesting. What is, and I'll color code it, what is log base eight? What is log base eight of two? Now this is interesting. I'll give you a few seconds to think about it."}, {"video_title": "Fancier logarithm expressions Logarithms Algebra II Khan Academy.mp3", "Sentence": "What is, and I'll color code it, what is log base eight? What is log base eight of two? Now this is interesting. I'll give you a few seconds to think about it. Well, we're asking ourselves, or this will evaluate to, the exponent that I have to raise eight to, to get to two. So let's think about that in another way. So we could say eight, eight to some power, to some power, and that exponent that I'm raising eight to is essentially what this logarithm would evaluate to."}, {"video_title": "Fancier logarithm expressions Logarithms Algebra II Khan Academy.mp3", "Sentence": "I'll give you a few seconds to think about it. Well, we're asking ourselves, or this will evaluate to, the exponent that I have to raise eight to, to get to two. So let's think about that in another way. So we could say eight, eight to some power, to some power, and that exponent that I'm raising eight to is essentially what this logarithm would evaluate to. Eight to some power is going to be equal to two, is going to be equal to two. Well, if two to the third power is eight, eight to the one third power is equal to two. So x is equal to one third."}, {"video_title": "Fancier logarithm expressions Logarithms Algebra II Khan Academy.mp3", "Sentence": "So we could say eight, eight to some power, to some power, and that exponent that I'm raising eight to is essentially what this logarithm would evaluate to. Eight to some power is going to be equal to two, is going to be equal to two. Well, if two to the third power is eight, eight to the one third power is equal to two. So x is equal to one third. Eight to the one third power is equal to two. Or you could say the cube root of eight is two. So in this case, x is one third."}, {"video_title": "Fancier logarithm expressions Logarithms Algebra II Khan Academy.mp3", "Sentence": "So x is equal to one third. Eight to the one third power is equal to two. Or you could say the cube root of eight is two. So in this case, x is one third. This logarithm right over here will evaluate to, will evaluate to one third. Fascinating. Let's mix it up a little bit more."}, {"video_title": "Fancier logarithm expressions Logarithms Algebra II Khan Academy.mp3", "Sentence": "So in this case, x is one third. This logarithm right over here will evaluate to, will evaluate to one third. Fascinating. Let's mix it up a little bit more. Let's say we had the log base two. Instead of eight, let's put a one eighth, one eighth right over here. So I'll give you a few seconds to think about that."}, {"video_title": "Fancier logarithm expressions Logarithms Algebra II Khan Academy.mp3", "Sentence": "Let's mix it up a little bit more. Let's say we had the log base two. Instead of eight, let's put a one eighth, one eighth right over here. So I'll give you a few seconds to think about that. Well, it's asking us, or this will evaluate to, the exponent that I have to raise two to, to get to one eighth. So if we said that this is equal to x, we're essentially saying two to the x power, two to the x power is equal to one eighth, is equal to one eighth. Well, we know two to the third power, let me write this down, we already know that two to the third power is equal to eight."}, {"video_title": "Fancier logarithm expressions Logarithms Algebra II Khan Academy.mp3", "Sentence": "So I'll give you a few seconds to think about that. Well, it's asking us, or this will evaluate to, the exponent that I have to raise two to, to get to one eighth. So if we said that this is equal to x, we're essentially saying two to the x power, two to the x power is equal to one eighth, is equal to one eighth. Well, we know two to the third power, let me write this down, we already know that two to the third power is equal to eight. If we want to get to one eighth, which is a reciprocal of eight, we just have to raise two to the negative three power. Two to the negative three power is one over two to the third power, which is the same thing as one over eight. So, if we're asking ourselves, what exponent do we have to raise two to to get to one eighth?"}, {"video_title": "Fancier logarithm expressions Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, we know two to the third power, let me write this down, we already know that two to the third power is equal to eight. If we want to get to one eighth, which is a reciprocal of eight, we just have to raise two to the negative three power. Two to the negative three power is one over two to the third power, which is the same thing as one over eight. So, if we're asking ourselves, what exponent do we have to raise two to to get to one eighth? Well, we have to raise it to the negative three power. So x is equal to negative three. This logarithm evaluates to negative three."}, {"video_title": "Fancier logarithm expressions Logarithms Algebra II Khan Academy.mp3", "Sentence": "So, if we're asking ourselves, what exponent do we have to raise two to to get to one eighth? Well, we have to raise it to the negative three power. So x is equal to negative three. This logarithm evaluates to negative three. Now let's really, really mix it up. What would be the log, what would be the log base eight, base eight of one, one half? What does this evaluate to?"}, {"video_title": "Fancier logarithm expressions Logarithms Algebra II Khan Academy.mp3", "Sentence": "This logarithm evaluates to negative three. Now let's really, really mix it up. What would be the log, what would be the log base eight, base eight of one, one half? What does this evaluate to? Let me clean this up so that we have some space to work with. So, as always, we're saying what power do I have to raise eight to to get to one half? So let's think about that a little bit."}, {"video_title": "Fancier logarithm expressions Logarithms Algebra II Khan Academy.mp3", "Sentence": "What does this evaluate to? Let me clean this up so that we have some space to work with. So, as always, we're saying what power do I have to raise eight to to get to one half? So let's think about that a little bit. We already know that eight to the one third power is equal to two. If we want the reciprocal of two right over here, we have to just raise eight to the negative one third. So let me write that down."}, {"video_title": "Fancier logarithm expressions Logarithms Algebra II Khan Academy.mp3", "Sentence": "So let's think about that a little bit. We already know that eight to the one third power is equal to two. If we want the reciprocal of two right over here, we have to just raise eight to the negative one third. So let me write that down. Eight to the negative one third power is going to be equal to one over eight to the one third power and we already know the cube root of eight or eight to the one third power is equal to two. This is equal to one half. So the log base eight of one half is equal to, well, what power do I have to raise eight to to get to one half is negative one third."}, {"video_title": "Function inverse example 1 Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So we have f of x is equal to negative x plus 4, and it's f of x is graphed right here on our coordinate plane. Let's try to figure out what the inverse of f is. And to figure out the inverse, what I like to do is I set y, I set the variable y equal to f of x, or we could write that y is equal to negative x plus 4. Right now we've solved for y in terms of x. To solve for the inverse, we do the opposite. We solve for x in terms of y. So let's subtract 4 from both sides."}, {"video_title": "Function inverse example 1 Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Right now we've solved for y in terms of x. To solve for the inverse, we do the opposite. We solve for x in terms of y. So let's subtract 4 from both sides. You get y minus 4 is equal to negative x. And then to solve for x, we can multiply both sides of this equation times negative 1. And so you get negative y plus 4 is equal to x."}, {"video_title": "Function inverse example 1 Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So let's subtract 4 from both sides. You get y minus 4 is equal to negative x. And then to solve for x, we can multiply both sides of this equation times negative 1. And so you get negative y plus 4 is equal to x. Or just because we're always used to writing the dependent variable on the left-hand side, we could rewrite this as x is equal to negative y plus 4. Or another way to write it is we could say that f inverse of y is equal to negative y plus 4. So this is the inverse function right here, and we've written it as a function of y."}, {"video_title": "Function inverse example 1 Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And so you get negative y plus 4 is equal to x. Or just because we're always used to writing the dependent variable on the left-hand side, we could rewrite this as x is equal to negative y plus 4. Or another way to write it is we could say that f inverse of y is equal to negative y plus 4. So this is the inverse function right here, and we've written it as a function of y. But we can just rename the y as x, so it's a function of x. So let's do that. So if we just rename this y as x, we get f inverse of x is equal to negative x plus 4."}, {"video_title": "Function inverse example 1 Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So this is the inverse function right here, and we've written it as a function of y. But we can just rename the y as x, so it's a function of x. So let's do that. So if we just rename this y as x, we get f inverse of x is equal to negative x plus 4. These two functions are identical. Here we just use y as the independent variable or as the input variable. Here we just use x, but they're identical functions."}, {"video_title": "Function inverse example 1 Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So if we just rename this y as x, we get f inverse of x is equal to negative x plus 4. These two functions are identical. Here we just use y as the independent variable or as the input variable. Here we just use x, but they're identical functions. Now, just out of interest, let's graph the inverse function and see how it might relate to this one right over here. So if you look at it, it actually looks fairly identical. It's negative x plus 4."}, {"video_title": "Function inverse example 1 Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Here we just use x, but they're identical functions. Now, just out of interest, let's graph the inverse function and see how it might relate to this one right over here. So if you look at it, it actually looks fairly identical. It's negative x plus 4. It's the exact same function. So let's see. If the y-intercept is 4, it's going to be the exact same thing."}, {"video_title": "Function inverse example 1 Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "It's negative x plus 4. It's the exact same function. So let's see. If the y-intercept is 4, it's going to be the exact same thing. The function is its own inverse. So if we were to graph it, we would put it right on top of this. So there's a couple of ways to think about it."}, {"video_title": "Function inverse example 1 Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "If the y-intercept is 4, it's going to be the exact same thing. The function is its own inverse. So if we were to graph it, we would put it right on top of this. So there's a couple of ways to think about it. In the first inverse function video, I talked about how a function and their inverse, they are the reflection over the line y equals x. So where's the line y equals x here? The line y equals x looks like this."}, {"video_title": "Function inverse example 1 Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So there's a couple of ways to think about it. In the first inverse function video, I talked about how a function and their inverse, they are the reflection over the line y equals x. So where's the line y equals x here? The line y equals x looks like this. Negative x plus 4 is actually perpendicular to y is equal to x. So when you reflect it, you're just kind of flipping it over, but it's going to be the same line. It is its own reflection."}, {"video_title": "Function inverse example 1 Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "The line y equals x looks like this. Negative x plus 4 is actually perpendicular to y is equal to x. So when you reflect it, you're just kind of flipping it over, but it's going to be the same line. It is its own reflection. Let's make sure that that actually makes sense. When we're dealing with the standard function right there, if you input a 2, it gets mapped to a 2. If you input a 4, it gets mapped to 0."}, {"video_title": "Function inverse example 1 Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "It is its own reflection. Let's make sure that that actually makes sense. When we're dealing with the standard function right there, if you input a 2, it gets mapped to a 2. If you input a 4, it gets mapped to 0. What happens if you go the other way? If you input a 2, well, 2 gets mapped to 2 either way, so that makes sense. For the regular function, 4 gets mapped to 0."}, {"video_title": "Function inverse example 1 Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "If you input a 4, it gets mapped to 0. What happens if you go the other way? If you input a 2, well, 2 gets mapped to 2 either way, so that makes sense. For the regular function, 4 gets mapped to 0. For the inverse function, 0 gets mapped to 4. So it actually makes complete sense. Let's think about it another way."}, {"video_title": "Function inverse example 1 Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "For the regular function, 4 gets mapped to 0. For the inverse function, 0 gets mapped to 4. So it actually makes complete sense. Let's think about it another way. For the regular function, let me write it explicitly down. This might be obvious to you, but just in case it's not, it might be helpful. Let's pick f of 5. f of 5 is equal to negative 1."}, {"video_title": "Function inverse example 1 Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Let's think about it another way. For the regular function, let me write it explicitly down. This might be obvious to you, but just in case it's not, it might be helpful. Let's pick f of 5. f of 5 is equal to negative 1. Or we could say the function f maps us from 5 to negative 1. Now what does f inverse do? What's f inverse?"}, {"video_title": "Function inverse example 1 Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Let's pick f of 5. f of 5 is equal to negative 1. Or we could say the function f maps us from 5 to negative 1. Now what does f inverse do? What's f inverse? f inverse of negative 1. f inverse of negative 1 is 5, is equal to 5. Or we could say that f maps us from negative 1 to 5. Once again, if you think about the sets, there are domains and there are ranges."}, {"video_title": "Function inverse example 1 Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "What's f inverse? f inverse of negative 1. f inverse of negative 1 is 5, is equal to 5. Or we could say that f maps us from negative 1 to 5. Once again, if you think about the sets, there are domains and there are ranges. Let's say that this is the domain of f. This is the range of f. f will take us from 5 to negative 1. That's what the function f does. We see that f inverse takes us back from negative 1 to 5. f inverse takes us back from negative 1 to 5."}, {"video_title": "Function inverse example 1 Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Once again, if you think about the sets, there are domains and there are ranges. Let's say that this is the domain of f. This is the range of f. f will take us from 5 to negative 1. That's what the function f does. We see that f inverse takes us back from negative 1 to 5. f inverse takes us back from negative 1 to 5. Just like it's supposed to do. Let's do one more of these. Here I have g of x is equal to negative 2x minus 1."}, {"video_title": "Function inverse example 1 Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "We see that f inverse takes us back from negative 1 to 5. f inverse takes us back from negative 1 to 5. Just like it's supposed to do. Let's do one more of these. Here I have g of x is equal to negative 2x minus 1. Just like the last problem, I like to just set y equal to this. We say y is equal to g of x, which is equal to negative 2x minus 1. Now we just solve for x. y plus 1 is equal to negative 2x."}, {"video_title": "Function inverse example 1 Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Here I have g of x is equal to negative 2x minus 1. Just like the last problem, I like to just set y equal to this. We say y is equal to g of x, which is equal to negative 2x minus 1. Now we just solve for x. y plus 1 is equal to negative 2x. We just added 1 to both sides. We can divide both sides of this equation by negative 2. You get negative y over 2 minus 1 half is equal to x."}, {"video_title": "Function inverse example 1 Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Now we just solve for x. y plus 1 is equal to negative 2x. We just added 1 to both sides. We can divide both sides of this equation by negative 2. You get negative y over 2 minus 1 half is equal to x. We could write x is equal to negative y over 2 minus 1 half. Or we could write f inverse as a function of y is equal to negative y over 2 minus 1 half. Or we could just rename y as x."}, {"video_title": "Function inverse example 1 Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "You get negative y over 2 minus 1 half is equal to x. We could write x is equal to negative y over 2 minus 1 half. Or we could write f inverse as a function of y is equal to negative y over 2 minus 1 half. Or we could just rename y as x. We could say that f inverse of y is equal to negative y over 2 minus 1 half. We started with a g of x, not an f of x. Make sure we get our notation right."}, {"video_title": "Function inverse example 1 Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Or we could just rename y as x. We could say that f inverse of y is equal to negative y over 2 minus 1 half. We started with a g of x, not an f of x. Make sure we get our notation right. Or we could just rename the y and say g inverse of x is equal to negative x over 2 minus 1 half. Now let's graph it. Its y intercept is negative 1 half."}, {"video_title": "Function inverse example 1 Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Make sure we get our notation right. Or we could just rename the y and say g inverse of x is equal to negative x over 2 minus 1 half. Now let's graph it. Its y intercept is negative 1 half. It's right over there. It has a slope of negative 1 half. Let's see."}, {"video_title": "Function inverse example 1 Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Its y intercept is negative 1 half. It's right over there. It has a slope of negative 1 half. Let's see. If we start at negative 1 half, if we move over to 1 in the positive direction, we'll go down half. If we move over 1 again, we'll go down half again. If we move back, we'll go like that."}, {"video_title": "Function inverse example 1 Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Let's see. If we start at negative 1 half, if we move over to 1 in the positive direction, we'll go down half. If we move over 1 again, we'll go down half again. If we move back, we'll go like that. The line, I'll try my best to draw it, will look something like that. It'll just keep going. It'll look something like that."}, {"video_title": "Function inverse example 1 Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "If we move back, we'll go like that. The line, I'll try my best to draw it, will look something like that. It'll just keep going. It'll look something like that. It'll keep going in both directions. Let's see if this really is a reflection over y equals x. y equals x looks like that. You can see they are reflections."}, {"video_title": "Function inverse example 1 Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "It'll look something like that. It'll keep going in both directions. Let's see if this really is a reflection over y equals x. y equals x looks like that. You can see they are reflections. If you reflect this guy, if you reflect this blue line, it becomes this orange line. The general idea, you literally just, the function is originally expressed, it's solved for y in terms of x. You just do some algebra, solve for x in terms of y."}, {"video_title": "Number of solutions to linear equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "And they give us three equations right over here. And before I deal with these equations in particular, let's just remind ourselves about when we might have one or infinite or no solutions. You're going to have one solution if you can, by solving the equation, come up with something like x is equal to some number. Let's say x is equal to, if I want to say in the abstract, x is equal to a. Or if we actually were to solve it, we get something like x equals 5 or 10 or negative pi, whatever it might be. But if you could actually solve for a specific x, then you have one solution. So this is one solution, just like that."}, {"video_title": "Number of solutions to linear equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "Let's say x is equal to, if I want to say in the abstract, x is equal to a. Or if we actually were to solve it, we get something like x equals 5 or 10 or negative pi, whatever it might be. But if you could actually solve for a specific x, then you have one solution. So this is one solution, just like that. Now, if you go and you try to manipulate these equations in completely legitimate ways, but you end up with something crazy like 3 equals 5, then you have no solutions. And if you just think about it reasonably, all of these equations are about finding an x that satisfies this. And if you were to just keep simplifying it and you were to get something like 3 equals 5, and you were to ask yourself the question, is there any x that can somehow magically make 3 equal 5?"}, {"video_title": "Number of solutions to linear equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "So this is one solution, just like that. Now, if you go and you try to manipulate these equations in completely legitimate ways, but you end up with something crazy like 3 equals 5, then you have no solutions. And if you just think about it reasonably, all of these equations are about finding an x that satisfies this. And if you were to just keep simplifying it and you were to get something like 3 equals 5, and you were to ask yourself the question, is there any x that can somehow magically make 3 equal 5? No. No x can magically make 3 equal 5. So there's no way that you can make this thing be actually true, no matter which x you pick."}, {"video_title": "Number of solutions to linear equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "And if you were to just keep simplifying it and you were to get something like 3 equals 5, and you were to ask yourself the question, is there any x that can somehow magically make 3 equal 5? No. No x can magically make 3 equal 5. So there's no way that you can make this thing be actually true, no matter which x you pick. So if you get something very strange like this, this means there's no solution. On the other hand, if you get something like 5 equals 5, and I'm just overusing the number 5. It didn't have to be the number 5."}, {"video_title": "Number of solutions to linear equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "So there's no way that you can make this thing be actually true, no matter which x you pick. So if you get something very strange like this, this means there's no solution. On the other hand, if you get something like 5 equals 5, and I'm just overusing the number 5. It didn't have to be the number 5. It could be 7 or 10 or 113, whatever. And actually, let me just not use 5 just to make sure that you don't think it's only for 5. If I just get something that something is equal to itself, which is just going to be true, no matter what x you pick, any x you pick, this would be true for, well, then you have an infinite, infinite solutions."}, {"video_title": "Number of solutions to linear equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "It didn't have to be the number 5. It could be 7 or 10 or 113, whatever. And actually, let me just not use 5 just to make sure that you don't think it's only for 5. If I just get something that something is equal to itself, which is just going to be true, no matter what x you pick, any x you pick, this would be true for, well, then you have an infinite, infinite solutions. So with that as a little bit of a primer, let's try to tackle these three equations. So over here, let's see. Maybe we could subtract."}, {"video_title": "Number of solutions to linear equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "If I just get something that something is equal to itself, which is just going to be true, no matter what x you pick, any x you pick, this would be true for, well, then you have an infinite, infinite solutions. So with that as a little bit of a primer, let's try to tackle these three equations. So over here, let's see. Maybe we could subtract. If we want to get rid of this 2 here on the left-hand side, we could subtract 2 from both sides. If we subtract 2 from both sides, we are going to be left with, on the left-hand side, negative 7x. And on the right-hand side, you're going to be left with 2x."}, {"video_title": "Number of solutions to linear equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "Maybe we could subtract. If we want to get rid of this 2 here on the left-hand side, we could subtract 2 from both sides. If we subtract 2 from both sides, we are going to be left with, on the left-hand side, negative 7x. And on the right-hand side, you're going to be left with 2x. This is going to cancel minus 9x. 2x minus 9x, if we simplify that, that's negative 7x. You get negative 7x is equal to negative 7x."}, {"video_title": "Number of solutions to linear equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "And on the right-hand side, you're going to be left with 2x. This is going to cancel minus 9x. 2x minus 9x, if we simplify that, that's negative 7x. You get negative 7x is equal to negative 7x. And you probably see where this is going. This is already true for any x that you pick. Negative 7 times that x is going to be equal to negative 7 times that x."}, {"video_title": "Number of solutions to linear equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "You get negative 7x is equal to negative 7x. And you probably see where this is going. This is already true for any x that you pick. Negative 7 times that x is going to be equal to negative 7 times that x. So we already are going into this scenario. But you're like, hey, so I don't see 13 equals 13. Well, what if you did something like you divide both sides by negative 7?"}, {"video_title": "Number of solutions to linear equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "Negative 7 times that x is going to be equal to negative 7 times that x. So we already are going into this scenario. But you're like, hey, so I don't see 13 equals 13. Well, what if you did something like you divide both sides by negative 7? At this point, what I'm doing is kind of unnecessary. You already understand that negative 7 times some number is always going to be negative 7 times that number. But if we were to do this, we would get x is equal to x."}, {"video_title": "Number of solutions to linear equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "Well, what if you did something like you divide both sides by negative 7? At this point, what I'm doing is kind of unnecessary. You already understand that negative 7 times some number is always going to be negative 7 times that number. But if we were to do this, we would get x is equal to x. And then we could subtract x from both sides. And then you would get 0 equals 0, which is true for any x that you pick. 0 is always going to be equal to 0."}, {"video_title": "Number of solutions to linear equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "But if we were to do this, we would get x is equal to x. And then we could subtract x from both sides. And then you would get 0 equals 0, which is true for any x that you pick. 0 is always going to be equal to 0. So any of these statements are going to be true for any x you pick. So for this equation right over here, we have an infinite number of solutions. Let's think about this one right over here in the middle."}, {"video_title": "Number of solutions to linear equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "0 is always going to be equal to 0. So any of these statements are going to be true for any x you pick. So for this equation right over here, we have an infinite number of solutions. Let's think about this one right over here in the middle. I'll do it a little bit different. I'll add this 2x and this negative 9x right over there. So we will get negative 7x plus 3 is equal to negative 7x."}, {"video_title": "Number of solutions to linear equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "Let's think about this one right over here in the middle. I'll do it a little bit different. I'll add this 2x and this negative 9x right over there. So we will get negative 7x plus 3 is equal to negative 7x. So 2x plus 9x is negative 7x plus 2. Well, let's add. Actually, we do that in that green color."}, {"video_title": "Number of solutions to linear equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "So we will get negative 7x plus 3 is equal to negative 7x. So 2x plus 9x is negative 7x plus 2. Well, let's add. Actually, we do that in that green color. Let's do that in that green color. Plus 2. That's this 2."}, {"video_title": "Number of solutions to linear equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "Actually, we do that in that green color. Let's do that in that green color. Plus 2. That's this 2. Now let's add 7x to both sides. Well, if you add 7x to the left-hand side, you're just going to be left with a 3 there. And if you add 7x to the right-hand side, this is going to go away."}, {"video_title": "Number of solutions to linear equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "That's this 2. Now let's add 7x to both sides. Well, if you add 7x to the left-hand side, you're just going to be left with a 3 there. And if you add 7x to the right-hand side, this is going to go away. And you're just going to be left with a 2 there. So all I did is I added 7x to both sides of that equation. And now we got something nonsensical."}, {"video_title": "Number of solutions to linear equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "And if you add 7x to the right-hand side, this is going to go away. And you're just going to be left with a 2 there. So all I did is I added 7x to both sides of that equation. And now we got something nonsensical. I don't care what x you pick, how magical that x might be. There's no way that that x is going to make 3 equal to 2. So in this scenario right over here, we have no solutions."}, {"video_title": "Number of solutions to linear equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "And now we got something nonsensical. I don't care what x you pick, how magical that x might be. There's no way that that x is going to make 3 equal to 2. So in this scenario right over here, we have no solutions. There's no x in the universe that can satisfy this equation. Now let's try this third scenario. So once again, maybe we'll subtract 3 from both sides just to get rid of this constant term."}, {"video_title": "Number of solutions to linear equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "So in this scenario right over here, we have no solutions. There's no x in the universe that can satisfy this equation. Now let's try this third scenario. So once again, maybe we'll subtract 3 from both sides just to get rid of this constant term. So we're going to get negative 7x on the left-hand side. On the right-hand side, we're going to have 2x minus 1. And now we can subtract 2x from both sides."}, {"video_title": "Number of solutions to linear equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "So once again, maybe we'll subtract 3 from both sides just to get rid of this constant term. So we're going to get negative 7x on the left-hand side. On the right-hand side, we're going to have 2x minus 1. And now we can subtract 2x from both sides. So subtracting 2x, you're going to get negative 9x is equal to negative 1. Now you can divide both sides by negative 9. And you are left with x is equal to 1 ninth."}, {"video_title": "Number of solutions to linear equations Linear equations Algebra I Khan Academy.mp3", "Sentence": "And now we can subtract 2x from both sides. So subtracting 2x, you're going to get negative 9x is equal to negative 1. Now you can divide both sides by negative 9. And you are left with x is equal to 1 ninth. So we're in this scenario right over here. We very explicitly were able to find an x, x equals 1 ninth, that satisfies this equation. So this right over here has exactly one solution."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "It seems a little bit too long. But anyway, let's actually just tackle these together. So the first day of spring, an entire field of flowering trees blossoms. The population of locusts consuming these flowers rapidly increases as the trees blossom. The relationship between the elapsed time t in days since the beginning of spring and the total number of locusts, L of t, so the number of locusts is gonna be a function of the number of days that have elapsed since the beginning of spring, is modeled by the following function. So locusts as a function of time is gonna be 750 times 1.85 to the t-th power. Complete the following sentence about the daily rate of change of the locust population."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "The population of locusts consuming these flowers rapidly increases as the trees blossom. The relationship between the elapsed time t in days since the beginning of spring and the total number of locusts, L of t, so the number of locusts is gonna be a function of the number of days that have elapsed since the beginning of spring, is modeled by the following function. So locusts as a function of time is gonna be 750 times 1.85 to the t-th power. Complete the following sentence about the daily rate of change of the locust population. Every day the locust population, well, every day, think about what's going to happen. I'll draw a little table just to make it hopefully a little bit clearer. So let me draw a little bit of a table."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "Complete the following sentence about the daily rate of change of the locust population. Every day the locust population, well, every day, think about what's going to happen. I'll draw a little table just to make it hopefully a little bit clearer. So let me draw a little bit of a table. So we'll put t and L of t. So when t is zero, so when zero days have elapsed, well, this is gonna be 1.85 to the zero-th power. This is gonna be one. So you're going to have 750 locusts right from the get-go."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "So let me draw a little bit of a table. So we'll put t and L of t. So when t is zero, so when zero days have elapsed, well, this is gonna be 1.85 to the zero-th power. This is gonna be one. So you're going to have 750 locusts right from the get-go. Then when t equals one, what's going to happen? Well, then this is going to be 750 times 1.85 to the first power. So it's gonna be times 1.85."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "So you're going to have 750 locusts right from the get-go. Then when t equals one, what's going to happen? Well, then this is going to be 750 times 1.85 to the first power. So it's gonna be times 1.85. When t is equal to two, what's L of t? It's going to be 750 times 1.85 squared. Well, that's the same thing as 1.85 times 1.85."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "So it's gonna be times 1.85. When t is equal to two, what's L of t? It's going to be 750 times 1.85 squared. Well, that's the same thing as 1.85 times 1.85. So notice, and this just comes out of this being an exponential function, every day you have 1.85 times as many as you had the day before. 1.85. We essentially take what we had the day before and we multiply by 1.85."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "Well, that's the same thing as 1.85 times 1.85. So notice, and this just comes out of this being an exponential function, every day you have 1.85 times as many as you had the day before. 1.85. We essentially take what we had the day before and we multiply by 1.85. And since 1.85 is larger than one, that's going to grow the number of locusts we have. So this is going to grow. I'm actually not using, I'm not on the website right now, so that's why normally there would be a dropdown here."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "We essentially take what we had the day before and we multiply by 1.85. And since 1.85 is larger than one, that's going to grow the number of locusts we have. So this is going to grow. I'm actually not using, I'm not on the website right now, so that's why normally there would be a dropdown here. So I'm going to grow by a factor of, well, I'm going to grow by a factor of 1.85 every day. Let's do another one of these. All right, so this one tells us that Vera is an ecologist who studies the rate of change in the bear population of Siberia over time."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "I'm actually not using, I'm not on the website right now, so that's why normally there would be a dropdown here. So I'm going to grow by a factor of, well, I'm going to grow by a factor of 1.85 every day. Let's do another one of these. All right, so this one tells us that Vera is an ecologist who studies the rate of change in the bear population of Siberia over time. The relationship between the elapsed time t in years since Vera began studying the population and the total number of bears, n of t, is modeled by the following function. All right, fair enough. Got a little exponential thing going on."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "All right, so this one tells us that Vera is an ecologist who studies the rate of change in the bear population of Siberia over time. The relationship between the elapsed time t in years since Vera began studying the population and the total number of bears, n of t, is modeled by the following function. All right, fair enough. Got a little exponential thing going on. Complete the following sentence about the yearly rate of change of the bear population. What's interesting about every year that passes, t is in years now, every year that passes is going to be 2 3rds times the year before. I can do that same table that I just did just to make that clear."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "Got a little exponential thing going on. Complete the following sentence about the yearly rate of change of the bear population. What's interesting about every year that passes, t is in years now, every year that passes is going to be 2 3rds times the year before. I can do that same table that I just did just to make that clear. So let me do that. Whoops. Let me make this clear."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "I can do that same table that I just did just to make that clear. So let me do that. Whoops. Let me make this clear. So table. So this is t and this is n of t. When t is zero, n of t, you're going to have 2,187 bears. So that's the first year that she began studying that population."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "Let me make this clear. So table. So this is t and this is n of t. When t is zero, n of t, you're going to have 2,187 bears. So that's the first year that she began studying that population. Zero years since Vera began studying the population. The first year is going to be 2,187 times 2 3rds to the first power. So times 2 3rds."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "So that's the first year that she began studying that population. Zero years since Vera began studying the population. The first year is going to be 2,187 times 2 3rds to the first power. So times 2 3rds. The second year is going to be 2,187 times 2 3rds to the second power. So that's just 2 3rds times 2 3rds. So each successive year, you're going to have 2 3rds of the bear population of the year before."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "So times 2 3rds. The second year is going to be 2,187 times 2 3rds to the second power. So that's just 2 3rds times 2 3rds. So each successive year, you're going to have 2 3rds of the bear population of the year before. You're multiplying the year before by 2 3rds. So every year the bear population shrinks by a factor of 2 3rds. All right, let's do one more of these."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "So each successive year, you're going to have 2 3rds of the bear population of the year before. You're multiplying the year before by 2 3rds. So every year the bear population shrinks by a factor of 2 3rds. All right, let's do one more of these. So they tell us that Akiba started studying how the number of branches on his tree change over time. All right. The relationship between the elapsed time t in years since Akiba started studying his tree and the total number of its branches, n of t, is modeled by the following function."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "All right, let's do one more of these. So they tell us that Akiba started studying how the number of branches on his tree change over time. All right. The relationship between the elapsed time t in years since Akiba started studying his tree and the total number of its branches, n of t, is modeled by the following function. Complete the following sentence about the yearly percent change in the number of branches. Every year, blank percent of branches are added or subtracted from the total number of branches. Well, I'll draw another table, although you might get used to just being able to look at this and say, well, look, each year you're going to have 1.75 times the number of branches you had the year before."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "The relationship between the elapsed time t in years since Akiba started studying his tree and the total number of its branches, n of t, is modeled by the following function. Complete the following sentence about the yearly percent change in the number of branches. Every year, blank percent of branches are added or subtracted from the total number of branches. Well, I'll draw another table, although you might get used to just being able to look at this and say, well, look, each year you're going to have 1.75 times the number of branches you had the year before. And so if you have 1.75 times the number of branches the year before, you have grown by 75%. And I'll make that a little bit clearer. So 75% of branches, every year 75% of branches are added to the total number of branches."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "Well, I'll draw another table, although you might get used to just being able to look at this and say, well, look, each year you're going to have 1.75 times the number of branches you had the year before. And so if you have 1.75 times the number of branches the year before, you have grown by 75%. And I'll make that a little bit clearer. So 75% of branches, every year 75% of branches are added to the total number of branches. And I'll just draw that table again like I've done in the last two examples to make that hopefully clear. Okay, so this is t and this is n of t. So t equals zero, you have 42 branches. t equals one, it's going to be 42 times 1.75."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "So 75% of branches, every year 75% of branches are added to the total number of branches. And I'll just draw that table again like I've done in the last two examples to make that hopefully clear. Okay, so this is t and this is n of t. So t equals zero, you have 42 branches. t equals one, it's going to be 42 times 1.75. Times 1.75. When t equals two, it's going to be 42 times 1.75 squared. 42 times 1.75 times 1.75."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "t equals one, it's going to be 42 times 1.75. Times 1.75. When t equals two, it's going to be 42 times 1.75 squared. 42 times 1.75 times 1.75. So every year you are multiplying times 1.75. So times 1.75, something funky is happening with my pen right over there. But if you're multiplying by 1.75, if you're growing by a factor of 1.75, this is the same thing as adding 75%."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "42 times 1.75 times 1.75. So every year you are multiplying times 1.75. So times 1.75, something funky is happening with my pen right over there. But if you're multiplying by 1.75, if you're growing by a factor of 1.75, this is the same thing as adding 75%. Once again, you are adding 75%. Think about it this way. If you just grew by a factor of one, then you're not adding anything."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "But if you're multiplying by 1.75, if you're growing by a factor of 1.75, this is the same thing as adding 75%. Once again, you are adding 75%. Think about it this way. If you just grew by a factor of one, then you're not adding anything. You're staying constant. If you grow by 10%, then you're going to be 1.1 times as large. If you grow by 200%, then you're going to be two times as large."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "If you just grew by a factor of one, then you're not adding anything. You're staying constant. If you grow by 10%, then you're going to be 1.1 times as large. If you grow by 200%, then you're going to be two times as large. So this right over here, this right over here is, is, is, or if you, let me be careful what I just said. I think I just mistaken. If you grow by 200%, you're going to be three times as large as you were before."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "If you grow by 200%, then you're going to be two times as large. So this right over here, this right over here is, is, is, or if you, let me be careful what I just said. I think I just mistaken. If you grow by 200%, you're going to be three times as large as you were before. One is constant, and then another 200% would be another twofold. So that would make you three times as large. Don't want to confuse you."}, {"video_title": "Introduction to function composition Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So we have three different function definitions here. This is f of x in blue. Here we map between different values of t and what g of t would be. So you could do this as a definition of g of t. And here we map from x to h of x. So for example, when x is equal to three, h of x is equal to zero. When x is equal to one, h of x is equal to two. And actually let me number this."}, {"video_title": "Introduction to function composition Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So you could do this as a definition of g of t. And here we map from x to h of x. So for example, when x is equal to three, h of x is equal to zero. When x is equal to one, h of x is equal to two. And actually let me number this. One, two, three, just like that. Now what I want to do in this video is introduce you to the idea of composing functions. Now what does it mean to compose functions?"}, {"video_title": "Introduction to function composition Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And actually let me number this. One, two, three, just like that. Now what I want to do in this video is introduce you to the idea of composing functions. Now what does it mean to compose functions? Well, that means to build up a function by composing one function of other functions. Or I guess you can think of nesting them. What do I mean by that?"}, {"video_title": "Introduction to function composition Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Now what does it mean to compose functions? Well, that means to build up a function by composing one function of other functions. Or I guess you can think of nesting them. What do I mean by that? Well, let's think about what it means to evaluate. What does it mean to evaluate f of, not x, but we're going to evaluate f of, actually let's just start with a little warm up. Let's evaluate f of g of two."}, {"video_title": "Introduction to function composition Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "What do I mean by that? Well, let's think about what it means to evaluate. What does it mean to evaluate f of, not x, but we're going to evaluate f of, actually let's just start with a little warm up. Let's evaluate f of g of two. Now what do you think this is going to be? And I encourage you to pause this video and think about it on your own. Well, it seems kind of daunting at first if you're not very familiar with the notation, but we just have to remember what a function is."}, {"video_title": "Introduction to function composition Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Let's evaluate f of g of two. Now what do you think this is going to be? And I encourage you to pause this video and think about it on your own. Well, it seems kind of daunting at first if you're not very familiar with the notation, but we just have to remember what a function is. A function is just a mapping from one set of numbers to another. So for example, when we're saying g of two, that means take the number two, input it into the function g, and then you're going to get an output, which we are going to call g of two. G of two."}, {"video_title": "Introduction to function composition Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Well, it seems kind of daunting at first if you're not very familiar with the notation, but we just have to remember what a function is. A function is just a mapping from one set of numbers to another. So for example, when we're saying g of two, that means take the number two, input it into the function g, and then you're going to get an output, which we are going to call g of two. G of two. Now we're going to use that output, g of two, and then input it into the function f. So we're going to input it into the function f. We're going to input it into the function f, and what we're going to get is f of the thing that we inputted. F of g of two. So let's just take this step by step."}, {"video_title": "Introduction to function composition Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "G of two. Now we're going to use that output, g of two, and then input it into the function f. So we're going to input it into the function f. We're going to input it into the function f, and what we're going to get is f of the thing that we inputted. F of g of two. So let's just take this step by step. What is g of two? Well, when t is equal to two, g of two is negative three. So g of two is negative three."}, {"video_title": "Introduction to function composition Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So let's just take this step by step. What is g of two? Well, when t is equal to two, g of two is negative three. So g of two is negative three. And so when I put negative three into f, what am I going to get? Well, I'm going to get negative three squared, negative three squared minus one, which is nine minus one, which is going to be equal to eight. So this right over here is equal to eight."}, {"video_title": "Introduction to function composition Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So g of two is negative three. And so when I put negative three into f, what am I going to get? Well, I'm going to get negative three squared, negative three squared minus one, which is nine minus one, which is going to be equal to eight. So this right over here is equal to eight. F of g of two is going to be equal to eight. Now, what would, using that same exact logic, what would f of h of two be? F of h of two be?"}, {"video_title": "Introduction to function composition Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So this right over here is equal to eight. F of g of two is going to be equal to eight. Now, what would, using that same exact logic, what would f of h of two be? F of h of two be? And once again, I encourage you to pause the video and think about it on your own. Well, let's think about it this way. Instead of doing it using this little diagram, here, everywhere you see, the input is x."}, {"video_title": "Introduction to function composition Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "F of h of two be? And once again, I encourage you to pause the video and think about it on your own. Well, let's think about it this way. Instead of doing it using this little diagram, here, everywhere you see, the input is x. So whatever the input is, you square it and minus one. Here, the input is h of two. And so we're going to take the input, which is h of two, and we're going to square it, and we're going to square it, and then we're going to subtract, and we're going to subtract one."}, {"video_title": "Introduction to function composition Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Instead of doing it using this little diagram, here, everywhere you see, the input is x. So whatever the input is, you square it and minus one. Here, the input is h of two. And so we're going to take the input, which is h of two, and we're going to square it, and we're going to square it, and then we're going to subtract, and we're going to subtract one. So f of h of two is h of two squared minus one. Now, what is h of two? When x is equal to two, h of two is one."}, {"video_title": "Introduction to function composition Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And so we're going to take the input, which is h of two, and we're going to square it, and we're going to square it, and then we're going to subtract, and we're going to subtract one. So f of h of two is h of two squared minus one. Now, what is h of two? When x is equal to two, h of two is one. So h of two is one. So since h of two is equal to one, this simplifies to one squared minus one. Well, that's just going to be one minus one, which is equal to zero."}, {"video_title": "Introduction to function composition Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "When x is equal to two, h of two is one. So h of two is one. So since h of two is equal to one, this simplifies to one squared minus one. Well, that's just going to be one minus one, which is equal to zero. Now, we could have done it with the diagram way. We could have said, hey, we're going to input two into h. Input two into h. If you input two into h, you get one. So that is h of two right over here."}, {"video_title": "Introduction to function composition Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Well, that's just going to be one minus one, which is equal to zero. Now, we could have done it with the diagram way. We could have said, hey, we're going to input two into h. Input two into h. If you input two into h, you get one. So that is h of two right over here. So that is h of two. And then we're going to input that into f. And then we're going to input that into f, which is going to give us f of one. F of one is one squared minus one, which is zero."}, {"video_title": "Introduction to function composition Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So that is h of two right over here. So that is h of two. And then we're going to input that into f. And then we're going to input that into f, which is going to give us f of one. F of one is one squared minus one, which is zero. Which is zero. So this right over here is f of h of two. H of two is the input into f. So the output is going to be f of our input."}, {"video_title": "Introduction to function composition Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "F of one is one squared minus one, which is zero. Which is zero. So this right over here is f of h of two. H of two is the input into f. So the output is going to be f of our input. F of h of two. Now we can go even further. Let's do a composite."}, {"video_title": "Introduction to function composition Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "H of two is the input into f. So the output is going to be f of our input. F of h of two. Now we can go even further. Let's do a composite. Let's compose three of these. Let's compose three of these functions together. So let's take, oh, I don't know."}, {"video_title": "Introduction to function composition Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Let's do a composite. Let's compose three of these. Let's compose three of these functions together. So let's take, oh, I don't know. Let's take g of, let's take, and I'm doing this on the fly a little bit. So I hope it's a good result. G of, and let me switch the order."}, {"video_title": "Introduction to function composition Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So let's take, oh, I don't know. Let's take g of, let's take, and I'm doing this on the fly a little bit. So I hope it's a good result. G of, and let me switch the order. G of f of, g of f of, f of two. And let me just think about this for one second. So that's going to be g of f of two."}, {"video_title": "Introduction to function composition Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "G of, and let me switch the order. G of f of, g of f of, f of two. And let me just think about this for one second. So that's going to be g of f of two. And let's take h of g of f of two, just for fun. H of. So now we're really doing a triple composition."}, {"video_title": "Introduction to function composition Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So that's going to be g of f of two. And let's take h of g of f of two, just for fun. H of. So now we're really doing a triple composition. So there's a bunch of ways we could do this. One way is to just try to evaluate what is f of two. Well, f of two is going to be equal to two squared minus one."}, {"video_title": "Introduction to function composition Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So now we're really doing a triple composition. So there's a bunch of ways we could do this. One way is to just try to evaluate what is f of two. Well, f of two is going to be equal to two squared minus one. It's gonna be four minus one, or three. So this is going to be equal to three. Now what is g of three?"}, {"video_title": "Introduction to function composition Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Well, f of two is going to be equal to two squared minus one. It's gonna be four minus one, or three. So this is going to be equal to three. Now what is g of three? G of three is, when t is equal to three, g of three is four. So g of three, this whole thing, this whole thing is four. F of two is three, g of three is four."}, {"video_title": "Introduction to function composition Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Now what is g of three? G of three is, when t is equal to three, g of three is four. So g of three, this whole thing, this whole thing is four. F of two is three, g of three is four. What is h of four? Well, we can just look back to our original graph here. When x is four, h of four is negative one."}, {"video_title": "Example 5 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "we're asked to solve the quadratic equation. Negative 3x squared plus 10x minus 3 is equal to 0. And it's already written in standard form. And there's many ways to solve this, but in particular, I'll solve it using the quadratic formula. So let me just rewrite it. We have negative 3x squared plus 10x minus 3 is equal to 0. And actually, solve it twice using the quadratic formula to show you that as long as we manipulated this in the valid way, the quadratic formula will give us the exact same roots, the exact same solutions to this equation."}, {"video_title": "Example 5 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "And there's many ways to solve this, but in particular, I'll solve it using the quadratic formula. So let me just rewrite it. We have negative 3x squared plus 10x minus 3 is equal to 0. And actually, solve it twice using the quadratic formula to show you that as long as we manipulated this in the valid way, the quadratic formula will give us the exact same roots, the exact same solutions to this equation. So in this form right over here, what are our a, b, c's? Well, let's just remind ourselves what the quadratic formula even is, actually. That's a good place to start."}, {"video_title": "Example 5 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "And actually, solve it twice using the quadratic formula to show you that as long as we manipulated this in the valid way, the quadratic formula will give us the exact same roots, the exact same solutions to this equation. So in this form right over here, what are our a, b, c's? Well, let's just remind ourselves what the quadratic formula even is, actually. That's a good place to start. The quadratic formula tells us that if we have a quadratic equation in the form ax squared plus bx plus c is equal to 0, so in standard form, then the roots of this are x are equal to negative b plus or minus the square root of b squared minus 4ac, all of that over 2a. And this is derived from completing the square in a general way. So it's no magic here, and I've derived it in other videos."}, {"video_title": "Example 5 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "That's a good place to start. The quadratic formula tells us that if we have a quadratic equation in the form ax squared plus bx plus c is equal to 0, so in standard form, then the roots of this are x are equal to negative b plus or minus the square root of b squared minus 4ac, all of that over 2a. And this is derived from completing the square in a general way. So it's no magic here, and I've derived it in other videos. But this is the quadratic formula. This is actually giving you two solutions, because you have the positive square root here and the negative square root. So let's apply it here in the case where, in this case, a is equal to negative 3. a is equal to negative 3. b is equal to 10. b is equal to 10."}, {"video_title": "Example 5 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So it's no magic here, and I've derived it in other videos. But this is the quadratic formula. This is actually giving you two solutions, because you have the positive square root here and the negative square root. So let's apply it here in the case where, in this case, a is equal to negative 3. a is equal to negative 3. b is equal to 10. b is equal to 10. And c is equal to negative 3. c is equal to negative 3. So applying the quadratic formula right here, we get our solutions to be x is equal to negative b. b is 10. So negative b is 10, plus or minus the square root of b squared."}, {"video_title": "Example 5 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So let's apply it here in the case where, in this case, a is equal to negative 3. a is equal to negative 3. b is equal to 10. b is equal to 10. And c is equal to negative 3. c is equal to negative 3. So applying the quadratic formula right here, we get our solutions to be x is equal to negative b. b is 10. So negative b is 10, plus or minus the square root of b squared. b is 10, so b squared is 100. Minus 4 times a times c. So minus 4 times negative 3 times negative 3. Let me just write it down."}, {"video_title": "Example 5 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So negative b is 10, plus or minus the square root of b squared. b is 10, so b squared is 100. Minus 4 times a times c. So minus 4 times negative 3 times negative 3. Let me just write it down. Minus 4 times negative 3 times negative 3. All of that's under the radical sign. And then all of that is over 2a."}, {"video_title": "Example 5 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Let me just write it down. Minus 4 times negative 3 times negative 3. All of that's under the radical sign. And then all of that is over 2a. So 2 times a is negative 6. So this is going to be equal to negative 10, plus or minus the square root of 100 minus 3 times negative 3 times negative 3 is positive 9. Positive 9 times 4 is positive 36."}, {"video_title": "Example 5 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "And then all of that is over 2a. So 2 times a is negative 6. So this is going to be equal to negative 10, plus or minus the square root of 100 minus 3 times negative 3 times negative 3 is positive 9. Positive 9 times 4 is positive 36. We have a minus sign out here. So minus 36. All of that over negative 6."}, {"video_title": "Example 5 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Positive 9 times 4 is positive 36. We have a minus sign out here. So minus 36. All of that over negative 6. This is equal to 100 minus 36 is 64. So negative 10, plus or minus the square root of 64. All of that over negative 6."}, {"video_title": "Example 5 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "All of that over negative 6. This is equal to 100 minus 36 is 64. So negative 10, plus or minus the square root of 64. All of that over negative 6. The principal square root of 64 is 8. But we're taking the positive and negative square root. So this is negative 10, plus or minus 8 over negative 6."}, {"video_title": "Example 5 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "All of that over negative 6. The principal square root of 64 is 8. But we're taking the positive and negative square root. So this is negative 10, plus or minus 8 over negative 6. So if we take the positive version, we say x could be equal to negative 10, plus 8 is negative 2 over negative 6. So that was taking the plus version. That's this right over here."}, {"video_title": "Example 5 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So this is negative 10, plus or minus 8 over negative 6. So if we take the positive version, we say x could be equal to negative 10, plus 8 is negative 2 over negative 6. So that was taking the plus version. That's this right over here. And negative 2 over negative 6 is equal to 1 third. If we take the negative square root, negative 10 minus 8. So let's take negative 10 minus 8."}, {"video_title": "Example 5 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "That's this right over here. And negative 2 over negative 6 is equal to 1 third. If we take the negative square root, negative 10 minus 8. So let's take negative 10 minus 8. That would be x is equal to negative 10 minus 8 is negative 18. And that's going to be over negative 6. Negative 18 divided by negative 6 is positive 3."}, {"video_title": "Example 5 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So let's take negative 10 minus 8. That would be x is equal to negative 10 minus 8 is negative 18. And that's going to be over negative 6. Negative 18 divided by negative 6 is positive 3. So the two roots for this quadratic equation are positive 1 third and positive 3. And I want to show you that we'll get the same answer even if we manipulate this. Some people might not like the fact that our first coefficient here is a negative 3."}, {"video_title": "Example 5 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Negative 18 divided by negative 6 is positive 3. So the two roots for this quadratic equation are positive 1 third and positive 3. And I want to show you that we'll get the same answer even if we manipulate this. Some people might not like the fact that our first coefficient here is a negative 3. Maybe they want a positive 3. So to get rid of that negative 3, they can multiply both sides of this equation times negative 1. And then if you did that, you would get 3x squared minus 10x plus 3 is equal to 0 times negative 1, which is still equal to 0."}, {"video_title": "Example 5 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Some people might not like the fact that our first coefficient here is a negative 3. Maybe they want a positive 3. So to get rid of that negative 3, they can multiply both sides of this equation times negative 1. And then if you did that, you would get 3x squared minus 10x plus 3 is equal to 0 times negative 1, which is still equal to 0. So in this case, a is equal to 3, b is equal to negative 10, and c is equal to 3 again. And we can apply the quadratic formula. We get x is equal to negative b. b is negative 10."}, {"video_title": "Example 5 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "And then if you did that, you would get 3x squared minus 10x plus 3 is equal to 0 times negative 1, which is still equal to 0. So in this case, a is equal to 3, b is equal to negative 10, and c is equal to 3 again. And we can apply the quadratic formula. We get x is equal to negative b. b is negative 10. So negative negative 10 is positive 10 plus or minus the square root of b squared, which is negative 10 squared, which is 100, minus 4 times a times c. a times c is 9, times 4 is 36. So minus 36, all of that over 2 times a, all of that over 6. So this is equal to 10 plus or minus the square root of 64, or really that's just going to be 8, all of that over 6."}, {"video_title": "Example 5 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "We get x is equal to negative b. b is negative 10. So negative negative 10 is positive 10 plus or minus the square root of b squared, which is negative 10 squared, which is 100, minus 4 times a times c. a times c is 9, times 4 is 36. So minus 36, all of that over 2 times a, all of that over 6. So this is equal to 10 plus or minus the square root of 64, or really that's just going to be 8, all of that over 6. If we add 8 here, we get 10 plus 8 is 18 over 6. We get x could be equal to 3. Or if we take the negative square root of the negative 8 here, 10 minus 8 is 2."}, {"video_title": "Factoring difference of squares analyzing factorization High School Math Khan Academy.mp3", "Sentence": "Musa and Fatu were each asked to factor the quadratic expression 16x squared minus 64. Their responses are shown below. So Musa factored it this way, Fatu factored it this way. Which student wrote an expression that is equivalent to 16x squared minus 64? So I encourage you to pause the video and figure that out. Which student wrote an expression that is equivalent to our original one, 16x squared minus 64? Well let's work through it together."}, {"video_title": "Factoring difference of squares analyzing factorization High School Math Khan Academy.mp3", "Sentence": "Which student wrote an expression that is equivalent to 16x squared minus 64? So I encourage you to pause the video and figure that out. Which student wrote an expression that is equivalent to our original one, 16x squared minus 64? Well let's work through it together. So let's see if first we can factor this out somehow to get what Musa got. And it looks like Musa first factored out a 16, and then he was left with a difference of squares. So let's see if we can do that."}, {"video_title": "Factoring difference of squares analyzing factorization High School Math Khan Academy.mp3", "Sentence": "Well let's work through it together. So let's see if first we can factor this out somehow to get what Musa got. And it looks like Musa first factored out a 16, and then he was left with a difference of squares. So let's see if we can do that. So we can write our original expression, 16x squared minus 64, we can write that as 16 times x squared minus 16 times four. And when you write it like that, it's very clear that you can factor out a 16. So this is going to be equal to a 16 times what you have left over is x squared minus four."}, {"video_title": "Factoring difference of squares analyzing factorization High School Math Khan Academy.mp3", "Sentence": "So let's see if we can do that. So we can write our original expression, 16x squared minus 64, we can write that as 16 times x squared minus 16 times four. And when you write it like that, it's very clear that you can factor out a 16. So this is going to be equal to a 16 times what you have left over is x squared minus four. And then x squared minus four, that's the difference of squares right over there. So we can, that part we can factor as, so we have our original 16. And then this part right over here, we can write as x plus two times x minus two."}, {"video_title": "Factoring difference of squares analyzing factorization High School Math Khan Academy.mp3", "Sentence": "So this is going to be equal to a 16 times what you have left over is x squared minus four. And then x squared minus four, that's the difference of squares right over there. So we can, that part we can factor as, so we have our original 16. And then this part right over here, we can write as x plus two times x minus two. x plus two times x minus two. What I just did in this last step, going from x squared minus four to x plus two times x minus two doesn't make any sense. I encourage you to watch some of the introductory videos on factoring difference of squares."}, {"video_title": "Factoring difference of squares analyzing factorization High School Math Khan Academy.mp3", "Sentence": "And then this part right over here, we can write as x plus two times x minus two. x plus two times x minus two. What I just did in this last step, going from x squared minus four to x plus two times x minus two doesn't make any sense. I encourage you to watch some of the introductory videos on factoring difference of squares. But the basic idea, I have a form here of a squared minus b squared. So it's going to have the form of a plus b times a minus b. In this case, it's x squared minus two squared."}, {"video_title": "Factoring difference of squares analyzing factorization High School Math Khan Academy.mp3", "Sentence": "I encourage you to watch some of the introductory videos on factoring difference of squares. But the basic idea, I have a form here of a squared minus b squared. So it's going to have the form of a plus b times a minus b. In this case, it's x squared minus two squared. So it's going to be x plus two times x minus two. So that's exactly what Moussa got. So this one, so Moussa did get an expression that is equivalent to 16 x squared minus 64."}, {"video_title": "Factoring difference of squares analyzing factorization High School Math Khan Academy.mp3", "Sentence": "In this case, it's x squared minus two squared. So it's going to be x plus two times x minus two. So that's exactly what Moussa got. So this one, so Moussa did get an expression that is equivalent to 16 x squared minus 64. Now let's think about Fatou. So Fatou didn't factor out a 16 from the get-go. It looks like he just immediately recognized that what we, our original expression is itself a difference of squares, even if we don't factor out a 16."}, {"video_title": "Factoring difference of squares analyzing factorization High School Math Khan Academy.mp3", "Sentence": "So this one, so Moussa did get an expression that is equivalent to 16 x squared minus 64. Now let's think about Fatou. So Fatou didn't factor out a 16 from the get-go. It looks like he just immediately recognized that what we, our original expression is itself a difference of squares, even if we don't factor out a 16. And so let's rewrite it. So our original expression, we could write as, so we could write it, instead of writing, well, let me just write it like this. This is our original expression."}, {"video_title": "Factoring difference of squares analyzing factorization High School Math Khan Academy.mp3", "Sentence": "It looks like he just immediately recognized that what we, our original expression is itself a difference of squares, even if we don't factor out a 16. And so let's rewrite it. So our original expression, we could write as, so we could write it, instead of writing, well, let me just write it like this. This is our original expression. 16 x squared minus 64, that's the same thing as, 16 x squared is the same thing as four x, the whole thing squared, and then minus eight squared. So when you write it like this, it's clear that this is a difference of squares. So this is going to be four x plus eight times four x minus eight."}, {"video_title": "Factoring difference of squares analyzing factorization High School Math Khan Academy.mp3", "Sentence": "This is our original expression. 16 x squared minus 64, that's the same thing as, 16 x squared is the same thing as four x, the whole thing squared, and then minus eight squared. So when you write it like this, it's clear that this is a difference of squares. So this is going to be four x plus eight times four x minus eight. Four x plus eight times four x minus eight. Once again, if this last step that I did doesn't make a lot of sense, I encourage you to watch the video on factoring difference of squares. We'll go a lot more into the intuition of it."}, {"video_title": "Simplifying square-root expressions Mathematics I High School Math Khan Academy.mp3", "Sentence": "Let's get some practice simplifying radical expressions that involve variables. So let's say I have two times the square root of seven x times three times the square root of 14x squared. Pause the video and see if you can simplify it, taking any perfect squares out, multiplying, and then taking any perfect squares out of the radical sign. Well, let's first just multiply this thing. So we can change the order of multiplication. This is going to be the same thing as two times three times the square root of seven x times the square root of 14x squared. And so this is going to be equal to six times, and then the product of two radicals, you could view that as the square root of the product."}, {"video_title": "Simplifying square-root expressions Mathematics I High School Math Khan Academy.mp3", "Sentence": "Well, let's first just multiply this thing. So we can change the order of multiplication. This is going to be the same thing as two times three times the square root of seven x times the square root of 14x squared. And so this is going to be equal to six times, and then the product of two radicals, you could view that as the square root of the product. So six times the square root of, and I'll, actually I'll just leave it like this. Seven times x, and then let me actually factor 14. 14 is two times seven times x squared."}, {"video_title": "Simplifying square-root expressions Mathematics I High School Math Khan Academy.mp3", "Sentence": "And so this is going to be equal to six times, and then the product of two radicals, you could view that as the square root of the product. So six times the square root of, and I'll, actually I'll just leave it like this. Seven times x, and then let me actually factor 14. 14 is two times seven times x squared. Actually, let me, let me extend my radical sign a little bit. Alright, and the reason why I didn't multiply it out, obviously we could have multiplied in our head, x times x squared is x to the third, and we could have said, alright, seven times 14 is what, 98, we could have done that, but when you're trying to factor out perfect squares, it's actually easier if it's in this factored form here, especially because from a variable point of view, you could view this as a perfect square already, and then 14's not a perfect square, seven isn't a perfect square, but seven times seven is. Seven times seven is a perfect square, that is 49, of course."}, {"video_title": "Simplifying square-root expressions Mathematics I High School Math Khan Academy.mp3", "Sentence": "14 is two times seven times x squared. Actually, let me, let me extend my radical sign a little bit. Alright, and the reason why I didn't multiply it out, obviously we could have multiplied in our head, x times x squared is x to the third, and we could have said, alright, seven times 14 is what, 98, we could have done that, but when you're trying to factor out perfect squares, it's actually easier if it's in this factored form here, especially because from a variable point of view, you could view this as a perfect square already, and then 14's not a perfect square, seven isn't a perfect square, but seven times seven is. Seven times seven is a perfect square, that is 49, of course. So let's rewrite this a little bit to see what we can do. This is going to be six times, and I could write it like this, the square root of, and let's put all the perfect squares first. So seven times seven, that is 49, that's those two, x squared, 49 x squared, and then I could, once again, separate the two radicals right over here, so whatever else is left."}, {"video_title": "Simplifying square-root expressions Mathematics I High School Math Khan Academy.mp3", "Sentence": "Seven times seven is a perfect square, that is 49, of course. So let's rewrite this a little bit to see what we can do. This is going to be six times, and I could write it like this, the square root of, and let's put all the perfect squares first. So seven times seven, that is 49, that's those two, x squared, 49 x squared, and then I could, once again, separate the two radicals right over here, so whatever else is left. So I've already used a seven, a seven, the x squared, I have a two x left, times two x. Hopefully you appreciate that these two things are equivalent, I could have put one big radical sign over 49 x squared times two x, which would have been exactly what you have there, but if you're taking the radical of the product of things, that's the same thing as the product of the radicals, this comes straight out of our exponent properties. But what's valuable about this is we now see this is six times, now we could take the square root of 49 x squared, that's gonna be seven x, square root of 49 is seven, square root of x squared is going to be x, and then we multiply that times the square root of two x, times the square root of two x, and so now we're in the home stretch. Six times seven is 42 x times the square root of two x."}, {"video_title": "Simplifying square-root expressions Mathematics I High School Math Khan Academy.mp3", "Sentence": "So seven times seven, that is 49, that's those two, x squared, 49 x squared, and then I could, once again, separate the two radicals right over here, so whatever else is left. So I've already used a seven, a seven, the x squared, I have a two x left, times two x. Hopefully you appreciate that these two things are equivalent, I could have put one big radical sign over 49 x squared times two x, which would have been exactly what you have there, but if you're taking the radical of the product of things, that's the same thing as the product of the radicals, this comes straight out of our exponent properties. But what's valuable about this is we now see this is six times, now we could take the square root of 49 x squared, that's gonna be seven x, square root of 49 is seven, square root of x squared is going to be x, and then we multiply that times the square root of two x, times the square root of two x, and so now we're in the home stretch. Six times seven is 42 x times the square root of two x. And the key thing to appreciate is I keep using this property that a radical of products, or the square root of products, is the same thing as the product of the square roots. So even this step that I did here, if you wanted, you could have had an intermediary step. You could have said that the square root of 49 x squared is the same thing as square root of 49 times the square root of x squared, which would get us square root of 49 is seven, square root of x squared is x right over there."}, {"video_title": "Simplifying square-root expressions Mathematics I High School Math Khan Academy.mp3", "Sentence": "Six times seven is 42 x times the square root of two x. And the key thing to appreciate is I keep using this property that a radical of products, or the square root of products, is the same thing as the product of the square roots. So even this step that I did here, if you wanted, you could have had an intermediary step. You could have said that the square root of 49 x squared is the same thing as square root of 49 times the square root of x squared, which would get us square root of 49 is seven, square root of x squared is x right over there. Let's do another one of these. So let's say I have square root of two a times the square root of 14 a to the third times the square root of five a. So like always, pause this video and see if you can simplify this on your own."}, {"video_title": "Simplifying square-root expressions Mathematics I High School Math Khan Academy.mp3", "Sentence": "You could have said that the square root of 49 x squared is the same thing as square root of 49 times the square root of x squared, which would get us square root of 49 is seven, square root of x squared is x right over there. Let's do another one of these. So let's say I have square root of two a times the square root of 14 a to the third times the square root of five a. So like always, pause this video and see if you can simplify this on your own. Multiply them and then take all the perfect squares out of the radical. So let's multiply first. So this is gonna be the same thing as the square root of, let's see, two times 14 times five."}, {"video_title": "Simplifying square-root expressions Mathematics I High School Math Khan Academy.mp3", "Sentence": "So like always, pause this video and see if you can simplify this on your own. Multiply them and then take all the perfect squares out of the radical. So let's multiply first. So this is gonna be the same thing as the square root of, let's see, two times 14 times five. So let me actually just, I'm just gonna, two and five are prime. 14, I can factor it as two times seven. So this is gonna be two times, instead of 14, I'm gonna write two times seven and then times five."}, {"video_title": "Simplifying square-root expressions Mathematics I High School Math Khan Academy.mp3", "Sentence": "So this is gonna be the same thing as the square root of, let's see, two times 14 times five. So let me actually just, I'm just gonna, two and five are prime. 14, I can factor it as two times seven. So this is gonna be two times, instead of 14, I'm gonna write two times seven and then times five. And then we have a times a to the third times a. Well, actually, let me just write that as a to the fifth. We have a to the first times a to the third times a to the first, add the exponents, you get a to the fifth."}, {"video_title": "Simplifying square-root expressions Mathematics I High School Math Khan Academy.mp3", "Sentence": "So this is gonna be two times, instead of 14, I'm gonna write two times seven and then times five. And then we have a times a to the third times a. Well, actually, let me just write that as a to the fifth. We have a to the first times a to the third times a to the first, add the exponents, you get a to the fifth. Now, what perfect squares do we have here? Well, we already see a perfect square in terms of two times two. And then a to the fifth isn't a perfect square if you think in terms of the variable a, but you could view that as a perfect square, a to the fourth times a."}, {"video_title": "Simplifying square-root expressions Mathematics I High School Math Khan Academy.mp3", "Sentence": "We have a to the first times a to the third times a to the first, add the exponents, you get a to the fifth. Now, what perfect squares do we have here? Well, we already see a perfect square in terms of two times two. And then a to the fifth isn't a perfect square if you think in terms of the variable a, but you could view that as a perfect square, a to the fourth times a. So let's rearrange this a little bit. And so this is going to be equal to the square root of, let me put my perfect squares out front. The square root of four, two times two, times a to the fourth."}, {"video_title": "Simplifying square-root expressions Mathematics I High School Math Khan Academy.mp3", "Sentence": "And then a to the fifth isn't a perfect square if you think in terms of the variable a, but you could view that as a perfect square, a to the fourth times a. So let's rearrange this a little bit. And so this is going to be equal to the square root of, let me put my perfect squares out front. The square root of four, two times two, times a to the fourth. And then let me put my non-perfect squares. Times, I have a seven, a five, and an a that I haven't used yet. So seven times five is 35, so it's 35a."}, {"video_title": "Simplifying square-root expressions Mathematics I High School Math Khan Academy.mp3", "Sentence": "The square root of four, two times two, times a to the fourth. And then let me put my non-perfect squares. Times, I have a seven, a five, and an a that I haven't used yet. So seven times five is 35, so it's 35a. And now, just like we said before, we could, well, let me do it. We could say, hey, look, this is the same thing as the square root of four times the square root of a to the fourth, just using exponent properties, and then times the square root of 35a. Now, principal root of four is positive two."}, {"video_title": "Simplifying square-root expressions Mathematics I High School Math Khan Academy.mp3", "Sentence": "So seven times five is 35, so it's 35a. And now, just like we said before, we could, well, let me do it. We could say, hey, look, this is the same thing as the square root of four times the square root of a to the fourth, just using exponent properties, and then times the square root of 35a. Now, principal root of four is positive two. You could view this as a positive square root. And then square root of a to the fourth, the principal root is gonna be a squared. And then we're going to have that times the square root of 35a."}, {"video_title": "Simplifying square-root expressions Mathematics I High School Math Khan Academy.mp3", "Sentence": "Now, principal root of four is positive two. You could view this as a positive square root. And then square root of a to the fourth, the principal root is gonna be a squared. And then we're going to have that times the square root of 35a. And we're done. Let's do one more example. And this time, we're gonna involve two variables, which, as we'll see, isn't that much more complicated."}, {"video_title": "Simplifying square-root expressions Mathematics I High School Math Khan Academy.mp3", "Sentence": "And then we're going to have that times the square root of 35a. And we're done. Let's do one more example. And this time, we're gonna involve two variables, which, as we'll see, isn't that much more complicated. So let's simplify the square root of 72x to the third, z to the third. So the key is, can we factor, 72 is not a perfect square, but is there a perfect square someplace in there? And you immediately see that if you try to factor it."}, {"video_title": "Simplifying square-root expressions Mathematics I High School Math Khan Academy.mp3", "Sentence": "And this time, we're gonna involve two variables, which, as we'll see, isn't that much more complicated. So let's simplify the square root of 72x to the third, z to the third. So the key is, can we factor, 72 is not a perfect square, but is there a perfect square someplace in there? And you immediately see that if you try to factor it. You get 36 times two, and 36, of course, is a perfect square. And likewise, x to the third and z to the third are not each perfect squares, but they each have an x squared and z squared in them. So let me rewrite this."}, {"video_title": "Simplifying square-root expressions Mathematics I High School Math Khan Academy.mp3", "Sentence": "And you immediately see that if you try to factor it. You get 36 times two, and 36, of course, is a perfect square. And likewise, x to the third and z to the third are not each perfect squares, but they each have an x squared and z squared in them. So let me rewrite this. This is the same thing. This is the same thing as, I can write, well, let me put all my perfect squares out front. So I have 36, 36."}, {"video_title": "Simplifying square-root expressions Mathematics I High School Math Khan Academy.mp3", "Sentence": "So let me rewrite this. This is the same thing. This is the same thing as, I can write, well, let me put all my perfect squares out front. So I have 36, 36. I'm gonna take an x squared out. x squared, I'm gonna take a z squared out, z squared. And then what we're left with is, we took a 36 out, so we're left with a two times two."}, {"video_title": "Simplifying square-root expressions Mathematics I High School Math Khan Academy.mp3", "Sentence": "So I have 36, 36. I'm gonna take an x squared out. x squared, I'm gonna take a z squared out, z squared. And then what we're left with is, we took a 36 out, so we're left with a two times two. And if we took an x squared out of this, we're left with just an x. X to the third divided by x squared is x. Two x, and then z to the third divided by z squared is just z. And you can verify this."}, {"video_title": "Simplifying square-root expressions Mathematics I High School Math Khan Academy.mp3", "Sentence": "And then what we're left with is, we took a 36 out, so we're left with a two times two. And if we took an x squared out of this, we're left with just an x. X to the third divided by x squared is x. Two x, and then z to the third divided by z squared is just z. And you can verify this. Multiply this all out. You should be getting, you should be getting exactly what we have here. I drew that little line on the z to differentiate so it doesn't look like my twos."}, {"video_title": "Simplifying square-root expressions Mathematics I High School Math Khan Academy.mp3", "Sentence": "And you can verify this. Multiply this all out. You should be getting, you should be getting exactly what we have here. I drew that little line on the z to differentiate so it doesn't look like my twos. 36 times two is 72. X squared times x is x to the third. Z squared times z is z to the third."}, {"video_title": "Simplifying square-root expressions Mathematics I High School Math Khan Academy.mp3", "Sentence": "I drew that little line on the z to differentiate so it doesn't look like my twos. 36 times two is 72. X squared times x is x to the third. Z squared times z is z to the third. But now this is pretty straightforward to factor because let me just, I'll do more steps than you would probably do if you're doing it on your own. But that's because the whole point here is to learn. So two x z."}, {"video_title": "Simplifying square-root expressions Mathematics I High School Math Khan Academy.mp3", "Sentence": "Z squared times z is z to the third. But now this is pretty straightforward to factor because let me just, I'll do more steps than you would probably do if you're doing it on your own. But that's because the whole point here is to learn. So two x z. So that's just using exponent properties. And so everything here is a perfect square. This is gonna be the square root of 36 times the square root of x squared times the square root of z squared."}, {"video_title": "Simplifying square-root expressions Mathematics I High School Math Khan Academy.mp3", "Sentence": "So two x z. So that's just using exponent properties. And so everything here is a perfect square. This is gonna be the square root of 36 times the square root of x squared times the square root of z squared. Which is going to be square root of 36 is, principal root of 36 is six. Principal root of x squared is x. Principal root of z squared is z."}, {"video_title": "Factoring quadratics as (x+a)(x+b) (example 2) Mathematics II High School Math Khan Academy.mp3", "Sentence": "To better understand how we can factor second degree expressions like this, I'm going to go through some examples. We'll factor this expression, and we'll factor this expression. And hopefully, it'll give you a background on how you could generally factor expressions like this. And to think about it, let's think about what happens if I were to multiply x plus something times x plus something else. Well, if I were to multiply this out, what do I get? Well, you're going to get x squared plus ax plus bx, which is the same thing as a plus bx, plus a times b, plus ab. So if you wanted to go from this form, which is what we have in these two examples, back to this, you really just have to think about, well, what's our coefficient on our x term?"}, {"video_title": "Factoring quadratics as (x+a)(x+b) (example 2) Mathematics II High School Math Khan Academy.mp3", "Sentence": "And to think about it, let's think about what happens if I were to multiply x plus something times x plus something else. Well, if I were to multiply this out, what do I get? Well, you're going to get x squared plus ax plus bx, which is the same thing as a plus bx, plus a times b, plus ab. So if you wanted to go from this form, which is what we have in these two examples, back to this, you really just have to think about, well, what's our coefficient on our x term? And can I figure out two numbers that when I take their sum are equal to that coefficient? And what's my constant term? And can I think of two numbers, those same two numbers, that when I take the product equal that constant term?"}, {"video_title": "Factoring quadratics as (x+a)(x+b) (example 2) Mathematics II High School Math Khan Academy.mp3", "Sentence": "So if you wanted to go from this form, which is what we have in these two examples, back to this, you really just have to think about, well, what's our coefficient on our x term? And can I figure out two numbers that when I take their sum are equal to that coefficient? And what's my constant term? And can I think of two numbers, those same two numbers, that when I take the product equal that constant term? So let's do that over here. If we look at our x coefficient, the coefficient on x, can we think of an a plus ab that is equal to that number, negative 14? And can we think of the same a and b that if we were to take its product, it would be equal to 40?"}, {"video_title": "Factoring quadratics as (x+a)(x+b) (example 2) Mathematics II High School Math Khan Academy.mp3", "Sentence": "And can I think of two numbers, those same two numbers, that when I take the product equal that constant term? So let's do that over here. If we look at our x coefficient, the coefficient on x, can we think of an a plus ab that is equal to that number, negative 14? And can we think of the same a and b that if we were to take its product, it would be equal to 40? It would be equal to 40. So what's an a and a b that would work over here? Well, let's think about this a little bit."}, {"video_title": "Factoring quadratics as (x+a)(x+b) (example 2) Mathematics II High School Math Khan Academy.mp3", "Sentence": "And can we think of the same a and b that if we were to take its product, it would be equal to 40? It would be equal to 40. So what's an a and a b that would work over here? Well, let's think about this a little bit. If I have 4 times 10 is 40, but 4 plus 10 is equal to positive 14. So that wouldn't quite work. But what happens if we make them both negative?"}, {"video_title": "Factoring quadratics as (x+a)(x+b) (example 2) Mathematics II High School Math Khan Academy.mp3", "Sentence": "Well, let's think about this a little bit. If I have 4 times 10 is 40, but 4 plus 10 is equal to positive 14. So that wouldn't quite work. But what happens if we make them both negative? If we have negative 4 plus negative 10, well, that's going to be equal to negative 14. And negative 4 times negative 10 is equal to 40. The fact that this number right over here is positive tells you that these are going to be the same sign."}, {"video_title": "Factoring quadratics as (x+a)(x+b) (example 2) Mathematics II High School Math Khan Academy.mp3", "Sentence": "But what happens if we make them both negative? If we have negative 4 plus negative 10, well, that's going to be equal to negative 14. And negative 4 times negative 10 is equal to 40. The fact that this number right over here is positive tells you that these are going to be the same sign. So these are going to be the same sign. These are going to have the exact same sign. If this number right over here was negative, then we would have different signs."}, {"video_title": "Factoring quadratics as (x+a)(x+b) (example 2) Mathematics II High School Math Khan Academy.mp3", "Sentence": "The fact that this number right over here is positive tells you that these are going to be the same sign. So these are going to be the same sign. These are going to have the exact same sign. If this number right over here was negative, then we would have different signs. And so if you have two numbers that are going to be the same sign and they add up to a negative number, then that tells you that they're both going to be negative. So just going back to this, we know that a is going to be negative 4, b is equal to negative 10, and we are done factoring it. We can factor this expression as x plus negative 4 times x plus negative 10."}, {"video_title": "Factoring quadratics as (x+a)(x+b) (example 2) Mathematics II High School Math Khan Academy.mp3", "Sentence": "If this number right over here was negative, then we would have different signs. And so if you have two numbers that are going to be the same sign and they add up to a negative number, then that tells you that they're both going to be negative. So just going back to this, we know that a is going to be negative 4, b is equal to negative 10, and we are done factoring it. We can factor this expression as x plus negative 4 times x plus negative 10. Or another way to write that, that's x minus 4 times x minus 10. Now let's do the same thing over here. Can we think of an a plus b that's equal to the coefficient on the x term?"}, {"video_title": "Factoring quadratics as (x+a)(x+b) (example 2) Mathematics II High School Math Khan Academy.mp3", "Sentence": "We can factor this expression as x plus negative 4 times x plus negative 10. Or another way to write that, that's x minus 4 times x minus 10. Now let's do the same thing over here. Can we think of an a plus b that's equal to the coefficient on the x term? Well, the coefficient on the x term here, this is essentially negative 1 times x. So we could say the coefficient is negative 1. And can we think of an a times b where it's going to be equal to negative 12?"}, {"video_title": "Factoring quadratics as (x+a)(x+b) (example 2) Mathematics II High School Math Khan Academy.mp3", "Sentence": "Can we think of an a plus b that's equal to the coefficient on the x term? Well, the coefficient on the x term here, this is essentially negative 1 times x. So we could say the coefficient is negative 1. And can we think of an a times b where it's going to be equal to negative 12? Well, let's think about this a little bit. The product of the two numbers is negative. So that means that they have different signs."}, {"video_title": "Factoring quadratics as (x+a)(x+b) (example 2) Mathematics II High School Math Khan Academy.mp3", "Sentence": "And can we think of an a times b where it's going to be equal to negative 12? Well, let's think about this a little bit. The product of the two numbers is negative. So that means that they have different signs. So one will be positive and one will be negative. And so when I add them two together, I get to negative 1. Well, let's just think about the factors of negative 12."}, {"video_title": "Factoring quadratics as (x+a)(x+b) (example 2) Mathematics II High School Math Khan Academy.mp3", "Sentence": "So that means that they have different signs. So one will be positive and one will be negative. And so when I add them two together, I get to negative 1. Well, let's just think about the factors of negative 12. Well, what about if 1 is 3 and maybe 1 is negative 4? Well, that seems to work. And you really just have to try these numbers out."}, {"video_title": "Factoring quadratics as (x+a)(x+b) (example 2) Mathematics II High School Math Khan Academy.mp3", "Sentence": "Well, let's just think about the factors of negative 12. Well, what about if 1 is 3 and maybe 1 is negative 4? Well, that seems to work. And you really just have to try these numbers out. If a is 3, so if we're 3 plus negative 4, that indeed turns out to be negative 1. And if we have 3 times negative 4, that indeed is equal to negative 12. So that seems to work out."}, {"video_title": "Factoring quadratics as (x+a)(x+b) (example 2) Mathematics II High School Math Khan Academy.mp3", "Sentence": "And you really just have to try these numbers out. If a is 3, so if we're 3 plus negative 4, that indeed turns out to be negative 1. And if we have 3 times negative 4, that indeed is equal to negative 12. So that seems to work out. And it's really a matter of trial and error. You could try negative 3 plus 4, but then that wouldn't have worked out over here. You could have tried 2 and 6, but that wouldn't have worked out on this number."}, {"video_title": "Factoring quadratics as (x+a)(x+b) (example 2) Mathematics II High School Math Khan Academy.mp3", "Sentence": "So that seems to work out. And it's really a matter of trial and error. You could try negative 3 plus 4, but then that wouldn't have worked out over here. You could have tried 2 and 6, but that wouldn't have worked out on this number. Or 2 and negative 6, you wouldn't have gotten the sum to be equal to negative 1. But now that we've figured out what the a and b are, what is this expression factored? Well, it's going to be x plus 3 times x plus negative 4, or we could say x minus 4."}, {"video_title": "Introduction to intercepts Algebra I Khan Academy.mp3", "Sentence": "So let's set up a little table here, x comma y. And then we could just try out a couple of x values here and then figure out what the corresponding y values are. And I'm gonna pick x values where it's gonna be fairly easy to calculate the y values. So let's say when x is equal to zero, then you're gonna have 1 half times zero minus three, well then y is going to be negative three. When x is, let's see, let me try x is equal to two because then 1 half times two is just gonna be one. So when x is equal to two, you're gonna have 1 half times two is one minus three is negative two. When x is equal to, let's try four."}, {"video_title": "Introduction to intercepts Algebra I Khan Academy.mp3", "Sentence": "So let's say when x is equal to zero, then you're gonna have 1 half times zero minus three, well then y is going to be negative three. When x is, let's see, let me try x is equal to two because then 1 half times two is just gonna be one. So when x is equal to two, you're gonna have 1 half times two is one minus three is negative two. When x is equal to, let's try four. So 1 half times four is two and then minus three is negative one. And we could keep going, but actually all we do need, all we need is two points for a line. So we're ready to plot this line if we like."}, {"video_title": "Introduction to intercepts Algebra I Khan Academy.mp3", "Sentence": "When x is equal to, let's try four. So 1 half times four is two and then minus three is negative one. And we could keep going, but actually all we do need, all we need is two points for a line. So we're ready to plot this line if we like. The point zero comma negative three is on this line. Zero comma negative three. Actually let me do this in a slightly darker color so we can see it on this white background."}, {"video_title": "Introduction to intercepts Algebra I Khan Academy.mp3", "Sentence": "So we're ready to plot this line if we like. The point zero comma negative three is on this line. Zero comma negative three. Actually let me do this in a slightly darker color so we can see it on this white background. Zero comma negative three is on the line. Two comma negative two is on the line. So two comma negative two."}, {"video_title": "Introduction to intercepts Algebra I Khan Academy.mp3", "Sentence": "Actually let me do this in a slightly darker color so we can see it on this white background. Zero comma negative three is on the line. Two comma negative two is on the line. So two comma negative two. And then we have four comma negative one. So when x is four, y is negative one. And I could draw a line that connects all of these."}, {"video_title": "Introduction to intercepts Algebra I Khan Academy.mp3", "Sentence": "So two comma negative two. And then we have four comma negative one. So when x is four, y is negative one. And I could draw a line that connects all of these. So it would look something like, let's see if I can do this. It would look something like, it would look something like, like that. So this right over here, this is literally, this is the graph of y is equal to 1 half x minus three."}, {"video_title": "Introduction to intercepts Algebra I Khan Academy.mp3", "Sentence": "And I could draw a line that connects all of these. So it would look something like, let's see if I can do this. It would look something like, it would look something like, like that. So this right over here, this is literally, this is the graph of y is equal to 1 half x minus three. Now when we look at a graph like this, an interesting thing that we might wanna ask ourselves is where does the graph intersect our axes? So first we could say, well where does it intersect our x-axis? And when you look at this, it looks like it happens at this point right over here."}, {"video_title": "Introduction to intercepts Algebra I Khan Academy.mp3", "Sentence": "So this right over here, this is literally, this is the graph of y is equal to 1 half x minus three. Now when we look at a graph like this, an interesting thing that we might wanna ask ourselves is where does the graph intersect our axes? So first we could say, well where does it intersect our x-axis? And when you look at this, it looks like it happens at this point right over here. And this point where a graph intersects an axis, this is called an intercept. And this one in particular is called the x-intercept. Why is it called the x-intercept?"}, {"video_title": "Introduction to intercepts Algebra I Khan Academy.mp3", "Sentence": "And when you look at this, it looks like it happens at this point right over here. And this point where a graph intersects an axis, this is called an intercept. And this one in particular is called the x-intercept. Why is it called the x-intercept? Because that's where the graph is intersecting the x-axis. And the x-intercept, it looks like, this is at the point six comma zero. Now it's very interesting."}, {"video_title": "Introduction to intercepts Algebra I Khan Academy.mp3", "Sentence": "Why is it called the x-intercept? Because that's where the graph is intersecting the x-axis. And the x-intercept, it looks like, this is at the point six comma zero. Now it's very interesting. The x-intercept happens when y is equal to zero. Remember, you're on the x-axis when you haven't moved up or down from that axis, which means y is equal to zero. So your x-intercept happens at x equals six, y equals zero, it's this coordinate."}, {"video_title": "Introduction to intercepts Algebra I Khan Academy.mp3", "Sentence": "Now it's very interesting. The x-intercept happens when y is equal to zero. Remember, you're on the x-axis when you haven't moved up or down from that axis, which means y is equal to zero. So your x-intercept happens at x equals six, y equals zero, it's this coordinate. Now what about the y-intercept? Well the y-intercept is this point right over here. This is where you intersect, or I guess you could say, intercept the y-axis."}, {"video_title": "Introduction to intercepts Algebra I Khan Academy.mp3", "Sentence": "So your x-intercept happens at x equals six, y equals zero, it's this coordinate. Now what about the y-intercept? Well the y-intercept is this point right over here. This is where you intersect, or I guess you could say, intercept the y-axis. So this right over here, that over there is the y-intercept. And the y-intercept is at the coordinate that has a zero for the x-coordinate. X is zero here and y is negative three."}, {"video_title": "Introduction to intercepts Algebra I Khan Academy.mp3", "Sentence": "This is where you intersect, or I guess you could say, intercept the y-axis. So this right over here, that over there is the y-intercept. And the y-intercept is at the coordinate that has a zero for the x-coordinate. X is zero here and y is negative three. X is zero and y is negative three. This was actually one of the points, or one of the pairs that we first tried out. And you can validate that six comma zero satisfies this equation right over here."}, {"video_title": "Introduction to intercepts Algebra I Khan Academy.mp3", "Sentence": "X is zero here and y is negative three. X is zero and y is negative three. This was actually one of the points, or one of the pairs that we first tried out. And you can validate that six comma zero satisfies this equation right over here. If x is six, one half times six is three, minus three is indeed equal to zero. So now that we know what an x-intercept is, it's the point where a graph intersects the x-axis, or intercepts the x-axis, and the y-intercept is the point where a graph intercepts the y-axis, or intersects the y-axis, let's try to see if we can find the x and y-intercepts for a few other linear equations. So let's say that I have the linear equation."}, {"video_title": "Introduction to intercepts Algebra I Khan Academy.mp3", "Sentence": "And you can validate that six comma zero satisfies this equation right over here. If x is six, one half times six is three, minus three is indeed equal to zero. So now that we know what an x-intercept is, it's the point where a graph intersects the x-axis, or intercepts the x-axis, and the y-intercept is the point where a graph intercepts the y-axis, or intersects the y-axis, let's try to see if we can find the x and y-intercepts for a few other linear equations. So let's say that I have the linear equation. Let's say that I have five x plus six y is equal to 30. I encourage you to pause this video and figure out what are the x and y-intercepts for the graph that represents the solutions, all the xy pairs that satisfy this equation. Well, the easiest thing to do here, let's see what the y value is when x is equal to zero, and what the x value is when y is equal to zero."}, {"video_title": "Introduction to intercepts Algebra I Khan Academy.mp3", "Sentence": "So let's say that I have the linear equation. Let's say that I have five x plus six y is equal to 30. I encourage you to pause this video and figure out what are the x and y-intercepts for the graph that represents the solutions, all the xy pairs that satisfy this equation. Well, the easiest thing to do here, let's see what the y value is when x is equal to zero, and what the x value is when y is equal to zero. When x is equal to zero, this becomes six y equal 30, and so six times what is 30? Well, y would be equal to five here. So when x is zero, y is five."}, {"video_title": "Introduction to intercepts Algebra I Khan Academy.mp3", "Sentence": "Well, the easiest thing to do here, let's see what the y value is when x is equal to zero, and what the x value is when y is equal to zero. When x is equal to zero, this becomes six y equal 30, and so six times what is 30? Well, y would be equal to five here. So when x is zero, y is five. And what about when y is zero? Well, when y is zero, that's gonna be zero, and you have five x is equal to 30. Well, then x would be equal to six."}, {"video_title": "Introduction to intercepts Algebra I Khan Academy.mp3", "Sentence": "So when x is zero, y is five. And what about when y is zero? Well, when y is zero, that's gonna be zero, and you have five x is equal to 30. Well, then x would be equal to six. Then x would be equal to six. So we could plot those points, zero comma five. When x is zero, y is five."}, {"video_title": "Introduction to intercepts Algebra I Khan Academy.mp3", "Sentence": "Well, then x would be equal to six. Then x would be equal to six. So we could plot those points, zero comma five. When x is zero, y is five. And when x is six, y is zero. So those are both points on this graph. And then the actual graph is going to, or the actual line that represents the x and y pairs that satisfy this equation, is going to look like, it's going to look like this."}, {"video_title": "Introduction to intercepts Algebra I Khan Academy.mp3", "Sentence": "When x is zero, y is five. And when x is six, y is zero. So those are both points on this graph. And then the actual graph is going to, or the actual line that represents the x and y pairs that satisfy this equation, is going to look like, it's going to look like this. I'll just try, so I can make it go. It's going to look like it's gonna go through those two points. And so it's gonna, I can make it go the other way too."}, {"video_title": "Introduction to intercepts Algebra I Khan Academy.mp3", "Sentence": "And then the actual graph is going to, or the actual line that represents the x and y pairs that satisfy this equation, is going to look like, it's going to look like this. I'll just try, so I can make it go. It's going to look like it's gonna go through those two points. And so it's gonna, I can make it go the other way too. Let me see. It's gonna go through those two points. So it's gonna look something like that."}, {"video_title": "Introduction to intercepts Algebra I Khan Academy.mp3", "Sentence": "And so it's gonna, I can make it go the other way too. Let me see. It's gonna go through those two points. So it's gonna look something like that. Now what are its x and y intercepts? Well, we already kind of figured it out, but the intercepts themselves, these are the points on the graph where they intersect the axes. So this right over here, this is the y-intercept."}, {"video_title": "Introduction to intercepts Algebra I Khan Academy.mp3", "Sentence": "So it's gonna look something like that. Now what are its x and y intercepts? Well, we already kind of figured it out, but the intercepts themselves, these are the points on the graph where they intersect the axes. So this right over here, this is the y-intercept. That point is the y-intercept, and it happens, it's always going to happen when x is equal to zero, and when x is equal to zero, we know that y is equal to five. It's that point, the point zero comma five. And what is the y, so what is the x-intercept?"}, {"video_title": "Introduction to intercepts Algebra I Khan Academy.mp3", "Sentence": "So this right over here, this is the y-intercept. That point is the y-intercept, and it happens, it's always going to happen when x is equal to zero, and when x is equal to zero, we know that y is equal to five. It's that point, the point zero comma five. And what is the y, so what is the x-intercept? The x-intercept is the point, it's actually the same x-intercept for that other, for this equation right over here. It's the point six comma zero. That point right over there."}, {"video_title": "Identifying perfect square form Mathematics II High School Math Khan Academy.mp3", "Sentence": "So let's work through it. This is the same thing as multiplying AX plus B times AX plus B. So let me fill that in. This is AX there, another AX there. I just wrote it in that order to make the color switching a little bit easier. So AX plus B times AX plus B. Well, what's that going to be equal to?"}, {"video_title": "Identifying perfect square form Mathematics II High School Math Khan Academy.mp3", "Sentence": "This is AX there, another AX there. I just wrote it in that order to make the color switching a little bit easier. So AX plus B times AX plus B. Well, what's that going to be equal to? Well, if you take this AX and you multiply it times that AX, you're going to get AX squared, AX, the entire thing squared. And then if you take this AX and then multiply it times this B, you're going to get ABX. Then if you take this B and you multiply it times this AX, you're going to get another ABX."}, {"video_title": "Identifying perfect square form Mathematics II High School Math Khan Academy.mp3", "Sentence": "Well, what's that going to be equal to? Well, if you take this AX and you multiply it times that AX, you're going to get AX squared, AX, the entire thing squared. And then if you take this AX and then multiply it times this B, you're going to get ABX. Then if you take this B and you multiply it times this AX, you're going to get another ABX. A, B, X. And then last but not least, if you take this B and multiply it times the other B, it's going to be plus B squared. And so what are you left with?"}, {"video_title": "Identifying perfect square form Mathematics II High School Math Khan Academy.mp3", "Sentence": "Then if you take this B and you multiply it times this AX, you're going to get another ABX. A, B, X. And then last but not least, if you take this B and multiply it times the other B, it's going to be plus B squared. And so what are you left with? Well, you're going to be left with A, I'll write it like this, AX squared, which we actually, if we want, well, I'll write it in a different way in a second. And then you have plus, plus two, it's a slightly different color, let me do that other color, plus two ABX and then finally plus B squared. Plus B squared."}, {"video_title": "Identifying perfect square form Mathematics II High School Math Khan Academy.mp3", "Sentence": "And so what are you left with? Well, you're going to be left with A, I'll write it like this, AX squared, which we actually, if we want, well, I'll write it in a different way in a second. And then you have plus, plus two, it's a slightly different color, let me do that other color, plus two ABX and then finally plus B squared. Plus B squared. Now I said I could write it in a slightly different way. What I could do is just rewrite out AX squared as being the same thing, this is the same thing as A squared, X squared, and then I can write out everything else the same way. Plus two ABX and then plus B squared."}, {"video_title": "Identifying perfect square form Mathematics II High School Math Khan Academy.mp3", "Sentence": "Plus B squared. Now I said I could write it in a slightly different way. What I could do is just rewrite out AX squared as being the same thing, this is the same thing as A squared, X squared, and then I can write out everything else the same way. Plus two ABX and then plus B squared. Now why did I, what's interesting about doing this? Well, now we can see the pattern for the square of any binomial or binomial like this. So for example, if someone were to walk up to you and say, alright, I have a trinomial of the form, let's say they have a trinomial of the form 25X squared plus 20X plus four, and they were to tell you to factor this, well actually, let's just do that."}, {"video_title": "Identifying perfect square form Mathematics II High School Math Khan Academy.mp3", "Sentence": "Plus two ABX and then plus B squared. Now why did I, what's interesting about doing this? Well, now we can see the pattern for the square of any binomial or binomial like this. So for example, if someone were to walk up to you and say, alright, I have a trinomial of the form, let's say they have a trinomial of the form 25X squared plus 20X plus four, and they were to tell you to factor this, well actually, let's just do that. Why don't you pause the video and see if you could factor this as a product of two binomials. Well, when you look at this, you say, well look, you know, this 25X squared, that looks like a square, a perfect square. 25X squared, that's the same thing as five squared X squared, or you could write it as five X, five X squared."}, {"video_title": "Identifying perfect square form Mathematics II High School Math Khan Academy.mp3", "Sentence": "So for example, if someone were to walk up to you and say, alright, I have a trinomial of the form, let's say they have a trinomial of the form 25X squared plus 20X plus four, and they were to tell you to factor this, well actually, let's just do that. Why don't you pause the video and see if you could factor this as a product of two binomials. Well, when you look at this, you say, well look, you know, this 25X squared, that looks like a square, a perfect square. 25X squared, that's the same thing as five squared X squared, or you could write it as five X, five X squared. This four here, that's a perfect square. That's the same thing as two squared. And let's see, 20 right over here, if we wanted to fit this pattern, we would say that A is five and B is two, and so let's see, what would be two times AB?"}, {"video_title": "Identifying perfect square form Mathematics II High School Math Khan Academy.mp3", "Sentence": "25X squared, that's the same thing as five squared X squared, or you could write it as five X, five X squared. This four here, that's a perfect square. That's the same thing as two squared. And let's see, 20 right over here, if we wanted to fit this pattern, we would say that A is five and B is two, and so let's see, what would be two times AB? Well, five times two, AB would be 10, and then two times that would be 20. So this right over here is, that is, plus two times five, two times five times two, times two X, times two X, I'll do it in this color, times two X. So you see that this completely matches this pattern here, where A is equal to five, and B is equal to two."}, {"video_title": "Identifying perfect square form Mathematics II High School Math Khan Academy.mp3", "Sentence": "And let's see, 20 right over here, if we wanted to fit this pattern, we would say that A is five and B is two, and so let's see, what would be two times AB? Well, five times two, AB would be 10, and then two times that would be 20. So this right over here is, that is, plus two times five, two times five times two, times two X, times two X, I'll do it in this color, times two X. So you see that this completely matches this pattern here, where A is equal to five, and B is equal to two. Once again, this is AX, whole thing squared, then you have two times A times B, X, we see that there, and then finally you have the B squared. So if you wanted to factor this, you could say, well, this is just going to be the same thing, since we know what A and B are, this is going to be five X plus two, five X plus two, five X plus two, whole thing squared. So the whole point of doing this is to start recognizing when we actually have perfect squares, especially perfect squares where the leading coefficient isn't one."}, {"video_title": "Difference of squares intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "This is a difference, we're subtracting between two quantities that are each squares. This is literally x squared, let me do that in a different color. This is x squared minus three squared. It's the difference between two quantities that have been squared. And it turns out that this is pretty straightforward to factor. And to see how it can be factored, let me pause there for a second and get a little bit of a review of multiplying binomials. So put this on the back burner a little bit before I give you the answer of how you factor this."}, {"video_title": "Difference of squares intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "It's the difference between two quantities that have been squared. And it turns out that this is pretty straightforward to factor. And to see how it can be factored, let me pause there for a second and get a little bit of a review of multiplying binomials. So put this on the back burner a little bit before I give you the answer of how you factor this. Let's do a little bit of an exercise. Let's multiply x plus a times x minus a, where a is some number. And we can do that using either the FOIL method, but I like just thinking of this as a distributive property twice."}, {"video_title": "Difference of squares intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "So put this on the back burner a little bit before I give you the answer of how you factor this. Let's do a little bit of an exercise. Let's multiply x plus a times x minus a, where a is some number. And we can do that using either the FOIL method, but I like just thinking of this as a distributive property twice. We could take x plus a and distribute it onto the x and onto the a. So when we multiply it by x, we would get x times x is x squared, a times x is plus ax. And then when we multiply it by the negative a, well, it'll become negative a times x minus a squared."}, {"video_title": "Difference of squares intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "And we can do that using either the FOIL method, but I like just thinking of this as a distributive property twice. We could take x plus a and distribute it onto the x and onto the a. So when we multiply it by x, we would get x times x is x squared, a times x is plus ax. And then when we multiply it by the negative a, well, it'll become negative a times x minus a squared. So these middle two terms cancel out, and you are left with x squared minus a squared. You're left with a difference of squares. X squared minus a squared."}, {"video_title": "Difference of squares intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "And then when we multiply it by the negative a, well, it'll become negative a times x minus a squared. So these middle two terms cancel out, and you are left with x squared minus a squared. You're left with a difference of squares. X squared minus a squared. So we have an interesting result right over here, that x squared minus a squared is equal to, is equal to x plus a, x plus a times x minus a. And so we can use, and this is for any a. So we could use this pattern now to factor this."}, {"video_title": "Difference of squares intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "X squared minus a squared. So we have an interesting result right over here, that x squared minus a squared is equal to, is equal to x plus a, x plus a times x minus a. And so we can use, and this is for any a. So we could use this pattern now to factor this. Here, what is our a? Our a is three. This is x squared minus three squared, or we could say minus our a squared if we say three is a."}, {"video_title": "Difference of squares intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "So we could use this pattern now to factor this. Here, what is our a? Our a is three. This is x squared minus three squared, or we could say minus our a squared if we say three is a. And so to factor it, this is just going to be equal to x plus our a, which is three, times x minus our a, which is three. So x plus three times x minus three. Now let's do some examples to really reinforce this idea of factoring differences of squares."}, {"video_title": "Difference of squares intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "This is x squared minus three squared, or we could say minus our a squared if we say three is a. And so to factor it, this is just going to be equal to x plus our a, which is three, times x minus our a, which is three. So x plus three times x minus three. Now let's do some examples to really reinforce this idea of factoring differences of squares. So let's say we want to factor, let me say, y squared minus 25. It has to be a difference of squares. And it doesn't work with a sum of squares."}, {"video_title": "Difference of squares intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "Now let's do some examples to really reinforce this idea of factoring differences of squares. So let's say we want to factor, let me say, y squared minus 25. It has to be a difference of squares. And it doesn't work with a sum of squares. Well, in this case, this is going to be y, and you have to confirm, okay, yeah, 25 is five squared, and y squared is, well, y squared. So this is gonna be y plus something times y minus something. And what is that something?"}, {"video_title": "Difference of squares intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "And it doesn't work with a sum of squares. Well, in this case, this is going to be y, and you have to confirm, okay, yeah, 25 is five squared, and y squared is, well, y squared. So this is gonna be y plus something times y minus something. And what is that something? Well, this right here is five squared, so it's y plus five times y minus five. And the variable doesn't have to come first. We could write 121 minus, I'll introduce a new variable, minus b squared."}, {"video_title": "Difference of squares intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "And what is that something? Well, this right here is five squared, so it's y plus five times y minus five. And the variable doesn't have to come first. We could write 121 minus, I'll introduce a new variable, minus b squared. Well, this is a difference of squares, because 121 is 11 squared. So this is going to be 11 plus something times 11 minus something. And in this case, that something is going to be the thing that was squared."}, {"video_title": "Difference of squares intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "We could write 121 minus, I'll introduce a new variable, minus b squared. Well, this is a difference of squares, because 121 is 11 squared. So this is going to be 11 plus something times 11 minus something. And in this case, that something is going to be the thing that was squared. So 11 plus b times 11 minus b. So in general, if you see a difference of squares, one square being subtracted from another square, and it could be a numeric perfect square, or it could be a variable that has been squared that you could take the square root of, well, then you could say, all right, well, that's just gonna be the first thing that's squared plus the second thing that has been squared, times the first thing that was squared minus the second thing that was squared. Now, some common mistakes that we see that people do, including my son, when they first learn this, is they say, okay, it's easy to recognize the difference of squares, but then they say, oh, is this y squared plus 25 times y squared minus 25?"}, {"video_title": "Difference of squares intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "And in this case, that something is going to be the thing that was squared. So 11 plus b times 11 minus b. So in general, if you see a difference of squares, one square being subtracted from another square, and it could be a numeric perfect square, or it could be a variable that has been squared that you could take the square root of, well, then you could say, all right, well, that's just gonna be the first thing that's squared plus the second thing that has been squared, times the first thing that was squared minus the second thing that was squared. Now, some common mistakes that we see that people do, including my son, when they first learn this, is they say, okay, it's easy to recognize the difference of squares, but then they say, oh, is this y squared plus 25 times y squared minus 25? No, the important thing to realize is, is that what is getting squared? Over here, y is the thing getting squared, and over here, it is five that is getting squared. Those are the things that are getting squared."}, {"video_title": "Difference of squares intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "Now, some common mistakes that we see that people do, including my son, when they first learn this, is they say, okay, it's easy to recognize the difference of squares, but then they say, oh, is this y squared plus 25 times y squared minus 25? No, the important thing to realize is, is that what is getting squared? Over here, y is the thing getting squared, and over here, it is five that is getting squared. Those are the things that are getting squared. Those are the things that are getting squared in this difference of squares, and so it's gonna be y plus five times y minus five. I encourage you to just try this out. We have a whole practice section on Khan Academy where you can do many, many more of these to become familiar."}, {"video_title": "Dimensions from volume of box Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "It could have been in cubic feet or cubic meters or cubic centimeters or cubic miles. Who knows? They just want to keep it as units, keep it as general as possible. The length is x units. The width is x plus 4 units. And the height is 9 units. Let me draw this box here."}, {"video_title": "Dimensions from volume of box Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "The length is x units. The width is x plus 4 units. And the height is 9 units. Let me draw this box here. So we have a nice little visualization. So they tell us that the length is x. Maybe we could call this the length right there."}, {"video_title": "Dimensions from volume of box Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Let me draw this box here. So we have a nice little visualization. So they tell us that the length is x. Maybe we could call this the length right there. They say the width is x plus 4. And the height is 9 of this box. In units, what are the dimensions of the box?"}, {"video_title": "Dimensions from volume of box Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Maybe we could call this the length right there. They say the width is x plus 4. And the height is 9 of this box. In units, what are the dimensions of the box? Well, they also tell us that the volume is 405. So the volume, 405, let me do it this way. So if we wanted to calculate the volume, what would it be?"}, {"video_title": "Dimensions from volume of box Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "In units, what are the dimensions of the box? Well, they also tell us that the volume is 405. So the volume, 405, let me do it this way. So if we wanted to calculate the volume, what would it be? Well, it would be the width. It would be x plus 4 times the length times x times 9. That's literally the volume of the box."}, {"video_title": "Dimensions from volume of box Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So if we wanted to calculate the volume, what would it be? Well, it would be the width. It would be x plus 4 times the length times x times 9. That's literally the volume of the box. Now, they also tell us that the volume of the box is 405 cubic units. It's equal to 405. So now we just solve for x."}, {"video_title": "Dimensions from volume of box Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "That's literally the volume of the box. Now, they also tell us that the volume of the box is 405 cubic units. It's equal to 405. So now we just solve for x. So what do we get here? If we distribute this x into this x plus 4, actually, if we distribute a 9x, let me just rewrite it, this is the same thing as 9x times x plus 4 is equal to 405. 9x times x is equal to 9x squared."}, {"video_title": "Dimensions from volume of box Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So now we just solve for x. So what do we get here? If we distribute this x into this x plus 4, actually, if we distribute a 9x, let me just rewrite it, this is the same thing as 9x times x plus 4 is equal to 405. 9x times x is equal to 9x squared. 9x times 4 is equal to 36. x is equal to 405. Now, we want our quadratic expression to be equal to 0. So let's subtract 405 from both sides of this equation."}, {"video_title": "Dimensions from volume of box Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "9x times x is equal to 9x squared. 9x times 4 is equal to 36. x is equal to 405. Now, we want our quadratic expression to be equal to 0. So let's subtract 405 from both sides of this equation. So when you do that, your right-hand side equals 0, and your left-hand side is 9x squared plus 36x minus 405. Now, is there any common factor to these numbers right here? Well, let's see, 405."}, {"video_title": "Dimensions from volume of box Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So let's subtract 405 from both sides of this equation. So when you do that, your right-hand side equals 0, and your left-hand side is 9x squared plus 36x minus 405. Now, is there any common factor to these numbers right here? Well, let's see, 405. 4 plus 0 plus 5 is 9. So that is divisible by 9. So all of these are divisible by 9."}, {"video_title": "Dimensions from volume of box Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Well, let's see, 405. 4 plus 0 plus 5 is 9. So that is divisible by 9. So all of these are divisible by 9. Let's just figure out what 405 divided by 9 is. So 9 goes into 405. 9 goes into 40 4 times."}, {"video_title": "Dimensions from volume of box Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So all of these are divisible by 9. Let's just figure out what 405 divided by 9 is. So 9 goes into 405. 9 goes into 40 4 times. 4 times 9 is 36. Subtract, you get 45. 9 goes into 45 5 times."}, {"video_title": "Dimensions from volume of box Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "9 goes into 40 4 times. 4 times 9 is 36. Subtract, you get 45. 9 goes into 45 5 times. 5 times 9 is 45. Subtract, you get 0. So it goes 45 times."}, {"video_title": "Dimensions from volume of box Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "9 goes into 45 5 times. 5 times 9 is 45. Subtract, you get 0. So it goes 45 times. So if we factor out a 9 here, we get 9 times x squared. Actually, even better, you don't even have to factor out a 9. Think about it, you can divide both sides of this equation by 9."}, {"video_title": "Dimensions from volume of box Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So it goes 45 times. So if we factor out a 9 here, we get 9 times x squared. Actually, even better, you don't even have to factor out a 9. Think about it, you can divide both sides of this equation by 9. So if you can divide all of the terms by 9, it won't change the equation. You're doing the same thing to both sides of the equation, which we've learned long ago is a very valid thing to do. So here you get x squared."}, {"video_title": "Dimensions from volume of box Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Think about it, you can divide both sides of this equation by 9. So if you can divide all of the terms by 9, it won't change the equation. You're doing the same thing to both sides of the equation, which we've learned long ago is a very valid thing to do. So here you get x squared. If you just had this expression here and someone would have told you to factor it, then you'd have to factor out the 9. But because this is an equation, it equals 0. Let's just divide everything by 9."}, {"video_title": "Dimensions from volume of box Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So here you get x squared. If you just had this expression here and someone would have told you to factor it, then you'd have to factor out the 9. But because this is an equation, it equals 0. Let's just divide everything by 9. It'll simplify things. So you get x squared plus 4x minus 45 is equal to 0. And now we can try to factor this right here."}, {"video_title": "Dimensions from volume of box Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Let's just divide everything by 9. It'll simplify things. So you get x squared plus 4x minus 45 is equal to 0. And now we can try to factor this right here. And this fits the pattern where we don't have a leading 1 out here, so we don't even have to do it by grouping. You just have to think, what two numbers, when I take their product, I get negative 45, and when I take their sum, I get positive 4. They're 4 apart."}, {"video_title": "Dimensions from volume of box Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "And now we can try to factor this right here. And this fits the pattern where we don't have a leading 1 out here, so we don't even have to do it by grouping. You just have to think, what two numbers, when I take their product, I get negative 45, and when I take their sum, I get positive 4. They're 4 apart. One has to be positive and one has to be negative. Their positive versions have to be 4 apart. Because when you're taking the sum, you're really taking their difference, because one of them is negative."}, {"video_title": "Dimensions from volume of box Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "They're 4 apart. One has to be positive and one has to be negative. Their positive versions have to be 4 apart. Because when you're taking the sum, you're really taking their difference, because one of them is negative. So let's think about it. Well, if you have positive 9 and negative 5, I think that'll work, right? Positive 9 plus negative 5 is 4."}, {"video_title": "Dimensions from volume of box Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Because when you're taking the sum, you're really taking their difference, because one of them is negative. So let's think about it. Well, if you have positive 9 and negative 5, I think that'll work, right? Positive 9 plus negative 5 is 4. And when you take their product, you get negative 45. So you have x plus 9 times x minus 5 is equal to 0. Just factored it out."}, {"video_title": "Dimensions from volume of box Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Positive 9 plus negative 5 is 4. And when you take their product, you get negative 45. So you have x plus 9 times x minus 5 is equal to 0. Just factored it out. And then we've seen this before. If you have two numbers, when you take their product, they equal 0. That means one of these numbers at least has to be equal to 0."}, {"video_title": "Dimensions from volume of box Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Just factored it out. And then we've seen this before. If you have two numbers, when you take their product, they equal 0. That means one of these numbers at least has to be equal to 0. So this means that x plus 9 is equal to 0. Let me scroll down a little bit. x plus 9 is equal to 0."}, {"video_title": "Dimensions from volume of box Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "That means one of these numbers at least has to be equal to 0. So this means that x plus 9 is equal to 0. Let me scroll down a little bit. x plus 9 is equal to 0. Or x minus 5 is equal to 0. So if we subtract 9 from this equation right there, you get x is equal to negative 9. Or if you add 5 to both sides of this equation here, you get x is equal to 5."}, {"video_title": "Dimensions from volume of box Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "x plus 9 is equal to 0. Or x minus 5 is equal to 0. So if we subtract 9 from this equation right there, you get x is equal to negative 9. Or if you add 5 to both sides of this equation here, you get x is equal to 5. So these are both possible values of x right here. So the box, if you take x is equal to negative 9, well, x equal to negative 9 won't work. Because if you put negative 9 here, you're going to have a box that has a width of negative 5, a length of negative 9, and a height of 9."}, {"video_title": "Dimensions from volume of box Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Or if you add 5 to both sides of this equation here, you get x is equal to 5. So these are both possible values of x right here. So the box, if you take x is equal to negative 9, well, x equal to negative 9 won't work. Because if you put negative 9 here, you're going to have a box that has a width of negative 5, a length of negative 9, and a height of 9. And if we're talking about our reality, we don't have negative distances like this. That can't be the length or the width. So x equals negative 9 isn't appropriate for this problem."}, {"video_title": "Dimensions from volume of box Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Because if you put negative 9 here, you're going to have a box that has a width of negative 5, a length of negative 9, and a height of 9. And if we're talking about our reality, we don't have negative distances like this. That can't be the length or the width. So x equals negative 9 isn't appropriate for this problem. Because in this problem, we need to have positive dimensions. So let's see what happens with x equals 5. If x equals 5, x plus 4 is 9."}, {"video_title": "Dimensions from volume of box Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So x equals negative 9 isn't appropriate for this problem. Because in this problem, we need to have positive dimensions. So let's see what happens with x equals 5. If x equals 5, x plus 4 is 9. And this dimension right here is going to be 5. And that seems pretty reasonable for our reality. And let's verify that this does end up with a volume of 405."}, {"video_title": "Dimensions from volume of box Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "If x equals 5, x plus 4 is 9. And this dimension right here is going to be 5. And that seems pretty reasonable for our reality. And let's verify that this does end up with a volume of 405. 9 times 5 is 45, times 9 is indeed 405. We just figured that out over here, that 45 times 9 is 405. So we're done."}, {"video_title": "Scaling & reflecting absolute value functions graph High School Math Khan Academy.mp3", "Sentence": "So you can see f of x is equal to the absolute value of x here in blue. And then g of x, not only does it look stretched or compressed, but it also is flipped over the x-axis. So like always, pause this video and see if you can come up yourself with the equation for g of x. Alright, now let's work through this together. So there's a couple of ways we could do it. We could first try to flip f of x and then try to stretch or compress it. Or we could stretch or compress it first and then try to flip it. Well, let's flip it first."}, {"video_title": "Scaling & reflecting absolute value functions graph High School Math Khan Academy.mp3", "Sentence": "So there's a couple of ways we could do it. We could first try to flip f of x and then try to stretch or compress it. Or we could stretch or compress it first and then try to flip it. Well, let's flip it first. So let's say that we have a function that looks like this. It's just exactly what f of x is, but flipped over the x-axis. So it's just flipped over the x-axis."}, {"video_title": "Scaling & reflecting absolute value functions graph High School Math Khan Academy.mp3", "Sentence": "Well, let's flip it first. So let's say that we have a function that looks like this. It's just exactly what f of x is, but flipped over the x-axis. So it's just flipped over the x-axis. So all the values for any given x, whatever y you used to get, you're now getting the negative of that. So this graph right over here, this would be the graph, I'll call this y is equal to the negative absolute value of x. Whatever the absolute value of x would've gotten you before, you're now going to get the negative or the opposite of it."}, {"video_title": "Scaling & reflecting absolute value functions graph High School Math Khan Academy.mp3", "Sentence": "So it's just flipped over the x-axis. So all the values for any given x, whatever y you used to get, you're now getting the negative of that. So this graph right over here, this would be the graph, I'll call this y is equal to the negative absolute value of x. Whatever the absolute value of x would've gotten you before, you're now going to get the negative or the opposite of it. So this is getting us closer to our definition of g of x. The key here is how do we appropriately stretch or squeeze this green function? So let's think about what's happening."}, {"video_title": "Scaling & reflecting absolute value functions graph High School Math Khan Academy.mp3", "Sentence": "Whatever the absolute value of x would've gotten you before, you're now going to get the negative or the opposite of it. So this is getting us closer to our definition of g of x. The key here is how do we appropriately stretch or squeeze this green function? So let's think about what's happening. On this green function, when x is equal to one, the function itself is equal to negative one. But we want it, if we want it to be the same as g, we want it to be equal to negative four. So it's actually four times the value."}, {"video_title": "Scaling & reflecting absolute value functions graph High School Math Khan Academy.mp3", "Sentence": "So let's think about what's happening. On this green function, when x is equal to one, the function itself is equal to negative one. But we want it, if we want it to be the same as g, we want it to be equal to negative four. So it's actually four times the value. For given x, at least for x equals one, g is giving me something four times the value that my green function is giving. And it's not just true for positive x. It's also true for negative x's."}, {"video_title": "Scaling & reflecting absolute value functions graph High School Math Khan Academy.mp3", "Sentence": "So it's actually four times the value. For given x, at least for x equals one, g is giving me something four times the value that my green function is giving. And it's not just true for positive x. It's also true for negative x's. You can see it right over here. When x is equal to negative one, my green function gives me negative one, but g gives me negative four. So it's giving me four times the value."}, {"video_title": "Scaling & reflecting absolute value functions graph High School Math Khan Academy.mp3", "Sentence": "It's also true for negative x's. You can see it right over here. When x is equal to negative one, my green function gives me negative one, but g gives me negative four. So it's giving me four times the value. It's giving me four times the negative value, so it's going even more negative. So what you can see, to go from the green to g, you have to multiply this thing right over here by four. So that is what essentially stretches it down, stretches it down in the vertical direction."}, {"video_title": "Scaling & reflecting absolute value functions graph High School Math Khan Academy.mp3", "Sentence": "So it's giving me four times the value. It's giving me four times the negative value, so it's going even more negative. So what you can see, to go from the green to g, you have to multiply this thing right over here by four. So that is what essentially stretches it down, stretches it down in the vertical direction. So we could say that g of x is equal to, it's not negative absolute value of x, negative four times the absolute value of x. And you could have done it the other way. You could have said, hey, let's first stretch or compress f. And say, all right, before we even flip it over, if we were to unflip g, it would look like this."}, {"video_title": "Scaling & reflecting absolute value functions graph High School Math Khan Academy.mp3", "Sentence": "So that is what essentially stretches it down, stretches it down in the vertical direction. So we could say that g of x is equal to, it's not negative absolute value of x, negative four times the absolute value of x. And you could have done it the other way. You could have said, hey, let's first stretch or compress f. And say, all right, before we even flip it over, if we were to unflip g, it would look like this. If we were to unflip g, it would look, it would look like, it would look like this if we were to unflip g. So this thing right over here, this thing looks like four times f of x. We could write this as, we could write this as y is equal to four times f of x. Or you could say y is equal to four times the absolute value of x."}, {"video_title": "Scaling & reflecting absolute value functions graph High School Math Khan Academy.mp3", "Sentence": "You could have said, hey, let's first stretch or compress f. And say, all right, before we even flip it over, if we were to unflip g, it would look like this. If we were to unflip g, it would look, it would look like, it would look like this if we were to unflip g. So this thing right over here, this thing looks like four times f of x. We could write this as, we could write this as y is equal to four times f of x. Or you could say y is equal to four times the absolute value of x. And then we have a negative sign. Whatever positive value you were getting before, you now get the opposite value. And that would flip it over and get you to g, which is exactly what we already got."}, {"video_title": "Perfect square factorization intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "And if someone asks you, hey, can you factor this into two binomials? Well, using techniques we learned in other videos, say, okay, I need to find two numbers whose product is nine and whose sum is six. And so I encourage you to think of, to pause this video and say, well, what two numbers can I, can add up to six, and if I take their product, I get nine? Well, nine only has so many factors, really one, three, and nine, and one plus nine does not equal six, and so, and negative one plus negative nine does not equal six, but three times three equals nine, and three plus three does equal six. Three times three, three plus three. And so we can factor this as x plus three times x plus three, which is, of course, the same thing as x plus three squared. And so what was it about this expression that made us recognize, or maybe now we will start to recognize as it being a perfect square?"}, {"video_title": "Perfect square factorization intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "Well, nine only has so many factors, really one, three, and nine, and one plus nine does not equal six, and so, and negative one plus negative nine does not equal six, but three times three equals nine, and three plus three does equal six. Three times three, three plus three. And so we can factor this as x plus three times x plus three, which is, of course, the same thing as x plus three squared. And so what was it about this expression that made us recognize, or maybe now we will start to recognize as it being a perfect square? Well, I have, of course, some variable that is being squared, which we need. I have some perfect square as the constant, and that whatever is being squared there, I have two times that as the coefficient on this first degree term here. Let's see if that is generally true, and I'll switch up the variables just to show that we can."}, {"video_title": "Perfect square factorization intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "And so what was it about this expression that made us recognize, or maybe now we will start to recognize as it being a perfect square? Well, I have, of course, some variable that is being squared, which we need. I have some perfect square as the constant, and that whatever is being squared there, I have two times that as the coefficient on this first degree term here. Let's see if that is generally true, and I'll switch up the variables just to show that we can. So let's say that I have a squared plus 14a plus 49. So a few interesting things are happening here. All right, I have my variable squared."}, {"video_title": "Perfect square factorization intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "Let's see if that is generally true, and I'll switch up the variables just to show that we can. So let's say that I have a squared plus 14a plus 49. So a few interesting things are happening here. All right, I have my variable squared. I have a perfect square constant term, that is seven squared right over here, and my coefficient on my first degree term here, that is two times the thing that's being squared. That is two times seven, or you could say it's seven plus seven. So you can immediately say, okay, if I want to factor this, this is going to be a plus seven squared."}, {"video_title": "Perfect square factorization intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "All right, I have my variable squared. I have a perfect square constant term, that is seven squared right over here, and my coefficient on my first degree term here, that is two times the thing that's being squared. That is two times seven, or you could say it's seven plus seven. So you can immediately say, okay, if I want to factor this, this is going to be a plus seven squared. And you can, of course, verify that by multiplying out, by figuring out what a plus seven squared is. Sometimes when you're first learning, it's like, hey, isn't that just a squared plus seven squared? No, remember, this is the same thing as a plus seven times a plus seven, and you can calculate this by using the FOIL, F-O-I-L technique."}, {"video_title": "Perfect square factorization intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "So you can immediately say, okay, if I want to factor this, this is going to be a plus seven squared. And you can, of course, verify that by multiplying out, by figuring out what a plus seven squared is. Sometimes when you're first learning, it's like, hey, isn't that just a squared plus seven squared? No, remember, this is the same thing as a plus seven times a plus seven, and you can calculate this by using the FOIL, F-O-I-L technique. I don't like that so much, because you're not thinking mathematically about what's happening. Really, you just have to do the distributive property twice here. First, you can multiply a plus seven times a, so a plus seven times a, and then multiply a plus seven times seven, so plus a plus seven times seven, and so this is going to be a squared plus seven a, plus, now we distribute this seven, plus seven a plus 49."}, {"video_title": "Perfect square factorization intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "No, remember, this is the same thing as a plus seven times a plus seven, and you can calculate this by using the FOIL, F-O-I-L technique. I don't like that so much, because you're not thinking mathematically about what's happening. Really, you just have to do the distributive property twice here. First, you can multiply a plus seven times a, so a plus seven times a, and then multiply a plus seven times seven, so plus a plus seven times seven, and so this is going to be a squared plus seven a, plus, now we distribute this seven, plus seven a plus 49. So now you see where that 14a came from. It's from the seven a plus the seven a. You see where the a squared came from, and you see where the 49 came from."}, {"video_title": "Perfect square factorization intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "First, you can multiply a plus seven times a, so a plus seven times a, and then multiply a plus seven times seven, so plus a plus seven times seven, and so this is going to be a squared plus seven a, plus, now we distribute this seven, plus seven a plus 49. So now you see where that 14a came from. It's from the seven a plus the seven a. You see where the a squared came from, and you see where the 49 came from. And you can speak of this in more general terms. If I wanted to just take the expression a plus b and square it, that's just a plus b times a plus b, and we do exactly what we did just here, but here I'm just doing it in very general terms with a or b, and you can think of a as either a constant number or even a variable, and so this is going to be, if we distribute this, it's going to be a plus b times that a, plus a plus b times that b. And so this is going to be a squared, now I'm just doing the distributive property again, a squared plus a b plus a b plus b squared."}, {"video_title": "Perfect square factorization intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "You see where the a squared came from, and you see where the 49 came from. And you can speak of this in more general terms. If I wanted to just take the expression a plus b and square it, that's just a plus b times a plus b, and we do exactly what we did just here, but here I'm just doing it in very general terms with a or b, and you can think of a as either a constant number or even a variable, and so this is going to be, if we distribute this, it's going to be a plus b times that a, plus a plus b times that b. And so this is going to be a squared, now I'm just doing the distributive property again, a squared plus a b plus a b plus b squared. So it's a squared plus two a b plus b b squared. So this is going to be the general form. So if a is the variable, which was x or a in this case, then it's just going to be whatever squared in the constant term, it's going to be two times that times the variable."}, {"video_title": "Perfect square factorization intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "And so this is going to be a squared, now I'm just doing the distributive property again, a squared plus a b plus a b plus b squared. So it's a squared plus two a b plus b b squared. So this is going to be the general form. So if a is the variable, which was x or a in this case, then it's just going to be whatever squared in the constant term, it's going to be two times that times the variable. And I want to show that there's some variation that you can entertain here. So if you were to see 25 plus 10x plus x squared, and someone said, hey, why don't you factor that? You could say, look, this right here is a perfect square, it's five squared, I have the variable squared right over here, and then this coefficient on our first degree term is two times five."}, {"video_title": "Perfect square factorization intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "So if a is the variable, which was x or a in this case, then it's just going to be whatever squared in the constant term, it's going to be two times that times the variable. And I want to show that there's some variation that you can entertain here. So if you were to see 25 plus 10x plus x squared, and someone said, hey, why don't you factor that? You could say, look, this right here is a perfect square, it's five squared, I have the variable squared right over here, and then this coefficient on our first degree term is two times five. And so you might immediately recognize this as five plus x squared. Now of course you could just rewrite this polynomial as x squared plus 10x plus 25, in which case you might say, okay, variable squared, some number squared, five squared, two times that number as a coefficient here, so that's going to be x plus five squared. And that's good, because these two things are absolutely equivalent."}, {"video_title": "Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "So what we really want to do is isolate the p on one side of this inequality, and preferably the left. That just makes it a little bit easier to read. It doesn't have to be, but we just want to isolate the p. So a good step to that is to get rid of this p on the right-hand side. And the best way I can think of doing that is subtracting p from the right. But of course, if we want to make sure that this inequality is always going to be true, if we do anything to the right, we also have to do that to the left. So we also have to subtract p from the left. And so the left-hand side, negative 3p minus p, that's negative 4p, and then we still have a minus 7 up here, is going to be less than p minus p. Those cancel out."}, {"video_title": "Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "And the best way I can think of doing that is subtracting p from the right. But of course, if we want to make sure that this inequality is always going to be true, if we do anything to the right, we also have to do that to the left. So we also have to subtract p from the left. And so the left-hand side, negative 3p minus p, that's negative 4p, and then we still have a minus 7 up here, is going to be less than p minus p. Those cancel out. It is less than 9. Now, the next thing I'm in the mood to do is get rid of this negative 7 or this minus 7 here so that we can better isolate the p on the left-hand side. So the best way I can think of to get rid of a negative 7 is to add 7 to it."}, {"video_title": "Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "And so the left-hand side, negative 3p minus p, that's negative 4p, and then we still have a minus 7 up here, is going to be less than p minus p. Those cancel out. It is less than 9. Now, the next thing I'm in the mood to do is get rid of this negative 7 or this minus 7 here so that we can better isolate the p on the left-hand side. So the best way I can think of to get rid of a negative 7 is to add 7 to it. Then it'll just cancel out to 0. So let's add 7 to both sides of this inequality. Negative 7 plus 7 cancel out."}, {"video_title": "Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "So the best way I can think of to get rid of a negative 7 is to add 7 to it. Then it'll just cancel out to 0. So let's add 7 to both sides of this inequality. Negative 7 plus 7 cancel out. All we're left with is negative 4p. On the right-hand side, we have 9 plus 7 is 16. And it's still less than."}, {"video_title": "Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "Negative 7 plus 7 cancel out. All we're left with is negative 4p. On the right-hand side, we have 9 plus 7 is 16. And it's still less than. Now, the last step to isolate the p is to get rid of this negative 4 coefficient. And the easiest way I can think of to get rid of this negative 4 coefficient is to divide both sides by negative 4. So if we divide this side by negative 4, these guys are going to cancel out."}, {"video_title": "Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "And it's still less than. Now, the last step to isolate the p is to get rid of this negative 4 coefficient. And the easiest way I can think of to get rid of this negative 4 coefficient is to divide both sides by negative 4. So if we divide this side by negative 4, these guys are going to cancel out. We're just going to be left with p. We also have to do it to the right-hand side. Now, there's one thing that you really have to remember since this is an inequality. This is not an equation."}, {"video_title": "Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "So if we divide this side by negative 4, these guys are going to cancel out. We're just going to be left with p. We also have to do it to the right-hand side. Now, there's one thing that you really have to remember since this is an inequality. This is not an equation. If you're dealing with an inequality, and if you multiply or divide both sides of the equation by a negative number, you have to swap the inequality. So in this case, the less than becomes greater than since we're dividing by a negative number. And so negative 4 divided by negative 4, those cancel out."}, {"video_title": "Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "This is not an equation. If you're dealing with an inequality, and if you multiply or divide both sides of the equation by a negative number, you have to swap the inequality. So in this case, the less than becomes greater than since we're dividing by a negative number. And so negative 4 divided by negative 4, those cancel out. We have p is greater than 16 divided by negative 4, which is negative 4. And we can plot this solution set right over here. And then we can try out some values to help us feel good about the idea of it working."}, {"video_title": "Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "And so negative 4 divided by negative 4, those cancel out. We have p is greater than 16 divided by negative 4, which is negative 4. And we can plot this solution set right over here. And then we can try out some values to help us feel good about the idea of it working. So let's say this is negative 5, negative 4, negative 3, negative 2, 1, negative 1, I should say, 0. Let me write that a little bit neater. Negative 1, 0."}, {"video_title": "Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "And then we can try out some values to help us feel good about the idea of it working. So let's say this is negative 5, negative 4, negative 3, negative 2, 1, negative 1, I should say, 0. Let me write that a little bit neater. Negative 1, 0. And then we can keep going to the right. And so our solution is p is not greater than or equal, so we have to exclude negative 4. p is greater than negative 4, so all the values above that. So negative 3.99999999 will work."}, {"video_title": "Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "Negative 1, 0. And then we can keep going to the right. And so our solution is p is not greater than or equal, so we have to exclude negative 4. p is greater than negative 4, so all the values above that. So negative 3.99999999 will work. Negative 4 will not work. And let's just try some values out to feel good that this is really the solution set. So first let's try out when p is equal to negative 3."}, {"video_title": "Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "So negative 3.99999999 will work. Negative 4 will not work. And let's just try some values out to feel good that this is really the solution set. So first let's try out when p is equal to negative 3. This should work. The way I've drawn it, this is in our solution set. p equals negative 3 is greater than negative 4."}, {"video_title": "Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "So first let's try out when p is equal to negative 3. This should work. The way I've drawn it, this is in our solution set. p equals negative 3 is greater than negative 4. So let's try that out. We have negative 3 times negative 3. The first negative 3 is this one, and then we're saying p is negative 3."}, {"video_title": "Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "p equals negative 3 is greater than negative 4. So let's try that out. We have negative 3 times negative 3. The first negative 3 is this one, and then we're saying p is negative 3. Minus 7 should be less than, instead of a p, we're going to put a negative 3, should be less than negative 3 plus 9. Negative 3 times negative 3 is 9. Minus 7 should be less than negative 3 plus 9 is 6."}, {"video_title": "Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "The first negative 3 is this one, and then we're saying p is negative 3. Minus 7 should be less than, instead of a p, we're going to put a negative 3, should be less than negative 3 plus 9. Negative 3 times negative 3 is 9. Minus 7 should be less than negative 3 plus 9 is 6. 9 minus 7 is 2. 2 should be less than 6, of which, of course, it is. Now let's try a value that definitely should not work."}, {"video_title": "Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "Minus 7 should be less than negative 3 plus 9 is 6. 9 minus 7 is 2. 2 should be less than 6, of which, of course, it is. Now let's try a value that definitely should not work. So let's try negative 5. Negative 5 is not in our solution set, so it should not work. So we have negative 3 times negative 5 minus 7."}, {"video_title": "Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "Now let's try a value that definitely should not work. So let's try negative 5. Negative 5 is not in our solution set, so it should not work. So we have negative 3 times negative 5 minus 7. Let's see whether it is less than negative 5 plus 9. Negative 3 times negative 5 is 15. Minus 7, it really should not be less than negative 5 plus 9."}, {"video_title": "Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "So we have negative 3 times negative 5 minus 7. Let's see whether it is less than negative 5 plus 9. Negative 3 times negative 5 is 15. Minus 7, it really should not be less than negative 5 plus 9. So we're just seeing if p equals negative 5 works. 15 minus 7 is 8. And so we get 8 is less than 4, which is definitely not the case."}, {"video_title": "Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "Minus 7, it really should not be less than negative 5 plus 9. So we're just seeing if p equals negative 5 works. 15 minus 7 is 8. And so we get 8 is less than 4, which is definitely not the case. So p equals negative 5 doesn't work, and it shouldn't work because that's not in our solution set. And now if we want to feel really good about it, we can actually try this boundary point. Negative 4 should not work, but it should satisfy the related equation."}, {"video_title": "Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "And so we get 8 is less than 4, which is definitely not the case. So p equals negative 5 doesn't work, and it shouldn't work because that's not in our solution set. And now if we want to feel really good about it, we can actually try this boundary point. Negative 4 should not work, but it should satisfy the related equation. When I talk about the related equation, negative 4 should satisfy negative 3 minus 7 is equal to p plus 9. It'll satisfy this, but it won't satisfy this because when we get the same value on both sides, the same value is not less than the same value. So let's try it out."}, {"video_title": "Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "Negative 4 should not work, but it should satisfy the related equation. When I talk about the related equation, negative 4 should satisfy negative 3 minus 7 is equal to p plus 9. It'll satisfy this, but it won't satisfy this because when we get the same value on both sides, the same value is not less than the same value. So let's try it out. Let's see whether negative 4 at least satisfies the related equation. So if we get negative 3 times negative 4 minus 7, this should be equal to negative 4 plus 9. So this is 12 minus 7 should be equal to negative 4 plus 9 should be equal to 5."}, {"video_title": "Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "So let's try it out. Let's see whether negative 4 at least satisfies the related equation. So if we get negative 3 times negative 4 minus 7, this should be equal to negative 4 plus 9. So this is 12 minus 7 should be equal to negative 4 plus 9 should be equal to 5. And this, of course, is true. 5 is equal to 5. So it satisfies the related equation, but it should not satisfy this."}, {"video_title": "Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "So this is 12 minus 7 should be equal to negative 4 plus 9 should be equal to 5. And this, of course, is true. 5 is equal to 5. So it satisfies the related equation, but it should not satisfy this. If you put negative 4 for p here, and I encourage you to do so, actually we could do it over here instead of an equal sign. If you put it into the original inequality, let me delete all of that, it really just becomes this. The original inequality is this right over here."}, {"video_title": "Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "So it satisfies the related equation, but it should not satisfy this. If you put negative 4 for p here, and I encourage you to do so, actually we could do it over here instead of an equal sign. If you put it into the original inequality, let me delete all of that, it really just becomes this. The original inequality is this right over here. If you put negative 4, you have less than, less than, and then you get 5 is less than 5, which is not the case. And that's good because we did not include that in the solution set. We put an open circle."}, {"video_title": "Multi-step inequalities 3 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "The original inequality is this right over here. If you put negative 4, you have less than, less than, and then you get 5 is less than 5, which is not the case. And that's good because we did not include that in the solution set. We put an open circle. If negative 4 was included, we would fill that in. But the only reason I would include negative 4 is if this was greater than or equal. So it's good that this does not work because negative 4 is not part of our solution set."}, {"video_title": "Reflecting and scaling absolute value function.mp3", "Sentence": "What is the equation of the new graph? So pause the video and see if you can figure that out. All right, let's work through it together now. Now you might not need to draw it visually, but I will just so that we can all together visualize what is going on. So let's say that's my x-axis and that is my y-axis. My y-axis. Y equals the absolute value of x."}, {"video_title": "Reflecting and scaling absolute value function.mp3", "Sentence": "Now you might not need to draw it visually, but I will just so that we can all together visualize what is going on. So let's say that's my x-axis and that is my y-axis. My y-axis. Y equals the absolute value of x. So for non-negative values of x, y is going to be equal to x. Absolute value of zero is zero. Absolute value of one is one."}, {"video_title": "Reflecting and scaling absolute value function.mp3", "Sentence": "Y equals the absolute value of x. So for non-negative values of x, y is going to be equal to x. Absolute value of zero is zero. Absolute value of one is one. Absolute value of two is two. So it's gonna look like this. It's just gonna have a slope of one."}, {"video_title": "Reflecting and scaling absolute value function.mp3", "Sentence": "Absolute value of one is one. Absolute value of two is two. So it's gonna look like this. It's just gonna have a slope of one. And then for negative values, when you take the absolute value, you're gonna take the opposite. You're gonna get the positive. So it's gonna look like this."}, {"video_title": "Reflecting and scaling absolute value function.mp3", "Sentence": "It's just gonna have a slope of one. And then for negative values, when you take the absolute value, you're gonna take the opposite. You're gonna get the positive. So it's gonna look like this. Let me see if I can draw that a little bit cleaner. This is a hand-drawn sketch, so bear with me. But hopefully this is familiar."}, {"video_title": "Reflecting and scaling absolute value function.mp3", "Sentence": "So it's gonna look like this. Let me see if I can draw that a little bit cleaner. This is a hand-drawn sketch, so bear with me. But hopefully this is familiar. You've seen the graph of y is equal to absolute value of x before. Now let's think about the different transformations. So first they say, is reflected across the x-axis."}, {"video_title": "Reflecting and scaling absolute value function.mp3", "Sentence": "But hopefully this is familiar. You've seen the graph of y is equal to absolute value of x before. Now let's think about the different transformations. So first they say, is reflected across the x-axis. So for example, if I have some x value right over here, before I would take the absolute value of x and I would end up there. But now we wanna reflect across the x-axis. So we wanna essentially get the negative of that value associated with that corresponding x."}, {"video_title": "Reflecting and scaling absolute value function.mp3", "Sentence": "So first they say, is reflected across the x-axis. So for example, if I have some x value right over here, before I would take the absolute value of x and I would end up there. But now we wanna reflect across the x-axis. So we wanna essentially get the negative of that value associated with that corresponding x. And so for example, this x, before we would get the absolute value of x, but now we wanna flip across the x-axis and we wanna get the negative of it. So in general, what we are doing is we are getting the negative of the absolute value of x. In general, if you're flipping over the x-axis, you're getting the negative."}, {"video_title": "Reflecting and scaling absolute value function.mp3", "Sentence": "So we wanna essentially get the negative of that value associated with that corresponding x. And so for example, this x, before we would get the absolute value of x, but now we wanna flip across the x-axis and we wanna get the negative of it. So in general, what we are doing is we are getting the negative of the absolute value of x. In general, if you're flipping over the x-axis, you're getting the negative. You're scaling the expression or the function by a negative. So this is going to be y is equal to the negative of the absolute value of x. Once again, whatever absolute value of x was giving you before for a given x, we now wanna get the negative of it."}, {"video_title": "Reflecting and scaling absolute value function.mp3", "Sentence": "In general, if you're flipping over the x-axis, you're getting the negative. You're scaling the expression or the function by a negative. So this is going to be y is equal to the negative of the absolute value of x. Once again, whatever absolute value of x was giving you before for a given x, we now wanna get the negative of it. We now wanna get the negative of it. So that's what reflecting across the x-axis does for us. But then they say, scaled vertically by a factor of seven."}, {"video_title": "Reflecting and scaling absolute value function.mp3", "Sentence": "Once again, whatever absolute value of x was giving you before for a given x, we now wanna get the negative of it. We now wanna get the negative of it. So that's what reflecting across the x-axis does for us. But then they say, scaled vertically by a factor of seven. By a factor of seven. And the way I view that is if you're scaling it vertically by a factor of seven, whatever y value you got for a given x, you now wanna get seven times the y value. Seven times the y value for a given x."}, {"video_title": "Reflecting and scaling absolute value function.mp3", "Sentence": "But then they say, scaled vertically by a factor of seven. By a factor of seven. And the way I view that is if you're scaling it vertically by a factor of seven, whatever y value you got for a given x, you now wanna get seven times the y value. Seven times the y value for a given x. And so if you think about that algebraically, well, if I want seven times the y value, I have to multiply this thing by seven. So I would get y is equal to negative seven times the absolute value of x. And that's essentially what they're asking."}, {"video_title": "Reflecting and scaling absolute value function.mp3", "Sentence": "Seven times the y value for a given x. And so if you think about that algebraically, well, if I want seven times the y value, I have to multiply this thing by seven. So I would get y is equal to negative seven times the absolute value of x. And that's essentially what they're asking. What is the equation of the new graph? And so that's what it would be. The negative flips us over the x-axis and then the seven scales vertically by a factor of seven."}, {"video_title": "Reflecting and scaling absolute value function.mp3", "Sentence": "And that's essentially what they're asking. What is the equation of the new graph? And so that's what it would be. The negative flips us over the x-axis and then the seven scales vertically by a factor of seven. But just to understand what this would look like, well, you multiply zero times seven. It doesn't change anything. But whatever x this is, this was equal to negative x, but now we're gonna get to negative seven x."}, {"video_title": "Reflecting and scaling absolute value function.mp3", "Sentence": "The negative flips us over the x-axis and then the seven scales vertically by a factor of seven. But just to understand what this would look like, well, you multiply zero times seven. It doesn't change anything. But whatever x this is, this was equal to negative x, but now we're gonna get to negative seven x. So let's see, two, three, four, five, six, seven. So we'd put something around that. So our graph is now going to look like this."}, {"video_title": "Reflecting and scaling absolute value function.mp3", "Sentence": "But whatever x this is, this was equal to negative x, but now we're gonna get to negative seven x. So let's see, two, three, four, five, six, seven. So we'd put something around that. So our graph is now going to look like this. It's going to be stretched along the vertical axis. If we were scaling vertically by something that had an absolute value less than one, then it would make the graph less tall. It would make it look wider."}, {"video_title": "Reflecting and scaling absolute value function.mp3", "Sentence": "So our graph is now going to look like this. It's going to be stretched along the vertical axis. If we were scaling vertically by something that had an absolute value less than one, then it would make the graph less tall. It would make it look wider. Let me make it at least look a little bit more symmetric. So it's gonna look something like that. But the key issue, and the reason why I'm drawing it so you can see that it looks like it's being scaled vertically."}, {"video_title": "Reflecting and scaling absolute value function.mp3", "Sentence": "It would make it look wider. Let me make it at least look a little bit more symmetric. So it's gonna look something like that. But the key issue, and the reason why I'm drawing it so you can see that it looks like it's being scaled vertically. It's being stretched in the vertical direction by a factor of seven. And the way we do that algebraically is we multiply by seven. The negative here is what flipped us over the x-axis."}, {"video_title": "Rewriting roots as rational exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "Well, a good way to figure out if things are equivalent is to just try to get them all in the same form. So the seventh root of v to the third power, the seventh root of something is the same thing as raising it to the 1 7th power. So this is equivalent to v to the third power raised to the 1 7th power. And if I raise something to an exponent and then raise that to an exponent, well, then that's the same thing as raising it to the product of these two exponents. So this is going to be the same thing as v to the 3 times 1 7th power, which of course is 3 7ths. So we've written it in multiple forms now. Let's see which of these match."}, {"video_title": "Rewriting roots as rational exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "And if I raise something to an exponent and then raise that to an exponent, well, then that's the same thing as raising it to the product of these two exponents. So this is going to be the same thing as v to the 3 times 1 7th power, which of course is 3 7ths. So we've written it in multiple forms now. Let's see which of these match. So v to the third to the 1 7th power, well, that was a form that we have right over here. So that is equivalent. v to the 3 7ths, that's what we have right over here."}, {"video_title": "Rewriting roots as rational exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "Let's see which of these match. So v to the third to the 1 7th power, well, that was a form that we have right over here. So that is equivalent. v to the 3 7ths, that's what we have right over here. So that one is definitely equivalent. Now let's think about this one. This is the cube root of v to the seventh."}, {"video_title": "Rewriting roots as rational exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "v to the 3 7ths, that's what we have right over here. So that one is definitely equivalent. Now let's think about this one. This is the cube root of v to the seventh. Is this going to be equivalent? Well, one way to think about it, this is going to be the same thing as v to the 1 3rd power. Actually, no, this wasn't the cube root of v to the 7th."}, {"video_title": "Rewriting roots as rational exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "This is the cube root of v to the seventh. Is this going to be equivalent? Well, one way to think about it, this is going to be the same thing as v to the 1 3rd power. Actually, no, this wasn't the cube root of v to the 7th. This was the cube root of v and that to the 7th power. So that's the same thing as v to the 1 3rd power and then that to the 7th power. So that is the same thing as v to the 7 3rd power, which is clearly different to v to the 3 7ths power."}, {"video_title": "Rewriting roots as rational exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "Actually, no, this wasn't the cube root of v to the 7th. This was the cube root of v and that to the 7th power. So that's the same thing as v to the 1 3rd power and then that to the 7th power. So that is the same thing as v to the 7 3rd power, which is clearly different to v to the 3 7ths power. So this is not going to be equivalent for all v's for which this expression is defined. Let's do a few more of these or similar types of problems dealing with roots and fractional exponents. The following equation is true for g greater than or equal to 0 and d is a constant."}, {"video_title": "Rewriting roots as rational exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "So that is the same thing as v to the 7 3rd power, which is clearly different to v to the 3 7ths power. So this is not going to be equivalent for all v's for which this expression is defined. Let's do a few more of these or similar types of problems dealing with roots and fractional exponents. The following equation is true for g greater than or equal to 0 and d is a constant. What is the value of d? Well, if I'm taking the 6th root of something, that's the same thing as raising it to the 1 6th power. So the 6th root of g to the 5th is the same thing as g to the 5th raised to the 1 6th power."}, {"video_title": "Rewriting roots as rational exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "The following equation is true for g greater than or equal to 0 and d is a constant. What is the value of d? Well, if I'm taking the 6th root of something, that's the same thing as raising it to the 1 6th power. So the 6th root of g to the 5th is the same thing as g to the 5th raised to the 1 6th power. And just like we just saw in the last example, that's the same thing as g to the 5 times 1 6th power. This is just our exponent properties. I raise something to an exponent and then raise that whole thing to another exponent, I can just multiply the exponents."}, {"video_title": "Rewriting roots as rational exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "So the 6th root of g to the 5th is the same thing as g to the 5th raised to the 1 6th power. And just like we just saw in the last example, that's the same thing as g to the 5 times 1 6th power. This is just our exponent properties. I raise something to an exponent and then raise that whole thing to another exponent, I can just multiply the exponents. So that's the same thing as g to the 5 6th power. And so d is 5 6th, 5 over 6. The 6th root of g to the 5th is the same thing as g to the 5 6th power."}, {"video_title": "Rewriting roots as rational exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "I raise something to an exponent and then raise that whole thing to another exponent, I can just multiply the exponents. So that's the same thing as g to the 5 6th power. And so d is 5 6th, 5 over 6. The 6th root of g to the 5th is the same thing as g to the 5 6th power. Let's do one more of these. The following equation is true for x greater than 0 and d is a constant. What is the value of d?"}, {"video_title": "Rewriting roots as rational exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "The 6th root of g to the 5th is the same thing as g to the 5 6th power. Let's do one more of these. The following equation is true for x greater than 0 and d is a constant. What is the value of d? All right, this is interesting. And I forgot to tell you in the last one, but pause this video as well and see if you can work it out. Pause for this question as well and see if you can work it out."}, {"video_title": "Rewriting roots as rational exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "What is the value of d? All right, this is interesting. And I forgot to tell you in the last one, but pause this video as well and see if you can work it out. Pause for this question as well and see if you can work it out. Well, here, let's just start rewriting the root as an exponent. So I can rewrite the whole thing. This is the same thing as 1 over, instead of writing the 7th root of x, I'll write x to the 1 7th power is equal to x to the d. And if I have 1 over something to a power, that's the same thing as that something raised to the negative of that power."}, {"video_title": "Rewriting roots as rational exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "Pause for this question as well and see if you can work it out. Well, here, let's just start rewriting the root as an exponent. So I can rewrite the whole thing. This is the same thing as 1 over, instead of writing the 7th root of x, I'll write x to the 1 7th power is equal to x to the d. And if I have 1 over something to a power, that's the same thing as that something raised to the negative of that power. So that is the same thing as x to the negative 1 7th power. And so that is going to be equal to x to the d. And so d must be equal to negative 1 7th. So the key here is when you're taking the reciprocal of something, that's the same thing as raising it to the negative of that exponent."}, {"video_title": "Rewriting roots as rational exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "This is the same thing as 1 over, instead of writing the 7th root of x, I'll write x to the 1 7th power is equal to x to the d. And if I have 1 over something to a power, that's the same thing as that something raised to the negative of that power. So that is the same thing as x to the negative 1 7th power. And so that is going to be equal to x to the d. And so d must be equal to negative 1 7th. So the key here is when you're taking the reciprocal of something, that's the same thing as raising it to the negative of that exponent. Another way of thinking about it is you could view this as x to the 1 7th to the negative 1 power. And then if you multiply these exponents, you get what we have right over there. But either way, d is equal to negative 1 7th."}, {"video_title": "Writing equations of perpendicular lines (example 2) High School Math Khan Academy.mp3", "Sentence": "What does that tell us? Well, if it's perpendicular to this line, its slope has to be the negative inverse of 2 5ths. So its slope, the negative inverse of 2 5ths, the inverse of 2 5ths is 5. Let me do it in a better color, a nicer green. If this line's slope is negative 2 5ths, the equation of the line we have to figure out that's perpendicular, its slope is going to be the inverse. So instead of 2 5ths, it's going to be 5 halves. Instead of being a negative, it's going to be a positive."}, {"video_title": "Writing equations of perpendicular lines (example 2) High School Math Khan Academy.mp3", "Sentence": "Let me do it in a better color, a nicer green. If this line's slope is negative 2 5ths, the equation of the line we have to figure out that's perpendicular, its slope is going to be the inverse. So instead of 2 5ths, it's going to be 5 halves. Instead of being a negative, it's going to be a positive. So this is the negative inverse of negative 2 5ths. You take the negative sign, it becomes positive. You swap the 5 and the 2, you get 5 halves."}, {"video_title": "Writing equations of perpendicular lines (example 2) High School Math Khan Academy.mp3", "Sentence": "Instead of being a negative, it's going to be a positive. So this is the negative inverse of negative 2 5ths. You take the negative sign, it becomes positive. You swap the 5 and the 2, you get 5 halves. So that is going to have to be our slope. And we can actually use the point slope form right here. It goes through this point right there."}, {"video_title": "Writing equations of perpendicular lines (example 2) High School Math Khan Academy.mp3", "Sentence": "You swap the 5 and the 2, you get 5 halves. So that is going to have to be our slope. And we can actually use the point slope form right here. It goes through this point right there. So let's use point slope form. y minus this y value, which has to be on the line, is equal to our slope, 5 halves, times x minus this x value, the x value when y is equal to 8. And this is the equation of the line in point slope form."}, {"video_title": "Writing equations of perpendicular lines (example 2) High School Math Khan Academy.mp3", "Sentence": "It goes through this point right there. So let's use point slope form. y minus this y value, which has to be on the line, is equal to our slope, 5 halves, times x minus this x value, the x value when y is equal to 8. And this is the equation of the line in point slope form. If you want to put it in slope intercept form, you can just do a little bit of algebraic manipulation. y minus 8 is equal to, let's distribute the 5 halves. So 5 halves x minus 5 halves times 2 is just 5."}, {"video_title": "Writing equations of perpendicular lines (example 2) High School Math Khan Academy.mp3", "Sentence": "And this is the equation of the line in point slope form. If you want to put it in slope intercept form, you can just do a little bit of algebraic manipulation. y minus 8 is equal to, let's distribute the 5 halves. So 5 halves x minus 5 halves times 2 is just 5. And then add 8 to both sides. You get y is equal to 5 halves x. Add 8 to negative 5, so plus 3."}, {"video_title": "Example Solving simple quadratic Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So in this situation, it looks like we might be able to isolate the x squared pretty simply because there's only one term that involves an x here, it's only this x squared term. So let's try to do that. So let me just rewrite it. We have 2x squared plus 3 is equal to 75. I'm going to try to isolate this x squared over here. The best way to do that, or at least the first step, would be to subtract 3 from both sides of this equation. So let's subtract 3 from both sides."}, {"video_title": "Example Solving simple quadratic Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "We have 2x squared plus 3 is equal to 75. I'm going to try to isolate this x squared over here. The best way to do that, or at least the first step, would be to subtract 3 from both sides of this equation. So let's subtract 3 from both sides. The left-hand side, we're just left with 2x squared. That was the whole point of subtracting 3 from both sides. And on the right-hand side, 75 minus 3 is 72."}, {"video_title": "Example Solving simple quadratic Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So let's subtract 3 from both sides. The left-hand side, we're just left with 2x squared. That was the whole point of subtracting 3 from both sides. And on the right-hand side, 75 minus 3 is 72. Now I want to isolate this x squared. I have a 2x squared here, so I could have just an x squared here if I divide this side, or really both sides, by 2. But if I do it to one side, I have to do it to the other side if I want to maintain the equality."}, {"video_title": "Example Solving simple quadratic Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "And on the right-hand side, 75 minus 3 is 72. Now I want to isolate this x squared. I have a 2x squared here, so I could have just an x squared here if I divide this side, or really both sides, by 2. But if I do it to one side, I have to do it to the other side if I want to maintain the equality. So the left side just becomes x squared, and the right-hand side is 72 divided by 2, is 36. So we're left with x squared is equal to 36. And then to solve for x, we can take the positive, the plus or minus square root of both sides."}, {"video_title": "Example Solving simple quadratic Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "But if I do it to one side, I have to do it to the other side if I want to maintain the equality. So the left side just becomes x squared, and the right-hand side is 72 divided by 2, is 36. So we're left with x squared is equal to 36. And then to solve for x, we can take the positive, the plus or minus square root of both sides. So we could say the plus or, let me write it this way. If we take the square root of both sides, we would get x is equal to the plus or minus square root of 36, which is equal to plus or minus 6. Let me just write that on another line."}, {"video_title": "Example Solving simple quadratic Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "And then to solve for x, we can take the positive, the plus or minus square root of both sides. So we could say the plus or, let me write it this way. If we take the square root of both sides, we would get x is equal to the plus or minus square root of 36, which is equal to plus or minus 6. Let me just write that on another line. So x is equal to plus or minus 6. And remember here, if something squared is equal to 36, that something could be the negative version or the positive version. It could be the principal root or it could be the negative root."}, {"video_title": "Example Solving simple quadratic Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Let me just write that on another line. So x is equal to plus or minus 6. And remember here, if something squared is equal to 36, that something could be the negative version or the positive version. It could be the principal root or it could be the negative root. Both negative 6 squared is 36 and positive 6 squared is 36. So both of these work. And you could put them back into the original equation to verify it."}, {"video_title": "Example Solving simple quadratic Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "It could be the principal root or it could be the negative root. Both negative 6 squared is 36 and positive 6 squared is 36. So both of these work. And you could put them back into the original equation to verify it. Let's do that. If you say 2 times 6 squared plus 3, that's 2 times 36, which is 72, plus 3 is 75. So that works."}, {"video_title": "Recognizing features of functions (example 2) Algebra II Khan Academy.mp3", "Sentence": "Which of these functions is odd? And so just let's remind ourselves what it means for a function to be odd. So I have a function, well they've already used f, g, and h, so I'll use j. So a function j is odd if you evaluate j at some value, so let's say j of a, and if you evaluate that j at the negative of that value, and if these two things are the negative of each other, then my function is odd. If these two things were the same, if you didn't have this negative here, then it would be an even function. So let's see which of these meet the criteria of being odd. So let's look at f of x."}, {"video_title": "Recognizing features of functions (example 2) Algebra II Khan Academy.mp3", "Sentence": "So a function j is odd if you evaluate j at some value, so let's say j of a, and if you evaluate that j at the negative of that value, and if these two things are the negative of each other, then my function is odd. If these two things were the same, if you didn't have this negative here, then it would be an even function. So let's see which of these meet the criteria of being odd. So let's look at f of x. So we could pick a particular point, so let's say when x is equal to 2. So we get f of 2, f of 2 is equal to 2. Now what is f of negative 2?"}, {"video_title": "Recognizing features of functions (example 2) Algebra II Khan Academy.mp3", "Sentence": "So let's look at f of x. So we could pick a particular point, so let's say when x is equal to 2. So we get f of 2, f of 2 is equal to 2. Now what is f of negative 2? f of negative 2 looks like it is 6. f of negative 2 is equal to 6. So these aren't the negative of each other. In order for this to be odd, f of negative 2 would have had to be equal to the negative of this, would have had to be equal to negative 2."}, {"video_title": "Recognizing features of functions (example 2) Algebra II Khan Academy.mp3", "Sentence": "Now what is f of negative 2? f of negative 2 looks like it is 6. f of negative 2 is equal to 6. So these aren't the negative of each other. In order for this to be odd, f of negative 2 would have had to be equal to the negative of this, would have had to be equal to negative 2. So f of x is definitely not odd. So all I had to do was find even one case that violated this constraint to be odd, and so I can say it's definitely not odd. Now let's look at g of x."}, {"video_title": "Recognizing features of functions (example 2) Algebra II Khan Academy.mp3", "Sentence": "In order for this to be odd, f of negative 2 would have had to be equal to the negative of this, would have had to be equal to negative 2. So f of x is definitely not odd. So all I had to do was find even one case that violated this constraint to be odd, and so I can say it's definitely not odd. Now let's look at g of x. So I could use the same, let's see, when x is equal to 2, we get g of 2 is equal to negative 7. Now let's look at when g is negative 2. So we get g of negative 2 is also equal to negative 7."}, {"video_title": "Recognizing features of functions (example 2) Algebra II Khan Academy.mp3", "Sentence": "Now let's look at g of x. So I could use the same, let's see, when x is equal to 2, we get g of 2 is equal to negative 7. Now let's look at when g is negative 2. So we get g of negative 2 is also equal to negative 7. So here we have a situation, and it looks like that's the case for any x we pick, that g of x is going to be equal to g of negative x. So g of x is equal to g of negative x. It's symmetric around the y, or I should say the vertical axis right over here."}, {"video_title": "Recognizing features of functions (example 2) Algebra II Khan Academy.mp3", "Sentence": "So we get g of negative 2 is also equal to negative 7. So here we have a situation, and it looks like that's the case for any x we pick, that g of x is going to be equal to g of negative x. So g of x is equal to g of negative x. It's symmetric around the y, or I should say the vertical axis right over here. So g of x is even, not odd. So which of these functions is odd? Definitely not g of x."}, {"video_title": "Recognizing features of functions (example 2) Algebra II Khan Academy.mp3", "Sentence": "It's symmetric around the y, or I should say the vertical axis right over here. So g of x is even, not odd. So which of these functions is odd? Definitely not g of x. So our last hope is h of x. Let's see if h of x seems to meet the criteria. So if we, I'll do it in this green color."}, {"video_title": "Recognizing features of functions (example 2) Algebra II Khan Academy.mp3", "Sentence": "Definitely not g of x. So our last hope is h of x. Let's see if h of x seems to meet the criteria. So if we, I'll do it in this green color. So if we take h of 1, and we can look at it even visually. So h of 1 gets us right over here. h of negative 1 seems to get us an equal amount, an equal distance, negative."}, {"video_title": "Recognizing features of functions (example 2) Algebra II Khan Academy.mp3", "Sentence": "So if we, I'll do it in this green color. So if we take h of 1, and we can look at it even visually. So h of 1 gets us right over here. h of negative 1 seems to get us an equal amount, an equal distance, negative. So it seems to fit for 1, for 2. Well, 2 is at the x-axis, but that's definitely h of 2 is 0, h of negative 2 is 0, but those are the negatives of each other. 0 is equal to negative 0."}, {"video_title": "Recognizing features of functions (example 2) Algebra II Khan Academy.mp3", "Sentence": "h of negative 1 seems to get us an equal amount, an equal distance, negative. So it seems to fit for 1, for 2. Well, 2 is at the x-axis, but that's definitely h of 2 is 0, h of negative 2 is 0, but those are the negatives of each other. 0 is equal to negative 0. If we go to, say, h of 4, h of 4 is this negative number, and h of negative 4 seems to be a positive number of the same magnitude. So once again, this is the negative of this. So it looks like this is indeed an odd function."}, {"video_title": "Recognizing features of functions (example 2) Algebra II Khan Academy.mp3", "Sentence": "0 is equal to negative 0. If we go to, say, h of 4, h of 4 is this negative number, and h of negative 4 seems to be a positive number of the same magnitude. So once again, this is the negative of this. So it looks like this is indeed an odd function. And another way to visually spot an odd function is a function, it's going to go through the origin, and you could essentially flip it over on both axes. So if you flip this, the right half, over the left half, and then flip that over the horizontal axis, you are going to get this right over here. So you see here we're going up and to the right, here we're going to go down and to the left, and then you curve right over there, you curve up just like that."}, {"video_title": "Recognizing features of functions (example 2) Algebra II Khan Academy.mp3", "Sentence": "So it looks like this is indeed an odd function. And another way to visually spot an odd function is a function, it's going to go through the origin, and you could essentially flip it over on both axes. So if you flip this, the right half, over the left half, and then flip that over the horizontal axis, you are going to get this right over here. So you see here we're going up and to the right, here we're going to go down and to the left, and then you curve right over there, you curve up just like that. But the easiest way to test it is just to do what we did. Look at a given x. So for example, when x is equal to 8, h of 8 looks like this number right around 8. h of negative 8 looks like it's pretty close to negative 8."}, {"video_title": "Recognizing features of functions (example 2) Algebra II Khan Academy.mp3", "Sentence": "So you see here we're going up and to the right, here we're going to go down and to the left, and then you curve right over there, you curve up just like that. But the easiest way to test it is just to do what we did. Look at a given x. So for example, when x is equal to 8, h of 8 looks like this number right around 8. h of negative 8 looks like it's pretty close to negative 8. So they seem to be the negative of each other. It sounds like a car crash just happened outside. Anyway, hopefully you enjoyed that."}, {"video_title": "Solving quadratics by taking square roots examples High School Math Khan Academy.mp3", "Sentence": "Figure out which x values will satisfy this equation. All right, let's work through this. And the way I'm gonna do this is I'm gonna isolate the x plus three squared on one side. And the best way to do that is to add four to both sides. So adding four to both sides. We'll get rid of this four, this subtracting four, this negative four on the left-hand side. And so we're just left with x plus three squared."}, {"video_title": "Solving quadratics by taking square roots examples High School Math Khan Academy.mp3", "Sentence": "And the best way to do that is to add four to both sides. So adding four to both sides. We'll get rid of this four, this subtracting four, this negative four on the left-hand side. And so we're just left with x plus three squared. x plus three squared. And on the right-hand side, I'm just gonna have zero plus four. So x plus three squared is equal to four."}, {"video_title": "Solving quadratics by taking square roots examples High School Math Khan Academy.mp3", "Sentence": "And so we're just left with x plus three squared. x plus three squared. And on the right-hand side, I'm just gonna have zero plus four. So x plus three squared is equal to four. And so now I could take the square root of both sides. Or another way of thinking about it, if I have something squared equaling four, I could say that that something needs to either be positive or negative two. So one way of thinking about it is I'm saying that x plus three is going to be equal to the plus or minus square root of that four."}, {"video_title": "Solving quadratics by taking square roots examples High School Math Khan Academy.mp3", "Sentence": "So x plus three squared is equal to four. And so now I could take the square root of both sides. Or another way of thinking about it, if I have something squared equaling four, I could say that that something needs to either be positive or negative two. So one way of thinking about it is I'm saying that x plus three is going to be equal to the plus or minus square root of that four. And hopefully this makes intuitive sense for you. If something squared is equal to four, that means that the something, that means that this something right over here is going to be equal to the positive square root of four or the negative square root of four. Or it's gonna be equal to positive or negative two."}, {"video_title": "Solving quadratics by taking square roots examples High School Math Khan Academy.mp3", "Sentence": "So one way of thinking about it is I'm saying that x plus three is going to be equal to the plus or minus square root of that four. And hopefully this makes intuitive sense for you. If something squared is equal to four, that means that the something, that means that this something right over here is going to be equal to the positive square root of four or the negative square root of four. Or it's gonna be equal to positive or negative two. And so we could write that x plus three could be equal to positive two or x plus three could be equal to negative two. Notice, if x plus three was positive two, two squared is equal to four. If x plus three was negative two, negative two squared is equal to four."}, {"video_title": "Solving quadratics by taking square roots examples High School Math Khan Academy.mp3", "Sentence": "Or it's gonna be equal to positive or negative two. And so we could write that x plus three could be equal to positive two or x plus three could be equal to negative two. Notice, if x plus three was positive two, two squared is equal to four. If x plus three was negative two, negative two squared is equal to four. So either of these would satisfy our equation. So if x plus three is equal to two, we could just subtract three from both sides to solve for x. And we're left with x is equal to negative one."}, {"video_title": "Solving quadratics by taking square roots examples High School Math Khan Academy.mp3", "Sentence": "If x plus three was negative two, negative two squared is equal to four. So either of these would satisfy our equation. So if x plus three is equal to two, we could just subtract three from both sides to solve for x. And we're left with x is equal to negative one. Or over here we could subtract three from both sides to solve for x. So or x is equal to negative two minus three is negative five. So those are the two possible solutions."}, {"video_title": "Solving quadratics by taking square roots examples High School Math Khan Academy.mp3", "Sentence": "And we're left with x is equal to negative one. Or over here we could subtract three from both sides to solve for x. So or x is equal to negative two minus three is negative five. So those are the two possible solutions. And you can verify that. Take these x values, substitute it back in, and then you can see when you substitute it back in, if you substitute x equals negative one, then x plus three is equal to two. Two squared is four, minus four is zero."}, {"video_title": "Solving quadratics by taking square roots examples High School Math Khan Academy.mp3", "Sentence": "So those are the two possible solutions. And you can verify that. Take these x values, substitute it back in, and then you can see when you substitute it back in, if you substitute x equals negative one, then x plus three is equal to two. Two squared is four, minus four is zero. And when x is equal to negative five, negative five plus three is negative two squared is positive four, minus four is also equal to zero. So these are the two possible x values that satisfy the equation. Now let's do another one that's presented to us in a slightly different way."}, {"video_title": "Solving quadratics by taking square roots examples High School Math Khan Academy.mp3", "Sentence": "Two squared is four, minus four is zero. And when x is equal to negative five, negative five plus three is negative two squared is positive four, minus four is also equal to zero. So these are the two possible x values that satisfy the equation. Now let's do another one that's presented to us in a slightly different way. So we are told that f of x is equal to x minus two squared minus nine. And then we're asked at what x values does the graph of y equals f of x intersect the x-axis? So if I'm just generally talking about some graph, so I'm not necessarily gonna draw that y equals f of x."}, {"video_title": "Solving quadratics by taking square roots examples High School Math Khan Academy.mp3", "Sentence": "Now let's do another one that's presented to us in a slightly different way. So we are told that f of x is equal to x minus two squared minus nine. And then we're asked at what x values does the graph of y equals f of x intersect the x-axis? So if I'm just generally talking about some graph, so I'm not necessarily gonna draw that y equals f of x. So if I'm just, so that's our y-axis, this is our x-axis. And so if I just have the graph of some function, if I have the graph of some function that looks something like that, let's say that's the y is equal to some other function, not necessarily this f of x. Y is equal to g of x. The x values where you intersect, where you intersect the x-axis, well in order to intersect the x-axis, y must be equal to zero."}, {"video_title": "Solving quadratics by taking square roots examples High School Math Khan Academy.mp3", "Sentence": "So if I'm just generally talking about some graph, so I'm not necessarily gonna draw that y equals f of x. So if I'm just, so that's our y-axis, this is our x-axis. And so if I just have the graph of some function, if I have the graph of some function that looks something like that, let's say that's the y is equal to some other function, not necessarily this f of x. Y is equal to g of x. The x values where you intersect, where you intersect the x-axis, well in order to intersect the x-axis, y must be equal to zero. So y is equal to zero there. Notice our y-coordinate at either of those points are going to be equal to zero. And that means that our function is equal to zero."}, {"video_title": "Solving quadratics by taking square roots examples High School Math Khan Academy.mp3", "Sentence": "The x values where you intersect, where you intersect the x-axis, well in order to intersect the x-axis, y must be equal to zero. So y is equal to zero there. Notice our y-coordinate at either of those points are going to be equal to zero. And that means that our function is equal to zero. So figuring out the x values where the graph of y equals f of x intersects the x-axis, this is equivalent to saying for what x values does f of x equal zero? So we could just say for what x values does this thing right over here equal zero? So let me just write that down."}, {"video_title": "Solving quadratics by taking square roots examples High School Math Khan Academy.mp3", "Sentence": "And that means that our function is equal to zero. So figuring out the x values where the graph of y equals f of x intersects the x-axis, this is equivalent to saying for what x values does f of x equal zero? So we could just say for what x values does this thing right over here equal zero? So let me just write that down. So we could rewrite this as x minus two squared minus nine equals zero. Well we could add nine to both sides and so we could get x minus two squared is equal to nine. And just like we saw before, that means that x minus two is equal to the positive or negative square root of nine."}, {"video_title": "Solving quadratics by taking square roots examples High School Math Khan Academy.mp3", "Sentence": "So let me just write that down. So we could rewrite this as x minus two squared minus nine equals zero. Well we could add nine to both sides and so we could get x minus two squared is equal to nine. And just like we saw before, that means that x minus two is equal to the positive or negative square root of nine. So we could say x minus two is equal to positive three or x minus two is equal to negative three. Well you add two to both sides of this, you get x is equal to five or if we add two to both sides of this equation, you'll get x is equal to negative one. And you can verify that."}, {"video_title": "Solving quadratics by taking square roots examples High School Math Khan Academy.mp3", "Sentence": "And just like we saw before, that means that x minus two is equal to the positive or negative square root of nine. So we could say x minus two is equal to positive three or x minus two is equal to negative three. Well you add two to both sides of this, you get x is equal to five or if we add two to both sides of this equation, you'll get x is equal to negative one. And you can verify that. If x is equal to five, five minus two is three squared is nine, minus nine is zero. So the point five comma zero is going to be on this graph. And also if x is equal to negative one, negative one minus two, negative three squared is positive nine, minus nine is zero."}, {"video_title": "Recognize functions from tables Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "We're asked to look at the table below. From the information given, is there a functional relationship between each person and his or her height? So a good place to start is just think about what a functional relationship means. Now there's definitely a relationship. They say, hey, if you're Joel, you're 5' 6\". If you're Nathan, you're 4' 11\". If you're Stuart, you're 5' 11\"."}, {"video_title": "Recognize functions from tables Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Now there's definitely a relationship. They say, hey, if you're Joel, you're 5' 6\". If you're Nathan, you're 4' 11\". If you're Stuart, you're 5' 11\". That is a relationship. Now in order for it to be a functional relationship, for every instance or every example of the independent variable, you can only have one example of the value of the function for it. So if this is a height function, so if you say height is the function, in order for this to be a functional relationship, no matter whose name you put inside of the height function, you need to only be able to get one value."}, {"video_title": "Recognize functions from tables Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "If you're Stuart, you're 5' 11\". That is a relationship. Now in order for it to be a functional relationship, for every instance or every example of the independent variable, you can only have one example of the value of the function for it. So if this is a height function, so if you say height is the function, in order for this to be a functional relationship, no matter whose name you put inside of the height function, you need to only be able to get one value. If there were two values associated with one person's name, it would not be a functional relationship. So if I were to ask you, what is the height of Nathan? Well, you look at the table and say, well, Nathan's height is 4' 11\"."}, {"video_title": "Recognize functions from tables Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So if this is a height function, so if you say height is the function, in order for this to be a functional relationship, no matter whose name you put inside of the height function, you need to only be able to get one value. If there were two values associated with one person's name, it would not be a functional relationship. So if I were to ask you, what is the height of Nathan? Well, you look at the table and say, well, Nathan's height is 4' 11\". Nathan is 4' 11\". There are not two heights for Nathan. There's only one height."}, {"video_title": "Recognize functions from tables Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Well, you look at the table and say, well, Nathan's height is 4' 11\". Nathan is 4' 11\". There are not two heights for Nathan. There's only one height. And for any one of these people that we can input into the function, there's only one height associated with them. So it is a functional relationship. And we can even see that on a graph."}, {"video_title": "Recognize functions from tables Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "There's only one height. And for any one of these people that we can input into the function, there's only one height associated with them. So it is a functional relationship. And we can even see that on a graph. Let me graph that out for you. So let's graph out. So let's see."}, {"video_title": "Recognize functions from tables Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And we can even see that on a graph. Let me graph that out for you. So let's graph out. So let's see. The highest height here is 6' 1\". So if we start off with, well, we don't have to start. Well, let me just start at 1 foot, 2 feet, 3 feet, 4 feet, 5 feet, 5 feet, and 6 feet."}, {"video_title": "Recognize functions from tables Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So let's see. The highest height here is 6' 1\". So if we start off with, well, we don't have to start. Well, let me just start at 1 foot, 2 feet, 3 feet, 4 feet, 5 feet, 5 feet, and 6 feet. And then if I were to plot the different names, the different people that I could put into our height function, we have, I'll just put the first letters of their names. We have Joel. We have Nathan."}, {"video_title": "Recognize functions from tables Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Well, let me just start at 1 foot, 2 feet, 3 feet, 4 feet, 5 feet, 5 feet, and 6 feet. And then if I were to plot the different names, the different people that I could put into our height function, we have, I'll just put the first letters of their names. We have Joel. We have Nathan. We have Stuart. We have LJ. And then we have Tariq right there."}, {"video_title": "Recognize functions from tables Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "We have Nathan. We have Stuart. We have LJ. And then we have Tariq right there. So let's plot them. So you have Joel. Joel's height is 5' 6\"."}, {"video_title": "Recognize functions from tables Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And then we have Tariq right there. So let's plot them. So you have Joel. Joel's height is 5' 6\". So 5' 6 is right about there. Then you have Nathan. Nathan's height, let me do it in a different color."}, {"video_title": "Recognize functions from tables Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Joel's height is 5' 6\". So 5' 6 is right about there. Then you have Nathan. Nathan's height, let me do it in a different color. Nathan's height is 4' 11\". So Nathan's height is 4' 11\". We will plot him right over there."}, {"video_title": "Recognize functions from tables Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Nathan's height, let me do it in a different color. Nathan's height is 4' 11\". So Nathan's height is 4' 11\". We will plot him right over there. Then you have Stuart. Stuart's height is 5' 11\". So Stuart's height is 5' 11\"."}, {"video_title": "Recognize functions from tables Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "We will plot him right over there. Then you have Stuart. Stuart's height is 5' 11\". So Stuart's height is 5' 11\". He is pretty close to 6 feet. So Stuart's height is 5' 11\". Then you have LJ."}, {"video_title": "Recognize functions from tables Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So Stuart's height is 5' 11\". He is pretty close to 6 feet. So Stuart's height is 5' 11\". Then you have LJ. LJ's height is 5' 6\". So you have two people with a height of 5' 6\". But that's OK. As long as for each person, you only have one height."}, {"video_title": "Recognize functions from tables Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Then you have LJ. LJ's height is 5' 6\". So you have two people with a height of 5' 6\". But that's OK. As long as for each person, you only have one height. And then finally, Tariq is 6' 1\". He's the tallest guy here. Tariq is right up here at 6' 1\"."}, {"video_title": "Recognize functions from tables Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "But that's OK. As long as for each person, you only have one height. And then finally, Tariq is 6' 1\". He's the tallest guy here. Tariq is right up here at 6' 1\". So notice, for any one of the inputs into our function, we only have one value. So this is a functional relationship. Now, you might say, OK, well, isn't everything a functional relationship?"}, {"video_title": "Recognize functions from tables Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Tariq is right up here at 6' 1\". So notice, for any one of the inputs into our function, we only have one value. So this is a functional relationship. Now, you might say, OK, well, isn't everything a functional relationship? No. If I gave you the situation, if I also wrote here, let's say the table was like this, and I also wrote that Stuart is 5' 3\". If this was our table, then we would no longer have a functional relationship."}, {"video_title": "Recognize functions from tables Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Now, you might say, OK, well, isn't everything a functional relationship? No. If I gave you the situation, if I also wrote here, let's say the table was like this, and I also wrote that Stuart is 5' 3\". If this was our table, then we would no longer have a functional relationship. Because for the input of Stuart, we would have two different values. If we were to graph this, we have Stuart here at 5' 11\". And then all of a sudden, we would also have Stuart at 5' 3\"."}, {"video_title": "Recognize functions from tables Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "If this was our table, then we would no longer have a functional relationship. Because for the input of Stuart, we would have two different values. If we were to graph this, we have Stuart here at 5' 11\". And then all of a sudden, we would also have Stuart at 5' 3\". Now, this doesn't make a lot of sense. So we would plot it right over here. So for Stuart, you would have two values."}, {"video_title": "Recognize functions from tables Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And then all of a sudden, we would also have Stuart at 5' 3\". Now, this doesn't make a lot of sense. So we would plot it right over here. So for Stuart, you would have two values. And so this wouldn't be a valid functional relationship. Because you wouldn't know what value to give if you were to take the height of Stuart. In order for this to be a function, there can only be one value for this."}, {"video_title": "Recognize functions from tables Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So for Stuart, you would have two values. And so this wouldn't be a valid functional relationship. Because you wouldn't know what value to give if you were to take the height of Stuart. In order for this to be a function, there can only be one value for this. You don't know in this situation, when I add this, whether it's 5' 3\", or 5' 11\". Now, this wasn't the case. So that isn't there."}, {"video_title": "Recognize functions from tables Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "In order for this to be a function, there can only be one value for this. You don't know in this situation, when I add this, whether it's 5' 3\", or 5' 11\". Now, this wasn't the case. So that isn't there. And so we know that the height of Stuart is 5' 11\". And this is a functional relationship. I think to some level, it might be confusing."}, {"video_title": "Recognize functions from tables Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So that isn't there. And so we know that the height of Stuart is 5' 11\". And this is a functional relationship. I think to some level, it might be confusing. Because it's such a simple idea. Each of these values can only have one height associated with it. That's what makes it a function."}, {"video_title": "Factorization with substitution Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "And they say that we can factor the expression as u plus v squared, where u and v are either constant integers or single variable expressions. What are u and v? And then they ask us to actually factor the expression. So pause this video and see if you can work on that. All right, so let's go with the first part of it. So they say they can factor the expression as u plus v squared. So how do we see this expression in terms of u plus v squared?"}, {"video_title": "Factorization with substitution Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "So pause this video and see if you can work on that. All right, so let's go with the first part of it. So they say they can factor the expression as u plus v squared. So how do we see this expression in terms of u plus v squared? Well, one way is to just remind ourselves what u plus v squared even is. U plus v squared, this is just going to be the square of a binomial, and we've seen this in many, many other videos. This is going to be u squared plus two times the product of these two terms, so two uv plus v squared."}, {"video_title": "Factorization with substitution Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "So how do we see this expression in terms of u plus v squared? Well, one way is to just remind ourselves what u plus v squared even is. U plus v squared, this is just going to be the square of a binomial, and we've seen this in many, many other videos. This is going to be u squared plus two times the product of these two terms, so two uv plus v squared. If you've never seen this before and you're not sure where this came from, I encourage you to watch some of those early videos where we explain this out. But does this match this pattern? Well, can we express this term as u squared?"}, {"video_title": "Factorization with substitution Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "This is going to be u squared plus two times the product of these two terms, so two uv plus v squared. If you've never seen this before and you're not sure where this came from, I encourage you to watch some of those early videos where we explain this out. But does this match this pattern? Well, can we express this term as u squared? Well, if this is u squared, then u would have to be equal to x plus seven. And when I say, actually, let me be a little careful. Can we express this entire thing right over here as u squared?"}, {"video_title": "Factorization with substitution Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "Well, can we express this term as u squared? Well, if this is u squared, then u would have to be equal to x plus seven. And when I say, actually, let me be a little careful. Can we express this entire thing right over here as u squared? If u squared is equal to x plus seven squared, that means that u is going to be equal to x plus seven. And then this right over here would have to be v squared. If this is v squared, then that means that v is equal to y squared, because y squared squared is equal to y to the fourth."}, {"video_title": "Factorization with substitution Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "Can we express this entire thing right over here as u squared? If u squared is equal to x plus seven squared, that means that u is going to be equal to x plus seven. And then this right over here would have to be v squared. If this is v squared, then that means that v is equal to y squared, because y squared squared is equal to y to the fourth. So v is equal to y squared. Now, they already told us that this can be factored as the expression u plus v squared, but let's make sure that this actually works. Is this middle term right over here, is this truly equal to two times u times v, two uv?"}, {"video_title": "Factorization with substitution Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "If this is v squared, then that means that v is equal to y squared, because y squared squared is equal to y to the fourth. So v is equal to y squared. Now, they already told us that this can be factored as the expression u plus v squared, but let's make sure that this actually works. Is this middle term right over here, is this truly equal to two times u times v, two uv? Well, let's see. Two times u would be two times x plus seven times v times y squared. And that's exactly what we have right over here."}, {"video_title": "Factorization with substitution Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "Is this middle term right over here, is this truly equal to two times u times v, two uv? Well, let's see. Two times u would be two times x plus seven times v times y squared. And that's exactly what we have right over here. It's two y squared times x plus seven. So this kind of hairy-looking expression actually does fit this pattern right over here. So you can view it as u plus v squared, where u is equal to x plus seven, and v is equal to y squared."}, {"video_title": "Factorization with substitution Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "And that's exactly what we have right over here. It's two y squared times x plus seven. So this kind of hairy-looking expression actually does fit this pattern right over here. So you can view it as u plus v squared, where u is equal to x plus seven, and v is equal to y squared. Now, using that, we can now actually factor the expression. We can write this thing as being equal to u plus v squared. And we know what u and v are."}, {"video_title": "Factorization with substitution Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "So you can view it as u plus v squared, where u is equal to x plus seven, and v is equal to y squared. Now, using that, we can now actually factor the expression. We can write this thing as being equal to u plus v squared. And we know what u and v are. So this whole expression is going to be equal to u, which is x plus seven, and I'll put it in parentheses just so you see it very clearly, plus v plus y squared squared, because that's exactly what we wrote over there. And of course, you don't have to write these parentheses. You could rewrite this as x plus seven plus y squared squared."}, {"video_title": "Factorization with substitution Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "And we know what u and v are. So this whole expression is going to be equal to u, which is x plus seven, and I'll put it in parentheses just so you see it very clearly, plus v plus y squared squared, because that's exactly what we wrote over there. And of course, you don't have to write these parentheses. You could rewrite this as x plus seven plus y squared squared. Let's do another example. So here, once again, we are told that we want to factor the following expression, and they're saying that we can factor the expression as u plus v times u minus v, where u and v are either constant integers or single-variable expressions. So pause this video and try to figure out what u and v are, and then actually factor the expression."}, {"video_title": "Factorization with substitution Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "You could rewrite this as x plus seven plus y squared squared. Let's do another example. So here, once again, we are told that we want to factor the following expression, and they're saying that we can factor the expression as u plus v times u minus v, where u and v are either constant integers or single-variable expressions. So pause this video and try to figure out what u and v are, and then actually factor the expression. All right, well, let's just remind ourselves in general what u plus v times u minus v is equal to. Well, if this is unfamiliar to you, I encourage you to watch the videos on difference of squares. But when you multiply this all out, this is going to give you a difference of squares, u squared minus v squared."}, {"video_title": "Factorization with substitution Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "So pause this video and try to figure out what u and v are, and then actually factor the expression. All right, well, let's just remind ourselves in general what u plus v times u minus v is equal to. Well, if this is unfamiliar to you, I encourage you to watch the videos on difference of squares. But when you multiply this all out, this is going to give you a difference of squares, u squared minus v squared. If you actually take the trouble of multiplying this out, you're going to see that that middle term, that middle third terms, or the middle terms, I should say, cancel out, so you're just left with the u squared minus the v squared. And so does this fit this pattern? Well, in order for this to be u squared and for this to be v squared, that means u squared is equal to four x squared, so that means that u would have to be equal to the square root of that, which would be two times x."}, {"video_title": "Factorization with substitution Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "But when you multiply this all out, this is going to give you a difference of squares, u squared minus v squared. If you actually take the trouble of multiplying this out, you're going to see that that middle term, that middle third terms, or the middle terms, I should say, cancel out, so you're just left with the u squared minus the v squared. And so does this fit this pattern? Well, in order for this to be u squared and for this to be v squared, that means u squared is equal to four x squared, so that means that u would have to be equal to the square root of that, which would be two times x. Notice, u squared would be two x squared, which is four x squared, and then v would have to be equal to the square root of nine y to the sixth, which the square root of nine is three, and the square root of y to the sixth is going to be y to the third power. And then we could use that to factor the expression, because we could say, hey, this right over here is the same thing as u squared minus v squared, so it's going to be equal to, we can factor it out as, or factor it as, u minus or u plus v times u minus v. So what's that going to be equal to? So u plus v is going to be equal to two x plus three y to the third, and then u minus v is going to be equal to two x, which is our u right over here, minus our v, minus three y to the third."}, {"video_title": "Factorization with substitution Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "Well, in order for this to be u squared and for this to be v squared, that means u squared is equal to four x squared, so that means that u would have to be equal to the square root of that, which would be two times x. Notice, u squared would be two x squared, which is four x squared, and then v would have to be equal to the square root of nine y to the sixth, which the square root of nine is three, and the square root of y to the sixth is going to be y to the third power. And then we could use that to factor the expression, because we could say, hey, this right over here is the same thing as u squared minus v squared, so it's going to be equal to, we can factor it out as, or factor it as, u minus or u plus v times u minus v. So what's that going to be equal to? So u plus v is going to be equal to two x plus three y to the third, and then u minus v is going to be equal to two x, which is our u right over here, minus our v, minus three y to the third. So there you had it. We factored the expression. You might want to write it down here, but we just did it right up there, and we're done."}, {"video_title": "Fractional exponent expressions 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "So first we can look at this first expression right here, where we're taking this product to the second power. We know that instead we can take each of these, each of the terms in the product to the second power, and then take the product. So this is going to be the same thing as r to the 2 3rd squared times s to the 3rd squared. And now let's look at this radical over here. We have the square root, but that's the exact same thing as raising something to the 1 half power. So this is equal to, so times this part, let me do this in a different color, this part right here, that is the same thing as 20. And instead of just writing 20, let me write 20 as a product of a perfect square and a non-perfect square."}, {"video_title": "Fractional exponent expressions 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "And now let's look at this radical over here. We have the square root, but that's the exact same thing as raising something to the 1 half power. So this is equal to, so times this part, let me do this in a different color, this part right here, that is the same thing as 20. And instead of just writing 20, let me write 20 as a product of a perfect square and a non-perfect square. So 20 is the same thing as 4 times 5, that's the 20 part, times r to the 4th, times s to the 5th. Now let me write s to the 5th also as a product of a perfect square and a non-perfect square. r to the 4th is obviously a perfect square, its square root is r squared."}, {"video_title": "Fractional exponent expressions 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "And instead of just writing 20, let me write 20 as a product of a perfect square and a non-perfect square. So 20 is the same thing as 4 times 5, that's the 20 part, times r to the 4th, times s to the 5th. Now let me write s to the 5th also as a product of a perfect square and a non-perfect square. r to the 4th is obviously a perfect square, its square root is r squared. But let's write s to the 5th in a similar way. So s to the 5th we can rewrite as s to the 4th times s. Right, s to the 4th times s to the 1st, that is s to the 5th. And of course all of this has to be raised to the 1 half power."}, {"video_title": "Fractional exponent expressions 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "r to the 4th is obviously a perfect square, its square root is r squared. But let's write s to the 5th in a similar way. So s to the 5th we can rewrite as s to the 4th times s. Right, s to the 4th times s to the 1st, that is s to the 5th. And of course all of this has to be raised to the 1 half power. Now let's simplify this even more. If we're taking something to the 2 3rds power, and then to the 2nd power, we can just multiply the exponents. So this term right here, we can simplify this as r to the 4 3rds power."}, {"video_title": "Fractional exponent expressions 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "And of course all of this has to be raised to the 1 half power. Now let's simplify this even more. If we're taking something to the 2 3rds power, and then to the 2nd power, we can just multiply the exponents. So this term right here, we can simplify this as r to the 4 3rds power. And just as a bit of review, taking something to the 4 3rds power, you can view it as either finding its cube root, taking it to the 1 3rd power, and then taking its cube root to the 4th power, or you can view it as taking it to the 4th power and then finding the cube root of that. Those are both legitimate ways of something being raised to the 4 3rds power. So you have r to the 4 3rds times s to the 3 times 2, times s to the 6th power."}, {"video_title": "Fractional exponent expressions 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "So this term right here, we can simplify this as r to the 4 3rds power. And just as a bit of review, taking something to the 4 3rds power, you can view it as either finding its cube root, taking it to the 1 3rd power, and then taking its cube root to the 4th power, or you can view it as taking it to the 4th power and then finding the cube root of that. Those are both legitimate ways of something being raised to the 4 3rds power. So you have r to the 4 3rds times s to the 3 times 2, times s to the 6th power. And then we could raise each of these terms right here to the 1 half power. So times, let me color code it a little bit. Times, and we actually would need the parentheses once we do that, times 4 to the 1 half, times 5 to the 1 half, that's that term right there, times r to the 4th to the 1 half power, times, I might run out of colors, s to the 4th to the 1 half power, raising each of these terms to that 1 half power."}, {"video_title": "Fractional exponent expressions 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "So you have r to the 4 3rds times s to the 3 times 2, times s to the 6th power. And then we could raise each of these terms right here to the 1 half power. So times, let me color code it a little bit. Times, and we actually would need the parentheses once we do that, times 4 to the 1 half, times 5 to the 1 half, that's that term right there, times r to the 4th to the 1 half power, times, I might run out of colors, s to the 4th to the 1 half power, raising each of these terms to that 1 half power. Times s to the 1 half power. And there's a lot of ways we can go with this, but the one thing that might jump out is that there are some perfect squares here, and we're raising them to the 1 half power, we're taking their square root, so let's simplify those. So this 4 to the 1 half, that's the same thing as 2, we're taking the principle root of 4."}, {"video_title": "Fractional exponent expressions 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "Times, and we actually would need the parentheses once we do that, times 4 to the 1 half, times 5 to the 1 half, that's that term right there, times r to the 4th to the 1 half power, times, I might run out of colors, s to the 4th to the 1 half power, raising each of these terms to that 1 half power. Times s to the 1 half power. And there's a lot of ways we can go with this, but the one thing that might jump out is that there are some perfect squares here, and we're raising them to the 1 half power, we're taking their square root, so let's simplify those. So this 4 to the 1 half, that's the same thing as 2, we're taking the principle root of 4. 5 to the 1 half, well, we can't take the square root of that, so let's just write that as the square root of 5. Square root of 5. r to the 4th to the 1 half, 4 times 1 half, there's two ways you can think about it, 4 times 1 half is 2, so this is r squared, or you could say the square root of r to the 4th is r squared. So this is r squared."}, {"video_title": "Fractional exponent expressions 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "So this 4 to the 1 half, that's the same thing as 2, we're taking the principle root of 4. 5 to the 1 half, well, we can't take the square root of that, so let's just write that as the square root of 5. Square root of 5. r to the 4th to the 1 half, 4 times 1 half, there's two ways you can think about it, 4 times 1 half is 2, so this is r squared, or you could say the square root of r to the 4th is r squared. So this is r squared. Similarly, the square root of s to the 4th, or s to the 4th to the 1 half is also s squared. And then this s to the 1 half, let's just write that as the square root of s. Square root of s. Just like that. And let's see what else we can do here."}, {"video_title": "Fractional exponent expressions 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "So this is r squared. Similarly, the square root of s to the 4th, or s to the 4th to the 1 half is also s squared. And then this s to the 1 half, let's just write that as the square root of s. Square root of s. Just like that. And let's see what else we can do here. So we have, let me write these other terms, we have an r to the 4 thirds, times s to the 6th, times 2 times the square root of 5, times r squared, times s squared, times the square root of s. Now, a couple of things we can do here, we can combine these s terms, let's do that. Actually, let's write the 2 out front first. So let's write the 2 out front first, so you have 2 times, now let's look at these 2 s terms over here."}, {"video_title": "Fractional exponent expressions 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "And let's see what else we can do here. So we have, let me write these other terms, we have an r to the 4 thirds, times s to the 6th, times 2 times the square root of 5, times r squared, times s squared, times the square root of s. Now, a couple of things we can do here, we can combine these s terms, let's do that. Actually, let's write the 2 out front first. So let's write the 2 out front first, so you have 2 times, now let's look at these 2 s terms over here. We have s to the 6th times s squared. And when someone says to simplify it, there's multiple interpretations for it, but we'll just say, s to the 6th times s squared, that's s to the 8th, right? 6 plus 2 times s to the 8th power, times, now this one's interesting, and we might want to break it up depending on what we consider to be truly simplified."}, {"video_title": "Fractional exponent expressions 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "So let's write the 2 out front first, so you have 2 times, now let's look at these 2 s terms over here. We have s to the 6th times s squared. And when someone says to simplify it, there's multiple interpretations for it, but we'll just say, s to the 6th times s squared, that's s to the 8th, right? 6 plus 2 times s to the 8th power, times, now this one's interesting, and we might want to break it up depending on what we consider to be truly simplified. We have r to the 4 third times r squared. r to the 4 thirds is the same thing as r to the 1 and 1 third, that's what 4 thirds is. So 1 and 1 third, 1 and 1 third, plus 2 is 3 and 1 third."}, {"video_title": "Fractional exponent expressions 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "6 plus 2 times s to the 8th power, times, now this one's interesting, and we might want to break it up depending on what we consider to be truly simplified. We have r to the 4 third times r squared. r to the 4 thirds is the same thing as r to the 1 and 1 third, that's what 4 thirds is. So 1 and 1 third, 1 and 1 third, plus 2 is 3 and 1 third. So we could write this times r to the 3 and 1 third. And that's a little inconsistent over here, I'm adding a fraction over here, with the s I kind of left out the s to the 1 half from the s's here. But we could play around with it and all of those would be valid expressions."}, {"video_title": "Fractional exponent expressions 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "So 1 and 1 third, 1 and 1 third, plus 2 is 3 and 1 third. So we could write this times r to the 3 and 1 third. And that's a little inconsistent over here, I'm adding a fraction over here, with the s I kind of left out the s to the 1 half from the s's here. But we could play around with it and all of those would be valid expressions. So we've already dealt with the 2, we've already dealt with these 2 s's, we've already dealt with these r's, and then you have the square root of 5 times the square root of s. And we could merge them if we want, but I won't do it just yet. Times the square root of 5 times the square root of s. Now there's 2 ways we could do it, we might not like having a fractional exponent here, and then we could break it out, or we might want to take this guy and merge it with the 8th power. Because you know that this is the same thing as s to the 1 half, so let's do it both ways."}, {"video_title": "Fractional exponent expressions 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "But we could play around with it and all of those would be valid expressions. So we've already dealt with the 2, we've already dealt with these 2 s's, we've already dealt with these r's, and then you have the square root of 5 times the square root of s. And we could merge them if we want, but I won't do it just yet. Times the square root of 5 times the square root of s. Now there's 2 ways we could do it, we might not like having a fractional exponent here, and then we could break it out, or we might want to take this guy and merge it with the 8th power. Because you know that this is the same thing as s to the 1 half, so let's do it both ways. So if we wanted to merge all of the exponents, we could write this as 2 times s to the 8th times s to the 1 half. So s to the 8th and s to the 1 half, that would be 2 times s to the 8, I could even write it as a decimal, 8.5 right? 8 plus, you could imagine this is s to the.5 power."}, {"video_title": "Fractional exponent expressions 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "Because you know that this is the same thing as s to the 1 half, so let's do it both ways. So if we wanted to merge all of the exponents, we could write this as 2 times s to the 8th times s to the 1 half. So s to the 8th and s to the 1 half, that would be 2 times s to the 8, I could even write it as a decimal, 8.5 right? 8 plus, you could imagine this is s to the.5 power. So that's 8.5 times r to the 3 and 1 third, I'm kind of mixing notations here, I have just a decimal notation, then I have a fraction notation, mixed number notation, times the square root of 5. This is one simplification, I kind of have it in the fewest terms possible. The other simplification, if you don't want to have these fractional exponents out here, you could write it as, I'll do this in a different color, you could write this, and these are all equivalent statements, so it's up to debate what simplified really means."}, {"video_title": "Fractional exponent expressions 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "8 plus, you could imagine this is s to the.5 power. So that's 8.5 times r to the 3 and 1 third, I'm kind of mixing notations here, I have just a decimal notation, then I have a fraction notation, mixed number notation, times the square root of 5. This is one simplification, I kind of have it in the fewest terms possible. The other simplification, if you don't want to have these fractional exponents out here, you could write it as, I'll do this in a different color, you could write this, and these are all equivalent statements, so it's up to debate what simplified really means. So you could write this as 2 times s to the 8th, instead of writing r to the 3 and 1 third, we could write r to the 3rd, r to the 3rd, times the cube root of r, which is the same thing as r to the 1 third. We could write r to the 3rd times r to the 1 third, r to the 1 third is the same thing as the cube root of r, and then you have the square root of these two guys, both of these guys are being raised to the 1 half power, so you could then say times the square root of 5s. I like this one a little bit more, the one on the left, to me this is really simplified, we've merged all of the bases, we have these two numbers here, we've merged all the s terms, all the r terms, this is a little bit more complicated, you have a cube root, you have separated the s's and the r's, so I would go with this one if someone really wanted me, and said hey Sal, simplify it how you like."}, {"video_title": "Example visually evaluating discrete functions.mp3", "Sentence": "Now when a lot of people see function notation like this, they see it as somewhat intimidating until you realize what it's saying. All a function is is something that takes an input, in this case, it's taking x as an input, and then the function does something to it, and then it spits out some other value which is going to be equal to y. So for example, what is h of four based on this graph that you see right over here? Pause this video and think through that. Well, all h of four means is when I input four into my function h, what y am I spitting out? Or another way to think about it, when x is equal to four, what is y equal to? Well, when x is equal to four, my function spits out that y is equal to three."}, {"video_title": "Example visually evaluating discrete functions.mp3", "Sentence": "Pause this video and think through that. Well, all h of four means is when I input four into my function h, what y am I spitting out? Or another way to think about it, when x is equal to four, what is y equal to? Well, when x is equal to four, my function spits out that y is equal to three. We know that from this point right over here. So y is equal to three, so h of four is equal to three. Let's do another example."}, {"video_title": "Example visually evaluating discrete functions.mp3", "Sentence": "Well, when x is equal to four, my function spits out that y is equal to three. We know that from this point right over here. So y is equal to three, so h of four is equal to three. Let's do another example. What is h of zero? Pause the video, try to work that through. Well, all this is saying is if I input x equals zero into the function, what is going to be the corresponding y?"}, {"video_title": "Example visually evaluating discrete functions.mp3", "Sentence": "Let's do another example. What is h of zero? Pause the video, try to work that through. Well, all this is saying is if I input x equals zero into the function, what is going to be the corresponding y? Well, when x equals zero, we see that y is equal to four. So it's as simple as that. Given the input, what is going to be the output?"}, {"video_title": "Example visually evaluating discrete functions.mp3", "Sentence": "Well, all this is saying is if I input x equals zero into the function, what is going to be the corresponding y? Well, when x equals zero, we see that y is equal to four. So it's as simple as that. Given the input, what is going to be the output? And that's what these points represent. Each of these points represent a different output for a given input. Now, it's always good to keep in mind one of the things that makes it a function is that for a given x that you input, you only get one y."}, {"video_title": "Example visually evaluating discrete functions.mp3", "Sentence": "Given the input, what is going to be the output? And that's what these points represent. Each of these points represent a different output for a given input. Now, it's always good to keep in mind one of the things that makes it a function is that for a given x that you input, you only get one y. For example, if we had two dots here, then all of a sudden, we have two dots for x equals six. Now, all of a sudden, we have a problem at figuring out what h of six would be equal to, because it could be equal to one, or it could be equal to three. So if we had this extra dot here, then this would no longer be a function."}, {"video_title": "Example visually evaluating discrete functions.mp3", "Sentence": "Now, it's always good to keep in mind one of the things that makes it a function is that for a given x that you input, you only get one y. For example, if we had two dots here, then all of a sudden, we have two dots for x equals six. Now, all of a sudden, we have a problem at figuring out what h of six would be equal to, because it could be equal to one, or it could be equal to three. So if we had this extra dot here, then this would no longer be a function. In order for it to be a function for any given x, it has to output a unique value. It can't output two possible values. Now, the other way is possible."}, {"video_title": "Example visually evaluating discrete functions.mp3", "Sentence": "So if we had this extra dot here, then this would no longer be a function. In order for it to be a function for any given x, it has to output a unique value. It can't output two possible values. Now, the other way is possible. It is possible to have two different x's that output the same value. For example, if this was circled in, what would h of negative four be? Well, h of negative four, when x is equal to negative four, when you put that into our function, it looks like the function would output two."}, {"video_title": "Example visually evaluating discrete functions.mp3", "Sentence": "Now, the other way is possible. It is possible to have two different x's that output the same value. For example, if this was circled in, what would h of negative four be? Well, h of negative four, when x is equal to negative four, when you put that into our function, it looks like the function would output two. So h of negative four would be equal to two. But h of two is also equal to, we see very clearly there, when we input a two into the function, the corresponding y value is two as well. So it's okay for two different x values to map to the same y value."}, {"video_title": "Example visually evaluating discrete functions.mp3", "Sentence": "Well, h of negative four, when x is equal to negative four, when you put that into our function, it looks like the function would output two. So h of negative four would be equal to two. But h of two is also equal to, we see very clearly there, when we input a two into the function, the corresponding y value is two as well. So it's okay for two different x values to map to the same y value. That works. But if you had some type of an arrangement, some type of a relationship, where for a given x value, you had two different y values, then that would no longer be a function. But the example they gave us is a function, assuming I don't modify it."}, {"video_title": "How to evaluate a piecewise function (example) Functions Algebra I Khan Academy.mp3", "Sentence": "So if t is less than or equal to negative 10, we use this case. If t is between negative 10 and negative two, we use this case. And if t is greater than or equal to negative two, we use this case. And then they ask us, what is the value of f of negative 10? So t is going to be equal to negative 10. So which case do we use? So let's see, if t is less than or equal to negative 10, we use this top case right over here."}, {"video_title": "How to evaluate a piecewise function (example) Functions Algebra I Khan Academy.mp3", "Sentence": "And then they ask us, what is the value of f of negative 10? So t is going to be equal to negative 10. So which case do we use? So let's see, if t is less than or equal to negative 10, we use this top case right over here. And t is equal to negative 10. That's the one that we're trying to evaluate. So we want to use this case right over here."}, {"video_title": "How to evaluate a piecewise function (example) Functions Algebra I Khan Academy.mp3", "Sentence": "So let's see, if t is less than or equal to negative 10, we use this top case right over here. And t is equal to negative 10. That's the one that we're trying to evaluate. So we want to use this case right over here. So f of negative 10 is going to be equal to negative 10. Everywhere we see a t here, we substitute it with a negative 10. Negative 10 squared minus five times, actually I don't have a denominator there."}, {"video_title": "How to evaluate a piecewise function (example) Functions Algebra I Khan Academy.mp3", "Sentence": "So we want to use this case right over here. So f of negative 10 is going to be equal to negative 10. Everywhere we see a t here, we substitute it with a negative 10. Negative 10 squared minus five times, actually I don't have a denominator there. I don't know why I wrote it so high. So it's going to be negative 10 squared minus five times negative 10. So let's see, negative 10 squared, that's positive 100."}, {"video_title": "How to evaluate a piecewise function (example) Functions Algebra I Khan Academy.mp3", "Sentence": "Negative 10 squared minus five times, actually I don't have a denominator there. I don't know why I wrote it so high. So it's going to be negative 10 squared minus five times negative 10. So let's see, negative 10 squared, that's positive 100. And then negative, or subtracting five times negative 10, this is going to be subtracting negative 50, or you're going to add 50. So this is going to be equal to 150. F of negative 10 is 150, because we used this case up here, because t is negative 10."}, {"video_title": "How to evaluate a piecewise function (example) Functions Algebra I Khan Academy.mp3", "Sentence": "So let's see, negative 10 squared, that's positive 100. And then negative, or subtracting five times negative 10, this is going to be subtracting negative 50, or you're going to add 50. So this is going to be equal to 150. F of negative 10 is 150, because we used this case up here, because t is negative 10. Let's do another one of these examples. So here we have, consider the following piecewise function, all right? What is the value of h of negative three?"}, {"video_title": "How to evaluate a piecewise function (example) Functions Algebra I Khan Academy.mp3", "Sentence": "F of negative 10 is 150, because we used this case up here, because t is negative 10. Let's do another one of these examples. So here we have, consider the following piecewise function, all right? What is the value of h of negative three? See, when h is negative three, which case do we use? We use this case if our x is between negative infinity and zero. And negative three is between negative infinity and zero, so we're going to use this case right over here."}, {"video_title": "How to evaluate a piecewise function (example) Functions Algebra I Khan Academy.mp3", "Sentence": "What is the value of h of negative three? See, when h is negative three, which case do we use? We use this case if our x is between negative infinity and zero. And negative three is between negative infinity and zero, so we're going to use this case right over here. If it was positive three, we would use this case. If it was positive 30, we would use this case. So we're going to use the first case again."}, {"video_title": "How to evaluate a piecewise function (example) Functions Algebra I Khan Academy.mp3", "Sentence": "And negative three is between negative infinity and zero, so we're going to use this case right over here. If it was positive three, we would use this case. If it was positive 30, we would use this case. So we're going to use the first case again. And so we're going, so for h of negative three, we're going to take negative three to the third power. So let's see, h of negative three is going to be negative three to the third power, which is negative 27. And we're done, that's h of negative three."}, {"video_title": "How to evaluate a piecewise function (example) Functions Algebra I Khan Academy.mp3", "Sentence": "So we're going to use the first case again. And so we're going, so for h of negative three, we're going to take negative three to the third power. So let's see, h of negative three is going to be negative three to the third power, which is negative 27. And we're done, that's h of negative three. Because we are using this case, you could almost just ignore these second two cases right over here. Let's do one more example. This one's a little bit different."}, {"video_title": "How to evaluate a piecewise function (example) Functions Algebra I Khan Academy.mp3", "Sentence": "And we're done, that's h of negative three. Because we are using this case, you could almost just ignore these second two cases right over here. Let's do one more example. This one's a little bit different. Below is a graph of the step function g of x. So we can see g of x right over here. It starts when x equals negative nine, it's at three, and then it jumps up, and then it jumps down."}, {"video_title": "How to evaluate a piecewise function (example) Functions Algebra I Khan Academy.mp3", "Sentence": "This one's a little bit different. Below is a graph of the step function g of x. So we can see g of x right over here. It starts when x equals negative nine, it's at three, and then it jumps up, and then it jumps down. Match each expression with its value. So g of negative 3.0001, so negative 3.0001, so that's right over here, and g of that we see is equal to three. So this is going to be equal to three right over here."}, {"video_title": "How to evaluate a piecewise function (example) Functions Algebra I Khan Academy.mp3", "Sentence": "It starts when x equals negative nine, it's at three, and then it jumps up, and then it jumps down. Match each expression with its value. So g of negative 3.0001, so negative 3.0001, so that's right over here, and g of that we see is equal to three. So this is going to be equal to three right over here. G of 3.99999, 3.99999, almost four. So let's draw the dotted line right here, it's going to be almost four. Well, g of 3.99999 is going to be seven."}, {"video_title": "How to evaluate a piecewise function (example) Functions Algebra I Khan Academy.mp3", "Sentence": "So this is going to be equal to three right over here. G of 3.99999, 3.99999, almost four. So let's draw the dotted line right here, it's going to be almost four. Well, g of 3.99999 is going to be seven. We see that right over there. So that is equal to seven. G of 4.00001."}, {"video_title": "How to evaluate a piecewise function (example) Functions Algebra I Khan Academy.mp3", "Sentence": "Well, g of 3.99999 is going to be seven. We see that right over there. So that is equal to seven. G of 4.00001. So g of four is still seven, but as soon as we go above four, we drop down over here. So g of 4.00001 is going to be negative three. I want to, actually, let's focus on that a little bit more."}, {"video_title": "How to evaluate a piecewise function (example) Functions Algebra I Khan Academy.mp3", "Sentence": "G of 4.00001. So g of four is still seven, but as soon as we go above four, we drop down over here. So g of 4.00001 is going to be negative three. I want to, actually, let's focus on that a little bit more. How did I know that? Well, I know that g of four is seven and not negative three because we have this dot is circled in up here, and it's hollow down here. But as soon as we get any amount larger than four, then the function drops down to this."}, {"video_title": "How to evaluate a piecewise function (example) Functions Algebra I Khan Academy.mp3", "Sentence": "I want to, actually, let's focus on that a little bit more. How did I know that? Well, I know that g of four is seven and not negative three because we have this dot is circled in up here, and it's hollow down here. But as soon as we get any amount larger than four, then the function drops down to this. So 4.0000, or as many, just slightly above four, the value of our function is going to be negative three. Now let's do g of nine. So g of nine, when x is nine, we go down here."}, {"video_title": "How to evaluate a piecewise function (example) Functions Algebra I Khan Academy.mp3", "Sentence": "But as soon as we get any amount larger than four, then the function drops down to this. So 4.0000, or as many, just slightly above four, the value of our function is going to be negative three. Now let's do g of nine. So g of nine, when x is nine, we go down here. You might be tempted to say it's negative three, but you see at this point right over here, we have an open circle. So that means that, well, it's not, you can't say that the function is negative three right over there, and there's no other place where we have a filled in circle for x equals nine. So the function g actually isn't defined at x equals nine."}, {"video_title": "Complex conjugates example Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And what you're going to find in this video is finding the conjugate of a complex number is shockingly easy. It's really the same as this number. I should say be a little bit more particular. It has the same real part, so the conjugate of this is going to have the exact same real part, but its imaginary part is going to have the opposite sign. So instead of having a negative 5i, it will have a positive 5i. So that right there is the complex conjugate of 7 minus 5i. And sometimes the notation for doing that is you'll take 7 minus 5i."}, {"video_title": "Complex conjugates example Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "It has the same real part, so the conjugate of this is going to have the exact same real part, but its imaginary part is going to have the opposite sign. So instead of having a negative 5i, it will have a positive 5i. So that right there is the complex conjugate of 7 minus 5i. And sometimes the notation for doing that is you'll take 7 minus 5i. If you have 7 minus 5i, and you put a line over it like that, that means I want the conjugate of 7 minus 5i, and that will equal 7 plus 5i. Or sometimes someone will write, you'll see z is the variable that people often use for complex numbers. So if z is 7 minus 5i, then they'll say the complex conjugate of z, and you put that line over the z, is going to be 7 plus 5i."}, {"video_title": "Complex conjugates example Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And sometimes the notation for doing that is you'll take 7 minus 5i. If you have 7 minus 5i, and you put a line over it like that, that means I want the conjugate of 7 minus 5i, and that will equal 7 plus 5i. Or sometimes someone will write, you'll see z is the variable that people often use for complex numbers. So if z is 7 minus 5i, then they'll say the complex conjugate of z, and you put that line over the z, is going to be 7 plus 5i. Now you're probably saying, OK, fairly straightforward to find a conjugate of a complex number, but what is it good for? And the simplest reason, or the most basic place where this is useful, is when you multiply any imaginary, any complex number times its conjugate, you're going to get a real number. And I want to emphasize, this right here is the conjugate, 7 plus 5i is the conjugate of 7 minus 5i, but 7 minus 5i is also the conjugate of 7 plus 5i, for obvious reasons."}, {"video_title": "Complex conjugates example Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So if z is 7 minus 5i, then they'll say the complex conjugate of z, and you put that line over the z, is going to be 7 plus 5i. Now you're probably saying, OK, fairly straightforward to find a conjugate of a complex number, but what is it good for? And the simplest reason, or the most basic place where this is useful, is when you multiply any imaginary, any complex number times its conjugate, you're going to get a real number. And I want to emphasize, this right here is the conjugate, 7 plus 5i is the conjugate of 7 minus 5i, but 7 minus 5i is also the conjugate of 7 plus 5i, for obvious reasons. If you started with this, and you changed the sign of the imaginary part, you would get 7 minus 5i. They're conjugates of each other. But let me show you that when I multiply complex conjugates, that I get a real number."}, {"video_title": "Complex conjugates example Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And I want to emphasize, this right here is the conjugate, 7 plus 5i is the conjugate of 7 minus 5i, but 7 minus 5i is also the conjugate of 7 plus 5i, for obvious reasons. If you started with this, and you changed the sign of the imaginary part, you would get 7 minus 5i. They're conjugates of each other. But let me show you that when I multiply complex conjugates, that I get a real number. So let's multiply 7 minus 5i times 7 plus 5i. And I will do that in blue. 7 minus 5i times 7 plus 5i."}, {"video_title": "Complex conjugates example Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "But let me show you that when I multiply complex conjugates, that I get a real number. So let's multiply 7 minus 5i times 7 plus 5i. And I will do that in blue. 7 minus 5i times 7 plus 5i. And remember, whenever you multiply these expressions, you really just have to multiply every term times each other. You could do the distributive property twice. You could do something like FOIL to remind yourself to multiply every part of this complex number times every part of this complex number."}, {"video_title": "Complex conjugates example Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "7 minus 5i times 7 plus 5i. And remember, whenever you multiply these expressions, you really just have to multiply every term times each other. You could do the distributive property twice. You could do something like FOIL to remind yourself to multiply every part of this complex number times every part of this complex number. So let's just do it any which way. So you'd have 7 times 7, which is 49. 7 times 5i, which is 35i."}, {"video_title": "Complex conjugates example Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "You could do something like FOIL to remind yourself to multiply every part of this complex number times every part of this complex number. So let's just do it any which way. So you'd have 7 times 7, which is 49. 7 times 5i, which is 35i. Then you have negative 5i times 7, which is negative 35i. You can see the imaginary part is canceling out. Then you have negative 5i times positive 5i."}, {"video_title": "Complex conjugates example Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "7 times 5i, which is 35i. Then you have negative 5i times 7, which is negative 35i. You can see the imaginary part is canceling out. Then you have negative 5i times positive 5i. Well, that's negative 25i squared. And negative 25i squared, remember, i squared is negative 1. So negative 25i squared, let me write this down."}, {"video_title": "Complex conjugates example Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Then you have negative 5i times positive 5i. Well, that's negative 25i squared. And negative 25i squared, remember, i squared is negative 1. So negative 25i squared, let me write this down. Negative 5i times 5i is negative 25 times i squared. i squared is negative 1. So negative 25 times negative 1 is positive 25."}, {"video_title": "Complex conjugates example Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So negative 25i squared, let me write this down. Negative 5i times 5i is negative 25 times i squared. i squared is negative 1. So negative 25 times negative 1 is positive 25. And these two guys over here cancel out. And we're just left with 49 plus 25. See, 50 plus 25 is 75."}, {"video_title": "Complex conjugates example Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So negative 25 times negative 1 is positive 25. And these two guys over here cancel out. And we're just left with 49 plus 25. See, 50 plus 25 is 75. So this is just 74. So we are just left with the real number 74. Another way to do it, you don't even have to do all this distributive property, you might just recognize that this looks like this is something plus something times, or something minus something times that same something plus something."}, {"video_title": "Complex conjugates example Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "See, 50 plus 25 is 75. So this is just 74. So we are just left with the real number 74. Another way to do it, you don't even have to do all this distributive property, you might just recognize that this looks like this is something plus something times, or something minus something times that same something plus something. And we know this pattern from our early algebra that a plus b times a minus b is equal to a squared minus b squared. It's equal to a difference of squares. And so in this case, a is 7, a squared is 49."}, {"video_title": "Complex conjugates example Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Another way to do it, you don't even have to do all this distributive property, you might just recognize that this looks like this is something plus something times, or something minus something times that same something plus something. And we know this pattern from our early algebra that a plus b times a minus b is equal to a squared minus b squared. It's equal to a difference of squares. And so in this case, a is 7, a squared is 49. And b, b in this case is 5i. b squared is 5i squared, which is 25i squared, which is negative 25. And we're subtracting that, so it's going to be positive plus 25."}, {"video_title": "Solving equations with the distributive property Linear equations Algebra I Khan Academy.mp3", "Sentence": "Now a good place to start is to just get rid of these parentheses, and the best way to get rid of these parentheses is to kind of multiply them out. So this has a negative 1, you just see a minus here, but it's really the same thing as having a negative 1 times this quantity, and here you have a 3 times this quantity. So let's multiply it out using the distributive property. So the left hand side of our equation we have our negative 9, and then we want to multiply the negative 1 times each of these terms. So negative 1 times 9x is negative 9x, and then negative 1 times negative 6 is plus 6, or positive 6. And then that is going to be equal to, let's distribute the 3, 3 times 4x is 12x, and then 3 times 6 is 18. Now what we want to do, let's combine our constant terms if we can, we have a negative 9 and a 6 here, on this side we've combined all of our like terms, we can't combine a 12x and an 18."}, {"video_title": "Solving equations with the distributive property Linear equations Algebra I Khan Academy.mp3", "Sentence": "So the left hand side of our equation we have our negative 9, and then we want to multiply the negative 1 times each of these terms. So negative 1 times 9x is negative 9x, and then negative 1 times negative 6 is plus 6, or positive 6. And then that is going to be equal to, let's distribute the 3, 3 times 4x is 12x, and then 3 times 6 is 18. Now what we want to do, let's combine our constant terms if we can, we have a negative 9 and a 6 here, on this side we've combined all of our like terms, we can't combine a 12x and an 18. So let's combine this, let's combine the negative 9 and the 6, our two constant terms on the left hand side of the equation. So we're going to have this negative 9x, so we're going to have negative 9x plus, let's see we have a negative 9 and then plus 6, so negative 9 plus 6 is negative 3. So we're going to have a negative 9x and then we have a negative 3, so minus 3 right here, that's the negative 9 plus the 6, and that is equal to 12x plus 18."}, {"video_title": "Solving equations with the distributive property Linear equations Algebra I Khan Academy.mp3", "Sentence": "Now what we want to do, let's combine our constant terms if we can, we have a negative 9 and a 6 here, on this side we've combined all of our like terms, we can't combine a 12x and an 18. So let's combine this, let's combine the negative 9 and the 6, our two constant terms on the left hand side of the equation. So we're going to have this negative 9x, so we're going to have negative 9x plus, let's see we have a negative 9 and then plus 6, so negative 9 plus 6 is negative 3. So we're going to have a negative 9x and then we have a negative 3, so minus 3 right here, that's the negative 9 plus the 6, and that is equal to 12x plus 18. Now we want to group all of the x terms on one side of the equation and all of the constant terms, the negative 3 and the positive 18 on the other side. I like to always have my x terms on the left hand side if I can, you don't have to have them on the left, so let's do that. So if I want all my x terms on the left, I have to get rid of this 12x from the right, and the best way to do that is to subtract 12x from both sides of the equation."}, {"video_title": "Solving equations with the distributive property Linear equations Algebra I Khan Academy.mp3", "Sentence": "So we're going to have a negative 9x and then we have a negative 3, so minus 3 right here, that's the negative 9 plus the 6, and that is equal to 12x plus 18. Now we want to group all of the x terms on one side of the equation and all of the constant terms, the negative 3 and the positive 18 on the other side. I like to always have my x terms on the left hand side if I can, you don't have to have them on the left, so let's do that. So if I want all my x terms on the left, I have to get rid of this 12x from the right, and the best way to do that is to subtract 12x from both sides of the equation. So let me subtract 12x from the right, and subtract 12x from the left. Now on the left hand side I have negative 9x minus 12x, so negative 9 minus 12, that's negative 21x minus 3 is equal to 12x minus 12x, well that's just nothing, that's 0, so I could write a 0 here but I don't have to write anything, that was the whole point of subtracting the 12x from the left hand side. And that is going to be equal to, so on the right hand side we just are left with an 18, we are just left with that 18 here, these guys cancelled out."}, {"video_title": "Solving equations with the distributive property Linear equations Algebra I Khan Academy.mp3", "Sentence": "So if I want all my x terms on the left, I have to get rid of this 12x from the right, and the best way to do that is to subtract 12x from both sides of the equation. So let me subtract 12x from the right, and subtract 12x from the left. Now on the left hand side I have negative 9x minus 12x, so negative 9 minus 12, that's negative 21x minus 3 is equal to 12x minus 12x, well that's just nothing, that's 0, so I could write a 0 here but I don't have to write anything, that was the whole point of subtracting the 12x from the left hand side. And that is going to be equal to, so on the right hand side we just are left with an 18, we are just left with that 18 here, these guys cancelled out. Now let's get rid of this negative 3 from the left hand side, so on the left hand side we only have x terms, on the right hand side we only have constant terms. So the best way to cancel out a negative 3 is to add 3, so it cancels out to 0, so we're going to add 3 to the left, let's add 3 to the right. And we get the left hand side of the equation, we have the negative 21x, no other x term to add or subtract here, so we have negative 21x, the negative 3 and the plus 3 or the positive 3 cancel out, that was the whole point, equals, what's 18 plus 3?"}, {"video_title": "Solving equations with the distributive property Linear equations Algebra I Khan Academy.mp3", "Sentence": "And that is going to be equal to, so on the right hand side we just are left with an 18, we are just left with that 18 here, these guys cancelled out. Now let's get rid of this negative 3 from the left hand side, so on the left hand side we only have x terms, on the right hand side we only have constant terms. So the best way to cancel out a negative 3 is to add 3, so it cancels out to 0, so we're going to add 3 to the left, let's add 3 to the right. And we get the left hand side of the equation, we have the negative 21x, no other x term to add or subtract here, so we have negative 21x, the negative 3 and the plus 3 or the positive 3 cancel out, that was the whole point, equals, what's 18 plus 3? 18 plus 3 is 21, so now we have negative 21x is equal to 21 and we want to solve for x. So if you have something times x and you just want it to be an x, let's divide by that something, and in this case that something is negative 21, so let's divide both sides of this equation by negative 21. Divide both sides by negative 21, the left hand side, negative 21 divided by negative 21, you're just left with an x, that was the whole point behind dividing by negative 21."}, {"video_title": "Solving equations with the distributive property Linear equations Algebra I Khan Academy.mp3", "Sentence": "And we get the left hand side of the equation, we have the negative 21x, no other x term to add or subtract here, so we have negative 21x, the negative 3 and the plus 3 or the positive 3 cancel out, that was the whole point, equals, what's 18 plus 3? 18 plus 3 is 21, so now we have negative 21x is equal to 21 and we want to solve for x. So if you have something times x and you just want it to be an x, let's divide by that something, and in this case that something is negative 21, so let's divide both sides of this equation by negative 21. Divide both sides by negative 21, the left hand side, negative 21 divided by negative 21, you're just left with an x, that was the whole point behind dividing by negative 21. And we get x is equal to, what's 21 divided by negative 21? Well that's just negative 1, right? You have the positive version divided by the negative version of itself, so it's just negative 1."}, {"video_title": "Solving equations with the distributive property Linear equations Algebra I Khan Academy.mp3", "Sentence": "Divide both sides by negative 21, the left hand side, negative 21 divided by negative 21, you're just left with an x, that was the whole point behind dividing by negative 21. And we get x is equal to, what's 21 divided by negative 21? Well that's just negative 1, right? You have the positive version divided by the negative version of itself, so it's just negative 1. So that is our answer. Now let's verify that this actually works for that original equation, so let's substitute negative 1 into that original equation. So we have negative 9, I'll do it over here, I'll do it in a different color than we've been using, we have negative 9 minus, that 1 wasn't there originally, it's there implicitly, minus 9 times negative 1."}, {"video_title": "Solving equations with the distributive property Linear equations Algebra I Khan Academy.mp3", "Sentence": "You have the positive version divided by the negative version of itself, so it's just negative 1. So that is our answer. Now let's verify that this actually works for that original equation, so let's substitute negative 1 into that original equation. So we have negative 9, I'll do it over here, I'll do it in a different color than we've been using, we have negative 9 minus, that 1 wasn't there originally, it's there implicitly, minus 9 times negative 1. 9 times, I'll put negative 1 in parentheses, minus 6 is equal to, well actually let me just solve for the left hand side when I substitute a negative 1 there. So the left hand side becomes negative 9 minus, 9 times negative 1 is negative 9 minus 6, and so this is negative 9 minus, in parentheses, negative 9 minus 6 is negative 15. So this is equal to negative 15, and so we get negative 9, let me make sure I did that, negative 9 minus 6, yep, negative 15."}, {"video_title": "Solving equations with the distributive property Linear equations Algebra I Khan Academy.mp3", "Sentence": "So we have negative 9, I'll do it over here, I'll do it in a different color than we've been using, we have negative 9 minus, that 1 wasn't there originally, it's there implicitly, minus 9 times negative 1. 9 times, I'll put negative 1 in parentheses, minus 6 is equal to, well actually let me just solve for the left hand side when I substitute a negative 1 there. So the left hand side becomes negative 9 minus, 9 times negative 1 is negative 9 minus 6, and so this is negative 9 minus, in parentheses, negative 9 minus 6 is negative 15. So this is equal to negative 15, and so we get negative 9, let me make sure I did that, negative 9 minus 6, yep, negative 15. So I have negative 9 minus negative 15, that's the same thing as negative 9 plus 15, which is 6. So that's what we get on the left hand side of the equation when we substitute x is equal to negative 1, we get that it equals 6. So let's see what happens when we substitute negative 1 to the right hand side of the equation."}, {"video_title": "Solving equations with the distributive property Linear equations Algebra I Khan Academy.mp3", "Sentence": "So this is equal to negative 15, and so we get negative 9, let me make sure I did that, negative 9 minus 6, yep, negative 15. So I have negative 9 minus negative 15, that's the same thing as negative 9 plus 15, which is 6. So that's what we get on the left hand side of the equation when we substitute x is equal to negative 1, we get that it equals 6. So let's see what happens when we substitute negative 1 to the right hand side of the equation. I'll do it in green. We get 3 times 4 times negative 1, plus 6, so that is 3 times negative 4 plus 6, negative 4 plus 6 is 2, so it's 3 times 2, which is also 6. So when x is equal to negative 1, you substitute here, the left hand side becomes 6, and the right hand side becomes 6."}, {"video_title": "Creating new function from composition Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "What I want to do in this video is come up with expressions that define a function composition. So for example, I want to figure out what is f of g of x? And I encourage you to pause the video and try to think about it on your own. Well, g of x in this case is the input to f of x. So wherever we see the x in this definition, that's the input. So we're going to replace the input with g of x. We're going to replace the x with g of x."}, {"video_title": "Creating new function from composition Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Well, g of x in this case is the input to f of x. So wherever we see the x in this definition, that's the input. So we're going to replace the input with g of x. We're going to replace the x with g of x. So f of g of x is going to be equal to the square root of, instead of an x, we would write a g of x, g of x squared minus 1. Now what is g of x equal to? Well, g of x is this thing right over here."}, {"video_title": "Creating new function from composition Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "We're going to replace the x with g of x. So f of g of x is going to be equal to the square root of, instead of an x, we would write a g of x, g of x squared minus 1. Now what is g of x equal to? Well, g of x is this thing right over here. So this is going to be equal to the square root of g of x is x over 1 plus x. We're going to square that minus 1. So f of g of x is also a function of x."}, {"video_title": "Creating new function from composition Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Well, g of x is this thing right over here. So this is going to be equal to the square root of g of x is x over 1 plus x. We're going to square that minus 1. So f of g of x is also a function of x. So f of g of x is the square root of, and we could write this as x squared over 1 plus x squared, but we could just leave it like this. It's equal to the square root of this whole thing, x over 1 plus x squared minus 1. Now let's go the other way around."}, {"video_title": "Creating new function from composition Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So f of g of x is also a function of x. So f of g of x is the square root of, and we could write this as x squared over 1 plus x squared, but we could just leave it like this. It's equal to the square root of this whole thing, x over 1 plus x squared minus 1. Now let's go the other way around. What is g of f of x? And once again, I encourage you to pause the video and try to think about it on your own. Well, f of x is now the input into g of x."}, {"video_title": "Creating new function from composition Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Now let's go the other way around. What is g of f of x? And once again, I encourage you to pause the video and try to think about it on your own. Well, f of x is now the input into g of x. So everywhere we see the x here, we'll replace it with f of x. So this is going to be equal to f of x over 1 plus f of x. And what's that equal to?"}, {"video_title": "Creating new function from composition Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Well, f of x is now the input into g of x. So everywhere we see the x here, we'll replace it with f of x. So this is going to be equal to f of x over 1 plus f of x. And what's that equal to? Well, f of x is equal to the square root of x squared minus 1. So it's going to be that over 1 plus the square root of x squared minus 1. So this is the composition f of g of x."}, {"video_title": "Creating new function from composition Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And what's that equal to? Well, f of x is equal to the square root of x squared minus 1. So it's going to be that over 1 plus the square root of x squared minus 1. So this is the composition f of g of x. You get this thing. This is g of f of x. You get this thing."}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So this is a screenshot of Desmos. It's an online graphing calculator. And what we're going to do is use it to understand how we can go about scaling functions. And I encourage you to go to Desmos and try it on your own, either during this video or after. So let's start with a nice, interesting function. Let's say f of x is equal to the absolute value of x. So that's pretty straightforward."}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And I encourage you to go to Desmos and try it on your own, either during this video or after. So let's start with a nice, interesting function. Let's say f of x is equal to the absolute value of x. So that's pretty straightforward. Now let's try to create a scaled version of f of x. So we could say g of x is equal to, well, I'll start with just absolute value of x. So it's the same as f of x."}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So that's pretty straightforward. Now let's try to create a scaled version of f of x. So we could say g of x is equal to, well, I'll start with just absolute value of x. So it's the same as f of x. So it just traced g of x right on top of f. But now let's multiply it by some constant. Let's multiply it by two. So notice the difference between g of x and f of x."}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So it's the same as f of x. So it just traced g of x right on top of f. But now let's multiply it by some constant. Let's multiply it by two. So notice the difference between g of x and f of x. And you can see that g of x is just two times f of x. In fact, we can write it this way. We can write g of x is equal to two times f of x."}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So notice the difference between g of x and f of x. And you can see that g of x is just two times f of x. In fact, we can write it this way. We can write g of x is equal to two times f of x. We get to the exact same place. But you can see that as our x increases, g of x increases twice as fast, at least for positive x's, on the right-hand side. And actually, as x decreases, g of x also increases twice as fast."}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "We can write g of x is equal to two times f of x. We get to the exact same place. But you can see that as our x increases, g of x increases twice as fast, at least for positive x's, on the right-hand side. And actually, as x decreases, g of x also increases twice as fast. So is that just a coincidence that we have a two here and it increased twice as fast? Well, let's put a three here. Well, now it looks like it's increasing three times as fast."}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And actually, as x decreases, g of x also increases twice as fast. So is that just a coincidence that we have a two here and it increased twice as fast? Well, let's put a three here. Well, now it looks like it's increasing three times as fast. And it does that in both directions. Now what if we were to put a 0.5 here? 0.5."}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Well, now it looks like it's increasing three times as fast. And it does that in both directions. Now what if we were to put a 0.5 here? 0.5. Well, now it looks like it's increasing half as fast. And that makes sense because we are just multiplying, we are scaling how much our f of x is. So before, when x equals one, we got to one."}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "0.5. Well, now it looks like it's increasing half as fast. And that makes sense because we are just multiplying, we are scaling how much our f of x is. So before, when x equals one, we got to one. But now when x equals one, we only get to 1 1\u20442. Before, when x equals five, we got to five. Now when we get to x equals five, we only get to 2.5."}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So before, when x equals one, we got to one. But now when x equals one, we only get to 1 1\u20442. Before, when x equals five, we got to five. Now when we get to x equals five, we only get to 2.5. So we're increasing half as fast, or we have half the slope. Now an interesting question to think about is what would happen if, instead of it just being an absolute value of x, let's say we were to have a non-zero y-intercept. So let's say, I don't know, plus six."}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Now when we get to x equals five, we only get to 2.5. So we're increasing half as fast, or we have half the slope. Now an interesting question to think about is what would happen if, instead of it just being an absolute value of x, let's say we were to have a non-zero y-intercept. So let's say, I don't know, plus six. So notice, then when we change this constant out front, it not only changes the slope, but it changes the y-intercept because we're multiplying this entire expression by 0.5. So if you multiply it by one, we're back to where we got before. And now if we multiply it by two, this should increase the y-intercept because remember, we're multiplying both of these terms by two."}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So let's say, I don't know, plus six. So notice, then when we change this constant out front, it not only changes the slope, but it changes the y-intercept because we're multiplying this entire expression by 0.5. So if you multiply it by one, we're back to where we got before. And now if we multiply it by two, this should increase the y-intercept because remember, we're multiplying both of these terms by two. And we see that. It not only doubles the slope, but it also increases the y-intercept. If we go to 0.5, not only did it decrease the slope by a factor of 1\u20442, or I guess you could say multiply the slope by 1\u20442, but it also made our y-intercept be half of what it was before."}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And now if we multiply it by two, this should increase the y-intercept because remember, we're multiplying both of these terms by two. And we see that. It not only doubles the slope, but it also increases the y-intercept. If we go to 0.5, not only did it decrease the slope by a factor of 1\u20442, or I guess you could say multiply the slope by 1\u20442, but it also made our y-intercept be half of what it was before. And we can see this more generally if we just put a general constant here and we can add a slider. And actually, let me make the constant go from zero to 10 with a step of, I don't know, 0.05. That's just how much does it increase every time you change the slider."}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "If we go to 0.5, not only did it decrease the slope by a factor of 1\u20442, or I guess you could say multiply the slope by 1\u20442, but it also made our y-intercept be half of what it was before. And we can see this more generally if we just put a general constant here and we can add a slider. And actually, let me make the constant go from zero to 10 with a step of, I don't know, 0.05. That's just how much does it increase every time you change the slider. And notice, when we increase our constant, not only are we getting narrower because the magnitude of the slope is being scaled, but our y-intercept increases. And then as k decreases, our y-intercept is being scaled down and our slope is being scaled down. Now that's one way that we could go about scaling, but what if instead of multiplying our entire function by some constant, we instead just replace the x with a constant times x?"}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "That's just how much does it increase every time you change the slider. And notice, when we increase our constant, not only are we getting narrower because the magnitude of the slope is being scaled, but our y-intercept increases. And then as k decreases, our y-intercept is being scaled down and our slope is being scaled down. Now that's one way that we could go about scaling, but what if instead of multiplying our entire function by some constant, we instead just replace the x with a constant times x? So instead of k times f of x, what if we did f of k times x? Another way to think about it is g of x is now equal to the absolute value of kx plus six. What do you think is going to happen?"}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Now that's one way that we could go about scaling, but what if instead of multiplying our entire function by some constant, we instead just replace the x with a constant times x? So instead of k times f of x, what if we did f of k times x? Another way to think about it is g of x is now equal to the absolute value of kx plus six. What do you think is going to happen? Pause this video and think about it. Well, now when we increase k, notice it has no impact on our y-intercept because it's not scaling the y-intercept, but it does have an impact on slope. When k goes from one to two, once again, we are now increasing twice as fast."}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "What do you think is going to happen? Pause this video and think about it. Well, now when we increase k, notice it has no impact on our y-intercept because it's not scaling the y-intercept, but it does have an impact on slope. When k goes from one to two, once again, we are now increasing twice as fast. And then when k goes from one to 1 1\u20442, we're now increasing half as fast. Now this is with an absolute value function. What if we did it with a different type of function?"}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "When k goes from one to two, once again, we are now increasing twice as fast. And then when k goes from one to 1 1\u20442, we're now increasing half as fast. Now this is with an absolute value function. What if we did it with a different type of function? Let's say we did it with a quadratic. So two minus x squared, and let me scroll down a little bit. And so you can see when k equals one, these are the same."}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "What if we did it with a different type of function? Let's say we did it with a quadratic. So two minus x squared, and let me scroll down a little bit. And so you can see when k equals one, these are the same. And now if we increase our k, let's say we increase our k to two, notice our parabola is, in this case, decreasing as we get further and further from zero at a faster and faster rate. That's because what you would have seen at x equals two, you're now seeing at x equals one because you are multiplying two times that. And so, and then if we go between zero and one, notice on either side of zero, our parabola is decreasing at a lower rate."}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And so you can see when k equals one, these are the same. And now if we increase our k, let's say we increase our k to two, notice our parabola is, in this case, decreasing as we get further and further from zero at a faster and faster rate. That's because what you would have seen at x equals two, you're now seeing at x equals one because you are multiplying two times that. And so, and then if we go between zero and one, notice on either side of zero, our parabola is decreasing at a lower rate. It's a changing rate, but it's a lower changing rate. I guess you could put it that way. And we could also try just to see what happens with our parabola here."}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And so, and then if we go between zero and one, notice on either side of zero, our parabola is decreasing at a lower rate. It's a changing rate, but it's a lower changing rate. I guess you could put it that way. And we could also try just to see what happens with our parabola here. If instead of doing kx, we once again put the k out front, what is that going to do? And notice, that is changing not only how fast the curve changes at different points, but it's now also changing the y-intercept because we are now scaling that y-intercept. So I'll leave you there."}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And we could also try just to see what happens with our parabola here. If instead of doing kx, we once again put the k out front, what is that going to do? And notice, that is changing not only how fast the curve changes at different points, but it's now also changing the y-intercept because we are now scaling that y-intercept. So I'll leave you there. This is just the beginning of thinking about scaling. I really want you to build an intuitive sense of what is going on here and really think about mathematically why it makes sense. And go on to Desmos and play around with it yourself and also try other types of functions and see what happens."}, {"video_title": "Imaginary roots of negative numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "That we can actually put input negative numbers in the domain of this function, that we can actually get imaginary or complex results. So we can rewrite negative 52 as negative 1 times 52. So this can be rewritten as the principal square root of negative 1 times 52. And then, if we assume that this is the principal branch of the complex square root function, we can rewrite this. This is going to be equal to the square root of negative 1 times the principal square root of negative 1, times the principal square root of 52. Now, I want to be very, very clear here. You can do what we just did."}, {"video_title": "Imaginary roots of negative numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And then, if we assume that this is the principal branch of the complex square root function, we can rewrite this. This is going to be equal to the square root of negative 1 times the principal square root of negative 1, times the principal square root of 52. Now, I want to be very, very clear here. You can do what we just did. If we have the principal square root of the product of two things, we can rewrite that as the principal square root of each, and then we take the product. But you can only do this if either both of these numbers are positive, or only one of them is negative. You cannot do this if both of these were negative."}, {"video_title": "Imaginary roots of negative numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "You can do what we just did. If we have the principal square root of the product of two things, we can rewrite that as the principal square root of each, and then we take the product. But you can only do this if either both of these numbers are positive, or only one of them is negative. You cannot do this if both of these were negative. For example, you could not do this. You could not say the principal square root of 52 is equal to negative 1 times negative 52. You could do this so far."}, {"video_title": "Imaginary roots of negative numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "You cannot do this if both of these were negative. For example, you could not do this. You could not say the principal square root of 52 is equal to negative 1 times negative 52. You could do this so far. I haven't said anything wrong. 52 is definitely negative 1 times negative 52. But then, since these are both negative, you cannot then say that this is equal to the square root of negative 1 times the square root of negative 52."}, {"video_title": "Imaginary roots of negative numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "You could do this so far. I haven't said anything wrong. 52 is definitely negative 1 times negative 52. But then, since these are both negative, you cannot then say that this is equal to the square root of negative 1 times the square root of negative 52. In fact, I invite you to continue on this train of reasoning. You're going to get a nonsensical answer. This is not okay."}, {"video_title": "Imaginary roots of negative numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "But then, since these are both negative, you cannot then say that this is equal to the square root of negative 1 times the square root of negative 52. In fact, I invite you to continue on this train of reasoning. You're going to get a nonsensical answer. This is not okay. You cannot do this right over here. And the reason why you cannot do this is that this property does not work when both of these numbers are negative. Now, with that said, we can do it if only one of them are negative, or both of them are positive, obviously."}, {"video_title": "Imaginary roots of negative numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "This is not okay. You cannot do this right over here. And the reason why you cannot do this is that this property does not work when both of these numbers are negative. Now, with that said, we can do it if only one of them are negative, or both of them are positive, obviously. Now, the principal square root of negative 1, if we're talking about the principal branch of the complex square root function, is i. So this right over here does simplify to i. And then let's think if we can simplify the square root of 52 any."}, {"video_title": "Imaginary roots of negative numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Now, with that said, we can do it if only one of them are negative, or both of them are positive, obviously. Now, the principal square root of negative 1, if we're talking about the principal branch of the complex square root function, is i. So this right over here does simplify to i. And then let's think if we can simplify the square root of 52 any. And to do that, we can think about its prime factorization, see if we have any perfect squares sitting in there. So let me, so 52 is 2 times 26, and 26 is 2 times 13. So we have 2 times 2 there, or 4 there, which is a perfect square."}, {"video_title": "Imaginary roots of negative numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And then let's think if we can simplify the square root of 52 any. And to do that, we can think about its prime factorization, see if we have any perfect squares sitting in there. So let me, so 52 is 2 times 26, and 26 is 2 times 13. So we have 2 times 2 there, or 4 there, which is a perfect square. So we can rewrite this as equal to, this is equal to, well we have our i now, the principal square root of negative 1 is i. The other square root of negative 1 is negative i. But the principal square root of negative 1 is i."}, {"video_title": "Imaginary roots of negative numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So we have 2 times 2 there, or 4 there, which is a perfect square. So we can rewrite this as equal to, this is equal to, well we have our i now, the principal square root of negative 1 is i. The other square root of negative 1 is negative i. But the principal square root of negative 1 is i. And then we're going to multiply that times the square root of 4 times 13. And this is equal to, and this is going to be equal to i times the square root of 4, i times the square root of 4, or the principal square root of 4 times the principal square root of 13. The principal square root of 4 is 2."}, {"video_title": "Imaginary roots of negative numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "But the principal square root of negative 1 is i. And then we're going to multiply that times the square root of 4 times 13. And this is equal to, and this is going to be equal to i times the square root of 4, i times the square root of 4, or the principal square root of 4 times the principal square root of 13. The principal square root of 4 is 2. So this all simplifies, and we can switch the order over here. This is equal to 2 times the square root of 13, 2 times the principal square root of 13 I should say, times i. And I just switched around the order, it makes it a little bit easier to read if I put the i after the numbers over here."}, {"video_title": "Imaginary roots of negative numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "The principal square root of 4 is 2. So this all simplifies, and we can switch the order over here. This is equal to 2 times the square root of 13, 2 times the principal square root of 13 I should say, times i. And I just switched around the order, it makes it a little bit easier to read if I put the i after the numbers over here. But I'm just multiplying i times 2 times the square root of 13. That's the same thing as multiplying 2 times the principal square root of 13 times i. And I think this is about as simplified as we can get here."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "So for example, if we look right over here, x is 1.585. b to the 1.585 is 3. So this is telling us that b to the 1.585 is equal to 3. Similarly, I can never say that, it's telling us that b to the 2.322 is 5. b to the 2.807 is 7. b to the 2.169 is 9. Now this table over here is telling us for any given value of y, what is going to be log base b of y? So this tells us that log base b of a is 0. Log base b of 2 is 1. Log base b of 2c is 1.585."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "Now this table over here is telling us for any given value of y, what is going to be log base b of y? So this tells us that log base b of a is 0. Log base b of 2 is 1. Log base b of 2c is 1.585. Log base b of 10d. So this is literally telling us that log base b of 10d is equal to 2.322. That's what this last column tells us."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "Log base b of 2c is 1.585. Log base b of 10d. So this is literally telling us that log base b of 10d is equal to 2.322. That's what this last column tells us. Now what I challenge you to do is pause this video. And using just the information here, and you don't need a calculator. In fact, you can't use a calculator."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "That's what this last column tells us. Now what I challenge you to do is pause this video. And using just the information here, and you don't need a calculator. In fact, you can't use a calculator. I forbid you. Try to figure out what a, b, c, and d are, just using your powers of reasoning. No calculator involved."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "In fact, you can't use a calculator. I forbid you. Try to figure out what a, b, c, and d are, just using your powers of reasoning. No calculator involved. Just use your powers of reasoning. Can you figure out what a, b, c, and d are? So I'm assuming you've given a go at it."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "No calculator involved. Just use your powers of reasoning. Can you figure out what a, b, c, and d are? So I'm assuming you've given a go at it. So let's see what we can deduce from this. So here we have just a bunch of numbers. We need to figure out what b is."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "So I'm assuming you've given a go at it. So let's see what we can deduce from this. So here we have just a bunch of numbers. We need to figure out what b is. So these are all kind of b to the 1.585 power is 3. I don't really know what to make sense of this stuff here. Maybe this table will help us."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "We need to figure out what b is. So these are all kind of b to the 1.585 power is 3. I don't really know what to make sense of this stuff here. Maybe this table will help us. So this first, let me do this in different colors. This first column right over here tells us that log base b of a, so now y is equal to a, that that is equal to 0. Now this is an equivalent statement to saying that b to the a power is equal to, oh sorry, not b to the a power."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "Maybe this table will help us. So this first, let me do this in different colors. This first column right over here tells us that log base b of a, so now y is equal to a, that that is equal to 0. Now this is an equivalent statement to saying that b to the a power is equal to, oh sorry, not b to the a power. This is an equivalent statement to saying b to the 0 power is equal to a. This is saying what exponent do I need to raise b to to get a? Well, you raise it to the 0 power."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "Now this is an equivalent statement to saying that b to the a power is equal to, oh sorry, not b to the a power. This is an equivalent statement to saying b to the 0 power is equal to a. This is saying what exponent do I need to raise b to to get a? Well, you raise it to the 0 power. This is saying b to the 0 power is equal to a. Now what is anything to the 0 power, assuming that it's not 0? If we're assuming that we're dealing with that b is not 0, if we're assuming that b is not 0, then so we're going to assume that, and we can't assume, and I think that's a safe assumption because we're raising b to all of these other powers, we're getting a non-zero value."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, you raise it to the 0 power. This is saying b to the 0 power is equal to a. Now what is anything to the 0 power, assuming that it's not 0? If we're assuming that we're dealing with that b is not 0, if we're assuming that b is not 0, then so we're going to assume that, and we can't assume, and I think that's a safe assumption because we're raising b to all of these other powers, we're getting a non-zero value. So since we know that b is not 0, anything to the 0 power is going to be 1. So this tells us that a is equal to 1. So we got 1 figured out."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "If we're assuming that we're dealing with that b is not 0, if we're assuming that b is not 0, then so we're going to assume that, and we can't assume, and I think that's a safe assumption because we're raising b to all of these other powers, we're getting a non-zero value. So since we know that b is not 0, anything to the 0 power is going to be 1. So this tells us that a is equal to 1. So we got 1 figured out. a is equal to 1. Now let's look at this next piece of information right over here. What does that tell us?"}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "So we got 1 figured out. a is equal to 1. Now let's look at this next piece of information right over here. What does that tell us? That tells us that log base b of 2 is equal to 1. This is equivalent to saying the power that I need to raise b to to get to 2 is 1. Or if I want to write it in exponential form, I could write this as saying that b to the first power is equal to 2."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "What does that tell us? That tells us that log base b of 2 is equal to 1. This is equivalent to saying the power that I need to raise b to to get to 2 is 1. Or if I want to write it in exponential form, I could write this as saying that b to the first power is equal to 2. So I'm raising something to the first power, and I'm getting 2. What is this thing? Well, that means that b must be 2."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "Or if I want to write it in exponential form, I could write this as saying that b to the first power is equal to 2. So I'm raising something to the first power, and I'm getting 2. What is this thing? Well, that means that b must be 2. 2 to the first power is 2. So b is equal to 2. b to the first power is equal to 2. You could say b to the first is equal to 2 to the first."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, that means that b must be 2. 2 to the first power is 2. So b is equal to 2. b to the first power is equal to 2. You could say b to the first is equal to 2 to the first. That's also equal to 2. So b must be equal to 2. So we've been able to figure that out."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "You could say b to the first is equal to 2 to the first. That's also equal to 2. So b must be equal to 2. So we've been able to figure that out. So this is a 2 right over here. It actually makes sense. 2 to the 1.585 power, yeah, that feels right."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "So we've been able to figure that out. So this is a 2 right over here. It actually makes sense. 2 to the 1.585 power, yeah, that feels right. That's about 3. Now let's see what else we can do. Let's see if we can figure out c. So let's look at this column."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "2 to the 1.585 power, yeah, that feels right. That's about 3. Now let's see what else we can do. Let's see if we can figure out c. So let's look at this column. Let's see what this column is telling us. That column, we could read it as log base b. Now our y is 2c."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "Let's see if we can figure out c. So let's look at this column. Let's see what this column is telling us. That column, we could read it as log base b. Now our y is 2c. Log base b of 2c is equal to 1.585. Or we could read this as b, if we write an exponential form, b to the 1.585 is equal to 2c. Now what's b to the 1.585?"}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "Now our y is 2c. Log base b of 2c is equal to 1.585. Or we could read this as b, if we write an exponential form, b to the 1.585 is equal to 2c. Now what's b to the 1.585? Well, they tell us right over here that b to the 1.585 is 3. So this right over here is equal to 3. So we get 2c is equal to 3."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "Now what's b to the 1.585? Well, they tell us right over here that b to the 1.585 is 3. So this right over here is equal to 3. So we get 2c is equal to 3. Or divide both sides by 2, we would get c is equal to 1.5. This is working out pretty well. And now we have this last column, which I will circle in purple."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "So we get 2c is equal to 3. Or divide both sides by 2, we would get c is equal to 1.5. This is working out pretty well. And now we have this last column, which I will circle in purple. And we can write this as log base b of 10d is equal to 2.322. So this is saying the power I need to raise b to get to 10d is 2.322. Or in exponential form, b to the 2.322 is equal to 10d."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "And now we have this last column, which I will circle in purple. And we can write this as log base b of 10d is equal to 2.322. So this is saying the power I need to raise b to get to 10d is 2.322. Or in exponential form, b to the 2.322 is equal to 10d. Now what is b to the 2.322? Well, they tell us over here b to the 2.322 is 5. So this is equal to 5."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "Or in exponential form, b to the 2.322 is equal to 10d. Now what is b to the 2.322? Well, they tell us over here b to the 2.322 is 5. So this is equal to 5. So we could write 10d is equal to 5. Or divide both sides by 10, d is equal to 0.5. And we're done."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And the general idea here is you can multiply these complex numbers like you would have multiplied any traditional binomial. You just have to remember that this isn't a variable. This is the imaginary unit i, or it's just i. But we could do that in two ways. We could just do the distributive property twice, which I like a little bit more just because it actually, you're doing it from a fundamental principle. There's nothing new. Or you could use FOIL, which you also use when you first multiply binomials."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "But we could do that in two ways. We could just do the distributive property twice, which I like a little bit more just because it actually, you're doing it from a fundamental principle. There's nothing new. Or you could use FOIL, which you also use when you first multiply binomials. And I'll do it both ways. So you could view, this is just a number, 1 minus 3i. And so we can distribute it over the two numbers inside of this expression."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Or you could use FOIL, which you also use when you first multiply binomials. And I'll do it both ways. So you could view, this is just a number, 1 minus 3i. And so we can distribute it over the two numbers inside of this expression. So when we're multiplying it times this entire expression, we can multiply 1 minus 3i times 2 and 1 minus 3i times 5i. So let's do that. So this can be rewritten as 1 minus 3i times 2."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And so we can distribute it over the two numbers inside of this expression. So when we're multiplying it times this entire expression, we can multiply 1 minus 3i times 2 and 1 minus 3i times 5i. So let's do that. So this can be rewritten as 1 minus 3i times 2. I'll write the 2 out front. Plus 1 minus 3i times 5i. All I did is a distributive property here."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So this can be rewritten as 1 minus 3i times 2. I'll write the 2 out front. Plus 1 minus 3i times 5i. All I did is a distributive property here. All I said is, look, if I have a times b plus c, this is the same thing as ab plus ac. I just distributed the a on the b or the c. I distributed the 1 minus 3i on the 2 and the 5i. And then I can do it again."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "All I did is a distributive property here. All I said is, look, if I have a times b plus c, this is the same thing as ab plus ac. I just distributed the a on the b or the c. I distributed the 1 minus 3i on the 2 and the 5i. And then I can do it again. I have a 2 now times 1 minus 3i. I can distribute it. 2 times 1 is 2."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And then I can do it again. I have a 2 now times 1 minus 3i. I can distribute it. 2 times 1 is 2. 2 times negative 3i is negative 6i. And over here, I'll do it again. 5i times 1."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "2 times 1 is 2. 2 times negative 3i is negative 6i. And over here, I'll do it again. 5i times 1. So it's plus 5i times 1 is 5i. And then 5i times negative 3i. So let's be careful here."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "5i times 1. So it's plus 5i times 1 is 5i. And then 5i times negative 3i. So let's be careful here. 5 times negative 3 is negative 15. And then I have an i times an i. I'm multiplying 5i. Let me do this over here."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So let's be careful here. 5 times negative 3 is negative 15. And then I have an i times an i. I'm multiplying 5i. Let me do this over here. 5i times negative 3i. This is the same thing as 5 times negative 3 times i. So the 5 times negative 3 is negative 15."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Let me do this over here. 5i times negative 3i. This is the same thing as 5 times negative 3 times i. So the 5 times negative 3 is negative 15. And then we have i times i, which is i squared. Now, we know what i squared is by definition. i squared is negative 1."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So the 5 times negative 3 is negative 15. And then we have i times i, which is i squared. Now, we know what i squared is by definition. i squared is negative 1. So you have negative 15 times negative 1. Well, that's the same thing as positive 15. So this can be rewritten as 2 minus 6i plus 5i."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "i squared is negative 1. So you have negative 15 times negative 1. Well, that's the same thing as positive 15. So this can be rewritten as 2 minus 6i plus 5i. Negative 15 times negative 1 is positive 15. Now, we can add the real parts. We have a 2 and we have a positive 15."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So this can be rewritten as 2 minus 6i plus 5i. Negative 15 times negative 1 is positive 15. Now, we can add the real parts. We have a 2 and we have a positive 15. So 2 plus 15. And we can add the imaginary parts. We have a negative 6."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "We have a 2 and we have a positive 15. So 2 plus 15. And we can add the imaginary parts. We have a negative 6. So we have a negative 6i, I should say. And then we have plus 5i. So plus 5i."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "We have a negative 6. So we have a negative 6i, I should say. And then we have plus 5i. So plus 5i. And 2 plus 15 is 17. And if I have negative 6 of something plus 5 of that something, what do I have? If I have 5 of that something and I take 6 of that something away, then I have negative 1 of that something."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So plus 5i. And 2 plus 15 is 17. And if I have negative 6 of something plus 5 of that something, what do I have? If I have 5 of that something and I take 6 of that something away, then I have negative 1 of that something. Negative 6i plus 5i is negative 1i. Or I could just say minus i. So in this way, I just multiplied these two expressions, or these two complex numbers, really."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "If I have 5 of that something and I take 6 of that something away, then I have negative 1 of that something. Negative 6i plus 5i is negative 1i. Or I could just say minus i. So in this way, I just multiplied these two expressions, or these two complex numbers, really. I multiplied them just using the distributive property twice. You could also do it using FOIL. And I'll do that right now really fast."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So in this way, I just multiplied these two expressions, or these two complex numbers, really. I multiplied them just using the distributive property twice. You could also do it using FOIL. And I'll do that right now really fast. It is a little bit faster, but it's a little bit mechanical. So you might forget why you're doing it in the first place. But at the end of the day, you are doing the same thing here."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And I'll do that right now really fast. It is a little bit faster, but it's a little bit mechanical. So you might forget why you're doing it in the first place. But at the end of the day, you are doing the same thing here. You're essentially multiplying every term of this first number, or every part of this first number, times every part of the second number. And FOIL just makes sure that we're doing it. And let me just write FOIL out here, which I'm not a huge fan of, but I'll do it, just in case that's the way you're learning it."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "But at the end of the day, you are doing the same thing here. You're essentially multiplying every term of this first number, or every part of this first number, times every part of the second number. And FOIL just makes sure that we're doing it. And let me just write FOIL out here, which I'm not a huge fan of, but I'll do it, just in case that's the way you're learning it. So FOIL says, let's do the first numbers. Let's multiply the first numbers. So that's going to be the 1 times the 2."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And let me just write FOIL out here, which I'm not a huge fan of, but I'll do it, just in case that's the way you're learning it. So FOIL says, let's do the first numbers. Let's multiply the first numbers. So that's going to be the 1 times the 2. So 1 times the 2. That is the F in FOIL. Then it says, let's multiply the outer numbers times each other."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So that's going to be the 1 times the 2. So 1 times the 2. That is the F in FOIL. Then it says, let's multiply the outer numbers times each other. So that's 1 times 5i. So plus 1 times 5i. This is the O in FOIL, the outer numbers."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Then it says, let's multiply the outer numbers times each other. So that's 1 times 5i. So plus 1 times 5i. This is the O in FOIL, the outer numbers. Then we do the inner numbers. Negative 3i times 2. So this is negative 3i times 2."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "This is the O in FOIL, the outer numbers. Then we do the inner numbers. Negative 3i times 2. So this is negative 3i times 2. Those are the inner numbers. And then we do the last numbers. Negative 3i times 5i."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So this is negative 3i times 2. Those are the inner numbers. And then we do the last numbers. Negative 3i times 5i. So negative 3i times 5i. These are the last numbers. And that's all that FOIL is telling us."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Negative 3i times 5i. So negative 3i times 5i. These are the last numbers. And that's all that FOIL is telling us. It's just making sure we're multiplying every part of this number times every part of that number. And then when we simplify it, 1 times 2 is 2. 1 times 5i is 5i."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And that's all that FOIL is telling us. It's just making sure we're multiplying every part of this number times every part of that number. And then when we simplify it, 1 times 2 is 2. 1 times 5i is 5i. Negative 3i times 2 is negative 6i. And negative 3i times 5i, well, we already figured out what that was. Negative 3i times 5i turns out to be 15."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "1 times 5i is 5i. Negative 3i times 2 is negative 6i. And negative 3i times 5i, well, we already figured out what that was. Negative 3i times 5i turns out to be 15. Negative 3 times 5 is negative 15. But i times i is negative 1. Negative 15 times negative 1 is positive 15."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Negative 3i times 5i turns out to be 15. Negative 3 times 5 is negative 15. But i times i is negative 1. Negative 15 times negative 1 is positive 15. Add the real parts. 2 plus 15. You get 17."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Negative 15 times negative 1 is positive 15. Add the real parts. 2 plus 15. You get 17. Add the imaginary parts. You have 5i minus 6i. You get negative i."}, {"video_title": "Equation special cases Linear equations Algebra I Khan Academy.mp3", "Sentence": "We have 8 times the quantity 3x plus 10 is equal to 28x minus 14 minus 4x. So like every equation we've done so far, we just want to isolate all of the x's on one side of this equation. But before we do that, we can actually simplify each of these sides. The left-hand side, we can multiply the quantity 3x plus 10 times 8. So we're essentially just distributing the 8, distributive property right here. So this is the same thing as 8 times 3x, which is 24x, plus 8 times 10, which is 80, is equal to... And over here we have 28x minus 14 minus 4x. So we can combine the 28x and the minus 4x."}, {"video_title": "Equation special cases Linear equations Algebra I Khan Academy.mp3", "Sentence": "The left-hand side, we can multiply the quantity 3x plus 10 times 8. So we're essentially just distributing the 8, distributive property right here. So this is the same thing as 8 times 3x, which is 24x, plus 8 times 10, which is 80, is equal to... And over here we have 28x minus 14 minus 4x. So we can combine the 28x and the minus 4x. If you have 28x minus 4x, that is 24x. And then you have the minus 14 right over here. Now the next thing we can do, and it's already looking a little bit suspicious, but just to confirm that it's as suspicious as it looks, let's try to subtract 24x from both sides of this equation."}, {"video_title": "Equation special cases Linear equations Algebra I Khan Academy.mp3", "Sentence": "So we can combine the 28x and the minus 4x. If you have 28x minus 4x, that is 24x. And then you have the minus 14 right over here. Now the next thing we can do, and it's already looking a little bit suspicious, but just to confirm that it's as suspicious as it looks, let's try to subtract 24x from both sides of this equation. And if we do that, we see that we actually remove the x's from both sides of the equation, because we have a 24x there and we have a 24x there. You might say, hey, let's put all the x's on the left-hand side, so let's get rid of this 24x. So you subtract 24x right over there, but you have to do it to the left-hand side as well."}, {"video_title": "Equation special cases Linear equations Algebra I Khan Academy.mp3", "Sentence": "Now the next thing we can do, and it's already looking a little bit suspicious, but just to confirm that it's as suspicious as it looks, let's try to subtract 24x from both sides of this equation. And if we do that, we see that we actually remove the x's from both sides of the equation, because we have a 24x there and we have a 24x there. You might say, hey, let's put all the x's on the left-hand side, so let's get rid of this 24x. So you subtract 24x right over there, but you have to do it to the left-hand side as well. And then we're left with, on the left-hand side, these guys cancel out. We're left with just 80 is equal to negative 14. Now this looks very bizarre."}, {"video_title": "Equation special cases Linear equations Algebra I Khan Academy.mp3", "Sentence": "So you subtract 24x right over there, but you have to do it to the left-hand side as well. And then we're left with, on the left-hand side, these guys cancel out. We're left with just 80 is equal to negative 14. Now this looks very bizarre. It's making a statement that 80 is equal to negative 14, which we know is not true. This does not happen. 80 is never equal to negative 14."}, {"video_title": "Equation special cases Linear equations Algebra I Khan Academy.mp3", "Sentence": "Now this looks very bizarre. It's making a statement that 80 is equal to negative 14, which we know is not true. This does not happen. 80 is never equal to negative 14. They are just inherently unequal. So this equation right here actually has no solution. This has no solution."}, {"video_title": "Powers of products & quotients (integer exponents) Mathematics I High School Math Khan Academy.mp3", "Sentence": "So let's say we have three to the negative eight times seven to the third, and we wanna raise that to the negative two power. And I want you to pause this video and see if you could simplify this on your own. So the key realization here, there's a couple of ways that you can tackle it, but the key thing to realize is if you have the product of two things and then you're raising that to some type of a exponent, that is going to be the same thing as raising each of these things to that exponent and then taking the product. So this is going to be the same thing as three to the negative eight and then that to the negative two times seven to the third to the negative two. So I'll do seven to the third right over here. And if I wanna simplify this, three to the negative eight to the negative two, we have the other exponent property that if you're raising to an exponent and then raising that whole thing to another exponent, then you can just multiply the exponents. So this is going to be three to the negative eight times negative two power."}, {"video_title": "Powers of products & quotients (integer exponents) Mathematics I High School Math Khan Academy.mp3", "Sentence": "So this is going to be the same thing as three to the negative eight and then that to the negative two times seven to the third to the negative two. So I'll do seven to the third right over here. And if I wanna simplify this, three to the negative eight to the negative two, we have the other exponent property that if you're raising to an exponent and then raising that whole thing to another exponent, then you can just multiply the exponents. So this is going to be three to the negative eight times negative two power. Well, negative eight times negative two is positive 16. So this is gonna be three to the 16th power right over there. And then this part right over here, seven to the third to the negative two, that's gonna be seven to the three times negative two, which is seven to the negative sixth power."}, {"video_title": "Powers of products & quotients (integer exponents) Mathematics I High School Math Khan Academy.mp3", "Sentence": "So this is going to be three to the negative eight times negative two power. Well, negative eight times negative two is positive 16. So this is gonna be three to the 16th power right over there. And then this part right over here, seven to the third to the negative two, that's gonna be seven to the three times negative two, which is seven to the negative sixth power. So that is seven to the negative sixth. And this would be about as much as you could simplify it. You could rewrite it different ways."}, {"video_title": "Powers of products & quotients (integer exponents) Mathematics I High School Math Khan Academy.mp3", "Sentence": "And then this part right over here, seven to the third to the negative two, that's gonna be seven to the three times negative two, which is seven to the negative sixth power. So that is seven to the negative sixth. And this would be about as much as you could simplify it. You could rewrite it different ways. Seven to the negative sixth, the same thing as one over seven to the sixth. So you could write it like three to the 16th. Let me use that same shade of blue."}, {"video_title": "Powers of products & quotients (integer exponents) Mathematics I High School Math Khan Academy.mp3", "Sentence": "You could rewrite it different ways. Seven to the negative sixth, the same thing as one over seven to the sixth. So you could write it like three to the 16th. Let me use that same shade of blue. Three to the 16th over seven to the sixth. But these two are equivalent. And there's other ways that you could have tackled this."}, {"video_title": "Powers of products & quotients (integer exponents) Mathematics I High School Math Khan Academy.mp3", "Sentence": "Let me use that same shade of blue. Three to the 16th over seven to the sixth. But these two are equivalent. And there's other ways that you could have tackled this. You could have said that this original thing right over here this is the same thing as three to the negative eight is the same thing as one over three to the eighth. So you could have said this is the same thing as seven to the third over three to the eighth. And then you're raising that to the negative two, in which case you would raise this numerator to the negative two and the denominator to negative two."}, {"video_title": "Powers of products & quotients (integer exponents) Mathematics I High School Math Khan Academy.mp3", "Sentence": "And there's other ways that you could have tackled this. You could have said that this original thing right over here this is the same thing as three to the negative eight is the same thing as one over three to the eighth. So you could have said this is the same thing as seven to the third over three to the eighth. And then you're raising that to the negative two, in which case you would raise this numerator to the negative two and the denominator to negative two. But you would have gotten to the exact same place. Let's do another one of these. So let's say, let me, so let's say that we have, we have got a to the negative two times eight to the seventh power."}, {"video_title": "Powers of products & quotients (integer exponents) Mathematics I High School Math Khan Academy.mp3", "Sentence": "And then you're raising that to the negative two, in which case you would raise this numerator to the negative two and the denominator to negative two. But you would have gotten to the exact same place. Let's do another one of these. So let's say, let me, so let's say that we have, we have got a to the negative two times eight to the seventh power. We wanna raise all of that to the second power. Well, like before, I can raise each of these things to the second power. So this is the same thing as a to the negative two to the second power times this thing to the second power, eight to the seventh to the second power."}, {"video_title": "Powers of products & quotients (integer exponents) Mathematics I High School Math Khan Academy.mp3", "Sentence": "So let's say, let me, so let's say that we have, we have got a to the negative two times eight to the seventh power. We wanna raise all of that to the second power. Well, like before, I can raise each of these things to the second power. So this is the same thing as a to the negative two to the second power times this thing to the second power, eight to the seventh to the second power. And then here, negative two times two is negative four. So that's a to the negative four times eight to the seven times two is 14, eight to the 14th power. In other videos, we go into more depth about why this should hopefully make intuitive sense."}, {"video_title": "Powers of products & quotients (integer exponents) Mathematics I High School Math Khan Academy.mp3", "Sentence": "So this is the same thing as a to the negative two to the second power times this thing to the second power, eight to the seventh to the second power. And then here, negative two times two is negative four. So that's a to the negative four times eight to the seven times two is 14, eight to the 14th power. In other videos, we go into more depth about why this should hopefully make intuitive sense. Here you have eight to the seventh times eight to the seventh where you would then add the two exponents and you would get to eight to the 14th. So however many times you have eight to the seventh, you would just keep adding the exponents or you would multiply by seven that many times. Hopefully that didn't sound too confusing."}, {"video_title": "Powers of products & quotients (integer exponents) Mathematics I High School Math Khan Academy.mp3", "Sentence": "In other videos, we go into more depth about why this should hopefully make intuitive sense. Here you have eight to the seventh times eight to the seventh where you would then add the two exponents and you would get to eight to the 14th. So however many times you have eight to the seventh, you would just keep adding the exponents or you would multiply by seven that many times. Hopefully that didn't sound too confusing. But the general idea is if you raise something to an exponent and then another exponent, you can multiply those exponents. Let's do one more example where we are dealing with quotients, which that first example could have even been perceived as. So let's say we have, let's say we have two to the negative 10 divided by four squared, and we're gonna raise all of that to the seventh power."}, {"video_title": "Powers of products & quotients (integer exponents) Mathematics I High School Math Khan Academy.mp3", "Sentence": "Hopefully that didn't sound too confusing. But the general idea is if you raise something to an exponent and then another exponent, you can multiply those exponents. Let's do one more example where we are dealing with quotients, which that first example could have even been perceived as. So let's say we have, let's say we have two to the negative 10 divided by four squared, and we're gonna raise all of that to the seventh power. Well, this is equivalent to two to the negative 10 raised to the seventh power over four squared raised to the seventh power. So if you have the difference of two things and you're raising it to some power, that's the same thing as the numerator raised to that power divided by the denominator raised to that power. Well, what's our numerator going to be?"}, {"video_title": "Powers of products & quotients (integer exponents) Mathematics I High School Math Khan Academy.mp3", "Sentence": "So let's say we have, let's say we have two to the negative 10 divided by four squared, and we're gonna raise all of that to the seventh power. Well, this is equivalent to two to the negative 10 raised to the seventh power over four squared raised to the seventh power. So if you have the difference of two things and you're raising it to some power, that's the same thing as the numerator raised to that power divided by the denominator raised to that power. Well, what's our numerator going to be? Well, we've done this drill before. It'd be two to the negative 10 times seventh power. So this would be equal to two to the negative 70th power."}, {"video_title": "Powers of products & quotients (integer exponents) Mathematics I High School Math Khan Academy.mp3", "Sentence": "Well, what's our numerator going to be? Well, we've done this drill before. It'd be two to the negative 10 times seventh power. So this would be equal to two to the negative 70th power. And then in the denominator, four to the second power, then that raised to the seventh power, well, two times seven is 14. So that's going to be four to the 17th, four to the 17th power. Now, we actually could think about simplifying this even more."}, {"video_title": "Powers of products & quotients (integer exponents) Mathematics I High School Math Khan Academy.mp3", "Sentence": "So this would be equal to two to the negative 70th power. And then in the denominator, four to the second power, then that raised to the seventh power, well, two times seven is 14. So that's going to be four to the 17th, four to the 17th power. Now, we actually could think about simplifying this even more. There's multiple ways that you could rewrite this. But one thing you could do is say, hey, look, four is a power of two. So you could rewrite this as, this is equal to two to the negative 70th power over, instead of writing four, instead of writing four to the 17th power, why did I write 17th power?"}, {"video_title": "Powers of products & quotients (integer exponents) Mathematics I High School Math Khan Academy.mp3", "Sentence": "Now, we actually could think about simplifying this even more. There's multiple ways that you could rewrite this. But one thing you could do is say, hey, look, four is a power of two. So you could rewrite this as, this is equal to two to the negative 70th power over, instead of writing four, instead of writing four to the 17th power, why did I write 17th power? It should be four to the 14th power. Let me correct that. Instead of writing four to the 14th power, I instead could write, so this is two, let me get the colors right."}, {"video_title": "Powers of products & quotients (integer exponents) Mathematics I High School Math Khan Academy.mp3", "Sentence": "So you could rewrite this as, this is equal to two to the negative 70th power over, instead of writing four, instead of writing four to the 17th power, why did I write 17th power? It should be four to the 14th power. Let me correct that. Instead of writing four to the 14th power, I instead could write, so this is two, let me get the colors right. This is two to the negative 70th over, over, instead of writing four, I could write two squared to the 14th power. Four is the same thing as two squared. And so now I can rewrite this whole thing as two to the negative 70th power over, well, two to the second, and then that to the 14th, well, that's two to the 28th power."}, {"video_title": "Powers of products & quotients (integer exponents) Mathematics I High School Math Khan Academy.mp3", "Sentence": "Instead of writing four to the 14th power, I instead could write, so this is two, let me get the colors right. This is two to the negative 70th over, over, instead of writing four, I could write two squared to the 14th power. Four is the same thing as two squared. And so now I can rewrite this whole thing as two to the negative 70th power over, well, two to the second, and then that to the 14th, well, that's two to the 28th power. Two to the 28th power. And so can I simplify this even more? Well, this is going to be equal to two to the, I can just, if I'm taking a quotient with the same base, I can subtract the exponents."}, {"video_title": "Powers of products & quotients (integer exponents) Mathematics I High School Math Khan Academy.mp3", "Sentence": "And so now I can rewrite this whole thing as two to the negative 70th power over, well, two to the second, and then that to the 14th, well, that's two to the 28th power. Two to the 28th power. And so can I simplify this even more? Well, this is going to be equal to two to the, I can just, if I'm taking a quotient with the same base, I can subtract the exponents. So it's gonna be negative 70. It's going to be negative 70 minus 28th power. Minus 28."}, {"video_title": "Interpreting points in context of graphs of systems.mp3", "Sentence": "We're told that Lauren uses a blend of dark roast beans and light roast beans to make coffee at her cafe. She needs 80 kilograms of beans in total for her next order. Dark roast beans cost $3 per kilogram. Light roast beans cost $2 per kilogram, and she wants to spend $220 in total. And they tell us here's a graph that shows a system of equations for this scenario where x is the number of kilograms of dark roast beans she buys and y is the number of kilograms of light roast beans she buys. All right, let me scroll down so we can take a look at this. And so sure enough, so this blue line, and I'll write it out in blue, this x is the number of kilograms of dark roast beans, y is the number of kilograms of light roast beans, and she wants to buy a total of 80 kilograms."}, {"video_title": "Interpreting points in context of graphs of systems.mp3", "Sentence": "Light roast beans cost $2 per kilogram, and she wants to spend $220 in total. And they tell us here's a graph that shows a system of equations for this scenario where x is the number of kilograms of dark roast beans she buys and y is the number of kilograms of light roast beans she buys. All right, let me scroll down so we can take a look at this. And so sure enough, so this blue line, and I'll write it out in blue, this x is the number of kilograms of dark roast beans, y is the number of kilograms of light roast beans, and she wants to buy a total of 80 kilograms. That's what they told us up here. We can go back to look at that. She needs, I'll underline this in blue, she needs 80 kilograms of beans."}, {"video_title": "Interpreting points in context of graphs of systems.mp3", "Sentence": "And so sure enough, so this blue line, and I'll write it out in blue, this x is the number of kilograms of dark roast beans, y is the number of kilograms of light roast beans, and she wants to buy a total of 80 kilograms. That's what they told us up here. We can go back to look at that. She needs, I'll underline this in blue, she needs 80 kilograms of beans. So that constraint that the sum of the kilograms of dark and light is equal to 80, that's represented by this equation, and if we were to graph it, that is this blue line right over here. And then this other constraint, 3x, well, the dark roast beans cost $3 per kilogram, so 3x is how much she spends on dark roast, 2y is how much she spends on light roast because it's $2 per kilogram, and 220 is the amount that she spends in total, and they tell us that up here. Dark roast beans cost $3 per kilogram, light roast beans cost $2 per kilogram, and she wants to spend $220."}, {"video_title": "Interpreting points in context of graphs of systems.mp3", "Sentence": "She needs, I'll underline this in blue, she needs 80 kilograms of beans. So that constraint that the sum of the kilograms of dark and light is equal to 80, that's represented by this equation, and if we were to graph it, that is this blue line right over here. And then this other constraint, 3x, well, the dark roast beans cost $3 per kilogram, so 3x is how much she spends on dark roast, 2y is how much she spends on light roast because it's $2 per kilogram, and 220 is the amount that she spends in total, and they tell us that up here. Dark roast beans cost $3 per kilogram, light roast beans cost $2 per kilogram, and she wants to spend $220. So this equation is another way of expressing what I just underlined up here in green. And the green line shows all of the xy combinations that would match these constraints. And so now let's do something interesting."}, {"video_title": "Interpreting points in context of graphs of systems.mp3", "Sentence": "Dark roast beans cost $3 per kilogram, light roast beans cost $2 per kilogram, and she wants to spend $220. So this equation is another way of expressing what I just underlined up here in green. And the green line shows all of the xy combinations that would match these constraints. And so now let's do something interesting. They've labeled some points here, point C, D, F, and E, and we're gonna think about what do each of these points represent? So for example, this point C that is on the green line, but it sits above the blue line, what does this tell us? What does this point C represent?"}, {"video_title": "Interpreting points in context of graphs of systems.mp3", "Sentence": "And so now let's do something interesting. They've labeled some points here, point C, D, F, and E, and we're gonna think about what do each of these points represent? So for example, this point C that is on the green line, but it sits above the blue line, what does this tell us? What does this point C represent? Pause this video and think about it. Well, if we're on the green line, that means that the amount that she spends on dark roast plus the amount that she spends on light roast is adding up to exactly $220. So she's definitely spending $220 at C, but how many total kilograms is she using?"}, {"video_title": "Interpreting points in context of graphs of systems.mp3", "Sentence": "What does this point C represent? Pause this video and think about it. Well, if we're on the green line, that means that the amount that she spends on dark roast plus the amount that she spends on light roast is adding up to exactly $220. So she's definitely spending $220 at C, but how many total kilograms is she using? Well, the fact that for this given x, we're sitting above the line, that means that she is not using exactly 80 kilograms. And we can see that over here. She's using, looks like 10 kilograms of dark, and it looks like something like 95 kilograms of light."}, {"video_title": "Interpreting points in context of graphs of systems.mp3", "Sentence": "So she's definitely spending $220 at C, but how many total kilograms is she using? Well, the fact that for this given x, we're sitting above the line, that means that she is not using exactly 80 kilograms. And we can see that over here. She's using, looks like 10 kilograms of dark, and it looks like something like 95 kilograms of light. If you were to add those two points together, it looks like she's using something closer to 105 kilograms. So point C is a situation where she is spending exactly $220, but she's using more than 80 kilograms because it's not meeting this second constraint. It's sitting above that line."}, {"video_title": "Interpreting points in context of graphs of systems.mp3", "Sentence": "She's using, looks like 10 kilograms of dark, and it looks like something like 95 kilograms of light. If you were to add those two points together, it looks like she's using something closer to 105 kilograms. So point C is a situation where she is spending exactly $220, but she's using more than 80 kilograms because it's not meeting this second constraint. It's sitting above that line. Now let's think about point D. What does that represent? Pause the video and try to figure that out. Well, because we sit on the blue line, that means that we are meeting this constraint that the kilograms of dark and light combined is equal to 80 kilograms."}, {"video_title": "Interpreting points in context of graphs of systems.mp3", "Sentence": "It's sitting above that line. Now let's think about point D. What does that represent? Pause the video and try to figure that out. Well, because we sit on the blue line, that means that we are meeting this constraint that the kilograms of dark and light combined is equal to 80 kilograms. So she's using exactly 80 kilograms here, but what about her spending? Well, because this point lies below the green line, that tells us that we are spending less than $220. And we could even try it out."}, {"video_title": "Interpreting points in context of graphs of systems.mp3", "Sentence": "Well, because we sit on the blue line, that means that we are meeting this constraint that the kilograms of dark and light combined is equal to 80 kilograms. So she's using exactly 80 kilograms here, but what about her spending? Well, because this point lies below the green line, that tells us that we are spending less than $220. And we could even try it out. Three times 20 plus two times 60 is what? 60 plus 120 is $180. And so this is a point where we're meeting this constraint, but we're not meeting this constraint."}, {"video_title": "Interpreting points in context of graphs of systems.mp3", "Sentence": "And we could even try it out. Three times 20 plus two times 60 is what? 60 plus 120 is $180. And so this is a point where we're meeting this constraint, but we're not meeting this constraint. We're underspending right over here. Now what about point F? Well, point F sits below both of these lines."}, {"video_title": "Interpreting points in context of graphs of systems.mp3", "Sentence": "And so this is a point where we're meeting this constraint, but we're not meeting this constraint. We're underspending right over here. Now what about point F? Well, point F sits below both of these lines. So pause your video and think about what that means. Well, if we're sitting below both of these lines, that means that neither are we spending $220 nor are we using 80 kilograms. And you could see that if you actually look at the numbers."}, {"video_title": "Interpreting points in context of graphs of systems.mp3", "Sentence": "Well, point F sits below both of these lines. So pause your video and think about what that means. Well, if we're sitting below both of these lines, that means that neither are we spending $220 nor are we using 80 kilograms. And you could see that if you actually look at the numbers. You don't have to do this, but this is just to make you feel good about it. It looks like she is using 30 kilograms of dark and 30 kilograms of light. So in total, she is using, so this is a situation where she's using 60 kilograms in total, not 80."}, {"video_title": "Interpreting points in context of graphs of systems.mp3", "Sentence": "And you could see that if you actually look at the numbers. You don't have to do this, but this is just to make you feel good about it. It looks like she is using 30 kilograms of dark and 30 kilograms of light. So in total, she is using, so this is a situation where she's using 60 kilograms in total, not 80. And so that's why we're not sitting on this blue line. And if you look at how much she's spending, she has 30 kilograms of each. So three times 30 plus two times 30, that's gonna be 90 plus 60."}, {"video_title": "Interpreting points in context of graphs of systems.mp3", "Sentence": "So in total, she is using, so this is a situation where she's using 60 kilograms in total, not 80. And so that's why we're not sitting on this blue line. And if you look at how much she's spending, she has 30 kilograms of each. So three times 30 plus two times 30, that's gonna be 90 plus 60. That's also less than 220. And so that's why we see this point is below these lines. And then last but not least, what does point E represent?"}, {"video_title": "Interpreting points in context of graphs of systems.mp3", "Sentence": "So three times 30 plus two times 30, that's gonna be 90 plus 60. That's also less than 220. And so that's why we see this point is below these lines. And then last but not least, what does point E represent? Well, point E sits on both of these lines. So that means that it meets both of these constraints. This is a situation where she is spending exactly $220, and the total number of kilograms she's using of dark and light is exactly 80."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "What I'd like to do in this video is use some geometric arguments to prove that the slopes of perpendicular lines are negative reciprocals of each other. And so just to start off, we have lines L and M, and we're going to assume that they are perpendicular, so they intersect at a right angle. We see that depicted right over here. And so I'm gonna now construct some other lines here to help us make our geometric argument. So let me draw a horizontal line that intersects at this point right over here. Let's call that point A. And so let me see if I can do that."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "And so I'm gonna now construct some other lines here to help us make our geometric argument. So let me draw a horizontal line that intersects at this point right over here. Let's call that point A. And so let me see if I can do that. There you go. So that's a horizontal line that intersects at A. And now I'm gonna drop some verticals from that."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "And so let me see if I can do that. There you go. So that's a horizontal line that intersects at A. And now I'm gonna drop some verticals from that. So I'm gonna drop a vertical line right over here, and I'm gonna drop a vertical line right over here. And so that is 90 degrees, and that is 90 degrees, and I've constructed it that way. This top line is perfectly horizontal, and then I've dropped two vertical things, so they're at 90 degree angles."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "And now I'm gonna drop some verticals from that. So I'm gonna drop a vertical line right over here, and I'm gonna drop a vertical line right over here. And so that is 90 degrees, and that is 90 degrees, and I've constructed it that way. This top line is perfectly horizontal, and then I've dropped two vertical things, so they're at 90 degree angles. And let me now set up some points. So that already said, that's point A. Let's call this point B."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "This top line is perfectly horizontal, and then I've dropped two vertical things, so they're at 90 degree angles. And let me now set up some points. So that already said, that's point A. Let's call this point B. Let's call this point C. Let's call this point D. And let's call this point E right over here. Now, let's think about what the slope of line L is. So slope of, let me move this over a little bit."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "Let's call this point B. Let's call this point C. Let's call this point D. And let's call this point E right over here. Now, let's think about what the slope of line L is. So slope of, let me move this over a little bit. So slope, slope of L is going to be what? Well, that's, you could view line L as line, the line that connects point CA. So it's the slope of CA, you could say."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "So slope of, let me move this over a little bit. So slope, slope of L is going to be what? Well, that's, you could view line L as line, the line that connects point CA. So it's the slope of CA, you could say. This is the same thing as slope of line CA. L is line CA. And so to find the slope, that's change in Y over change in X."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "So it's the slope of CA, you could say. This is the same thing as slope of line CA. L is line CA. And so to find the slope, that's change in Y over change in X. So our change in Y is going to be CB. So it's gonna be the length of segment CB. That is our change in Y."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "And so to find the slope, that's change in Y over change in X. So our change in Y is going to be CB. So it's gonna be the length of segment CB. That is our change in Y. So it is CB over our change in X, which is the length of segment BA, which is the length of segment BA right over here. So that is BA. Now, what is the slope of line M?"}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "That is our change in Y. So it is CB over our change in X, which is the length of segment BA, which is the length of segment BA right over here. So that is BA. Now, what is the slope of line M? So slope, slope of M, and we could also say slope of, we could call line M line AE, line AE, like that. Well, if we're going to go between point A and point E, once again, it's just change in Y over change in X. Well, what's our change in Y going to be?"}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "Now, what is the slope of line M? So slope, slope of M, and we could also say slope of, we could call line M line AE, line AE, like that. Well, if we're going to go between point A and point E, once again, it's just change in Y over change in X. Well, what's our change in Y going to be? Well, our change in Y, well, we're gonna go from this level down to this level as we go from A to E. We could have done it over here as well. We're gonna go from A to E. That is our change in Y. So we might be tempted to say, well, that's just going to be the length of segment DE."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "Well, what's our change in Y going to be? Well, our change in Y, well, we're gonna go from this level down to this level as we go from A to E. We could have done it over here as well. We're gonna go from A to E. That is our change in Y. So we might be tempted to say, well, that's just going to be the length of segment DE. But remember, our Y is decreasing. So we're gonna subtract that length as we go from this Y level to that Y level over there. And what is our change in X?"}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "So we might be tempted to say, well, that's just going to be the length of segment DE. But remember, our Y is decreasing. So we're gonna subtract that length as we go from this Y level to that Y level over there. And what is our change in X? So our change in X, we're going to go, as we go from A to E, our change in X is going to be the length of segment AD. So AD. So our slope of M is going to be negative DE."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "And what is our change in X? So our change in X, we're going to go, as we go from A to E, our change in X is going to be the length of segment AD. So AD. So our slope of M is going to be negative DE. It's going to be the negative of this length because we're dropping by that much. That's our change in Y over segment A, over segment AD. So some of you might already be quite inspired by what we've already written because now we just have to establish that these two are, these two triangles, triangles CBA and triangle ADE, are similar."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "So our slope of M is going to be negative DE. It's going to be the negative of this length because we're dropping by that much. That's our change in Y over segment A, over segment AD. So some of you might already be quite inspired by what we've already written because now we just have to establish that these two are, these two triangles, triangles CBA and triangle ADE, are similar. And then we're going to be able to show that these are the negative reciprocal of each other. So let's show that these two triangles are similar. So let's say that we have this angle right over here."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "So some of you might already be quite inspired by what we've already written because now we just have to establish that these two are, these two triangles, triangles CBA and triangle ADE, are similar. And then we're going to be able to show that these are the negative reciprocal of each other. So let's show that these two triangles are similar. So let's say that we have this angle right over here. And let's say that angle has measure X, just for kicks. And let's say that we have, let me do another color for, let's say we have this angle right over here. And let's say that the measure, that that has measure Y."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "So let's say that we have this angle right over here. And let's say that angle has measure X, just for kicks. And let's say that we have, let me do another color for, let's say we have this angle right over here. And let's say that the measure, that that has measure Y. Well we know X plus Y plus 90 is equal to 180 because together they are supplementary. So I could write, I could write that X plus 90, plus 90, plus Y, plus Y, is going to be equal to, is going to be equal to 180 degrees. If you want you could subtract 90 from both sides of that and you could say, look, X plus Y is going to be equal to 90 degrees."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "And let's say that the measure, that that has measure Y. Well we know X plus Y plus 90 is equal to 180 because together they are supplementary. So I could write, I could write that X plus 90, plus 90, plus Y, plus Y, is going to be equal to, is going to be equal to 180 degrees. If you want you could subtract 90 from both sides of that and you could say, look, X plus Y is going to be equal to 90 degrees. Is going to be equal to 90 degrees. These are algebraically equivalent statements. So is equal to 90 degrees."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "If you want you could subtract 90 from both sides of that and you could say, look, X plus Y is going to be equal to 90 degrees. Is going to be equal to 90 degrees. These are algebraically equivalent statements. So is equal to 90 degrees. And how can we use this to fill out some of the other angles in these triangles? Well, let's see. X plus this angle down here has to be equal to 90 degrees."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "So is equal to 90 degrees. And how can we use this to fill out some of the other angles in these triangles? Well, let's see. X plus this angle down here has to be equal to 90 degrees. Or you could say X plus 90 plus what is going to be equal to 180. I'm looking at triangle CBA right over here. The interior angles of a triangle add up to 180."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "X plus this angle down here has to be equal to 90 degrees. Or you could say X plus 90 plus what is going to be equal to 180. I'm looking at triangle CBA right over here. The interior angles of a triangle add up to 180. So X plus 90 plus what is equal to 180? Well X plus 90 plus Y is equal to 180. We already established that."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "The interior angles of a triangle add up to 180. So X plus 90 plus what is equal to 180? Well X plus 90 plus Y is equal to 180. We already established that. Similarly over here. Y plus 90 plus what is going to be equal to 180? Well, same argument."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "We already established that. Similarly over here. Y plus 90 plus what is going to be equal to 180? Well, same argument. We already know Y plus 90 plus X is equal to 180. So Y plus 90 plus X is equal to, is equal to 180. And so notice, we have now established that triangle ABC and triangle EDA, that all of their interior angles, their corresponding interior angles are the same or that their three different angle measures all correspond to each other."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "Well, same argument. We already know Y plus 90 plus X is equal to 180. So Y plus 90 plus X is equal to, is equal to 180. And so notice, we have now established that triangle ABC and triangle EDA, that all of their interior angles, their corresponding interior angles are the same or that their three different angle measures all correspond to each other. They both have an angle of X, they both have a measure X, they both have an angle of measure Y, and they're both right triangles. So just by angle, angle, angle, so we could say by angle, angle, angle, one of our similarity postulates, we know that triangle E, triangle EDA, EDA is similar to triangle, to triangle ABC, to triangle ABC. And so that tells us that the ratio of corresponding sides are going to be the same."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "And so notice, we have now established that triangle ABC and triangle EDA, that all of their interior angles, their corresponding interior angles are the same or that their three different angle measures all correspond to each other. They both have an angle of X, they both have a measure X, they both have an angle of measure Y, and they're both right triangles. So just by angle, angle, angle, so we could say by angle, angle, angle, one of our similarity postulates, we know that triangle E, triangle EDA, EDA is similar to triangle, to triangle ABC, to triangle ABC. And so that tells us that the ratio of corresponding sides are going to be the same. And so for example, we know, let's find the ratio of corresponding sides. We know that the ratio of, let's say CB to BA, so let's write this down. We know that the ratio, so this tells us that the ratio of corresponding sides are going to be the same."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "And so that tells us that the ratio of corresponding sides are going to be the same. And so for example, we know, let's find the ratio of corresponding sides. We know that the ratio of, let's say CB to BA, so let's write this down. We know that the ratio, so this tells us that the ratio of corresponding sides are going to be the same. So the ratio of CB over BA, over BA is going to be equal to, is going to be equal to, well the corresponding side to CB, it's the side opposite the X degree angle right over here. So the corresponding side to CB is side AD, so that's going to be equal to AD over, what's the corresponding side to BA? Well BA is opposite the Y degree angle, so over here the corresponding side is DE, AD over DE, let me do that same color, over DE."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "We know that the ratio, so this tells us that the ratio of corresponding sides are going to be the same. So the ratio of CB over BA, over BA is going to be equal to, is going to be equal to, well the corresponding side to CB, it's the side opposite the X degree angle right over here. So the corresponding side to CB is side AD, so that's going to be equal to AD over, what's the corresponding side to BA? Well BA is opposite the Y degree angle, so over here the corresponding side is DE, AD over DE, let me do that same color, over DE. And so this right over here, this right over here we saw from the beginning, this is the slope, this is the slope of L. So slope, slope of L. And how does this relate to the slope of M? Notice the slope of M is the negative reciprocal of this. You take the reciprocal, you're going to get DE over AD, and then you have to take this negative right over here."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "Well BA is opposite the Y degree angle, so over here the corresponding side is DE, AD over DE, let me do that same color, over DE. And so this right over here, this right over here we saw from the beginning, this is the slope, this is the slope of L. So slope, slope of L. And how does this relate to the slope of M? Notice the slope of M is the negative reciprocal of this. You take the reciprocal, you're going to get DE over AD, and then you have to take this negative right over here. So we could write this as the negative reciprocal of slope of M. Negative reciprocal, reciprocal of M's, of M's slope. And there you have it. We've just shown that if we start with, if we assume these L and M are perpendicular, and we set up these similar triangles, and we were able to show that the slope of L is the negative reciprocal of the slope of M."}, {"video_title": "Compound inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "So this is a compound inequality. We have two conditions here. So z can satisfy this, or z can satisfy this over here. So let's just solve each of these inequalities and just know that z can satisfy either of them. So let's just look at this. So if we look at just this one over here, we have 5z plus 7 is less than 27. Let's isolate the z's on the left-hand side."}, {"video_title": "Compound inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "So let's just solve each of these inequalities and just know that z can satisfy either of them. So let's just look at this. So if we look at just this one over here, we have 5z plus 7 is less than 27. Let's isolate the z's on the left-hand side. So let's subtract 7 from both sides to get rid of this 7 on the left-hand side. And so our left-hand side is just going to be 5z plus 7 minus 7. Those cancel out."}, {"video_title": "Compound inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "Let's isolate the z's on the left-hand side. So let's subtract 7 from both sides to get rid of this 7 on the left-hand side. And so our left-hand side is just going to be 5z plus 7 minus 7. Those cancel out. 5z is less than 27 minus 7 is 20. So we have 5z is less than 20. Now we can divide both sides of this inequality by 5."}, {"video_title": "Compound inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "Those cancel out. 5z is less than 27 minus 7 is 20. So we have 5z is less than 20. Now we can divide both sides of this inequality by 5. And we don't have to swap the inequality because we're dividing by a positive number. And so we get z is less than 20 over 5. z is less than 4. Now this was only one of the conditions."}, {"video_title": "Compound inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "Now we can divide both sides of this inequality by 5. And we don't have to swap the inequality because we're dividing by a positive number. And so we get z is less than 20 over 5. z is less than 4. Now this was only one of the conditions. Let's look at the other one over here. We have negative 3z is less than or equal to 18. Now to isolate the z, we can just divide both sides of this inequality by negative 3."}, {"video_title": "Compound inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "Now this was only one of the conditions. Let's look at the other one over here. We have negative 3z is less than or equal to 18. Now to isolate the z, we can just divide both sides of this inequality by negative 3. But remember, when you divide or multiply both sides of an inequality by a negative number, you have to swap the inequality. So we could write this negative 3z. We're going to divide it by negative 3."}, {"video_title": "Compound inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "Now to isolate the z, we can just divide both sides of this inequality by negative 3. But remember, when you divide or multiply both sides of an inequality by a negative number, you have to swap the inequality. So we could write this negative 3z. We're going to divide it by negative 3. And then you have 18. We're going to divide it by negative 3. But we're going to swap the inequality."}, {"video_title": "Compound inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "We're going to divide it by negative 3. And then you have 18. We're going to divide it by negative 3. But we're going to swap the inequality. So the less than or equal will become greater than or equal to. And so these guys cancel out. Negative 3 divided by negative 3 is 1."}, {"video_title": "Compound inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "But we're going to swap the inequality. So the less than or equal will become greater than or equal to. And so these guys cancel out. Negative 3 divided by negative 3 is 1. So we have z is greater than or equal to 18 over negative 3 is negative 6. And remember, it's this constraint or this constraint. And this constraint right over here boils down to this."}, {"video_title": "Compound inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "Negative 3 divided by negative 3 is 1. So we have z is greater than or equal to 18 over negative 3 is negative 6. And remember, it's this constraint or this constraint. And this constraint right over here boils down to this. And this one boils down to this. So our solution said z is less than 4 or z is greater than or equal to negative 6. So let me make this clear."}, {"video_title": "Compound inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "And this constraint right over here boils down to this. And this one boils down to this. So our solution said z is less than 4 or z is greater than or equal to negative 6. So let me make this clear. Let me rewrite it. So z could be less than 4 or z is greater than or equal to negative 6. It can satisfy either one of these."}, {"video_title": "Compound inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "So let me make this clear. Let me rewrite it. So z could be less than 4 or z is greater than or equal to negative 6. It can satisfy either one of these. And this is kind of interesting here. Because let's plot these. So there's a number line right over there."}, {"video_title": "Compound inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "It can satisfy either one of these. And this is kind of interesting here. Because let's plot these. So there's a number line right over there. Let's say that 0 is over here. We have 1, 2, 3, 4 is right over there. And then negative 6, we have 1, 2, 3, 4, 5, 6."}, {"video_title": "Compound inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "So there's a number line right over there. Let's say that 0 is over here. We have 1, 2, 3, 4 is right over there. And then negative 6, we have 1, 2, 3, 4, 5, 6. That's negative 6 over there. Now let's think about z being less than 4. z being less than 4, we would put a circle around 4 since we're not including 4. And it would be everything less than 4."}, {"video_title": "Compound inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "And then negative 6, we have 1, 2, 3, 4, 5, 6. That's negative 6 over there. Now let's think about z being less than 4. z being less than 4, we would put a circle around 4 since we're not including 4. And it would be everything less than 4. Now let's think about what z being greater than or equal to negative 6 would mean. That means you can include negative 6. And it's everything."}, {"video_title": "Compound inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "And it would be everything less than 4. Now let's think about what z being greater than or equal to negative 6 would mean. That means you can include negative 6. And it's everything. Let me do that in a different color. It means you can include negative 6. I want to do that."}, {"video_title": "Compound inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "And it's everything. Let me do that in a different color. It means you can include negative 6. I want to do that. It means you include negative 6. Let me do it in a more different color. I'll do it in orange."}, {"video_title": "Compound inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "I want to do that. It means you include negative 6. Let me do it in a more different color. I'll do it in orange. So z is greater than or equal to negative 6. It means you can include negative 6. And it's everything greater than that."}, {"video_title": "Compound inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "I'll do it in orange. So z is greater than or equal to negative 6. It means you can include negative 6. And it's everything greater than that. Including 4. So it's everything greater than that. What we've done is we've essentially shaded in the entire number line."}, {"video_title": "Compound inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "And it's everything greater than that. Including 4. So it's everything greater than that. What we've done is we've essentially shaded in the entire number line. Every number will meet either one of these constraints or both of them. If we're over here, we're going to meet both of the constraints. If we're a number out here, we're going to meet this constraint."}, {"video_title": "Compound inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "What we've done is we've essentially shaded in the entire number line. Every number will meet either one of these constraints or both of them. If we're over here, we're going to meet both of the constraints. If we're a number out here, we're going to meet this constraint. If we're a number down here, we're going to meet this constraint. You can just try it out with a bunch of numbers. 0 will work."}, {"video_title": "Compound inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "If we're a number out here, we're going to meet this constraint. If we're a number down here, we're going to meet this constraint. You can just try it out with a bunch of numbers. 0 will work. 0 plus 7 is 7, which is less than 27. And 3 times 0 is less than 18. So it meets both constraints."}, {"video_title": "Compound inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "0 will work. 0 plus 7 is 7, which is less than 27. And 3 times 0 is less than 18. So it meets both constraints. If we're over here, it should only meet one of the constraints. Negative 3 times 4 is negative 12, which is less than 18. So it meets this constraint."}, {"video_title": "Compound inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "So it meets both constraints. If we're over here, it should only meet one of the constraints. Negative 3 times 4 is negative 12, which is less than 18. So it meets this constraint. But it won't meet this constraint. Because you do 5 times 4 plus 7 is 27, which is not less than 27. It's equal to 27."}, {"video_title": "Compound inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "So it meets this constraint. But it won't meet this constraint. Because you do 5 times 4 plus 7 is 27, which is not less than 27. It's equal to 27. But remember, this is an or. So you just have to meet one of the constraints. So 4 meets this constraint."}, {"video_title": "Multiplying & dividing rational expressions monomials High School Math Khan Academy.mp3", "Sentence": "So up here we are multiplying two rational expressions, and here we're dividing one rational expression by another one. And what I encourage you to do is pause these videos and think about what these become when you multiply them out, and maybe you simplify it a little bit. And I also want you to think about what constraints do you have to put on the x values in order for your resulting expression to be algebraically equivalent to your original expression. So let's work it out together, just so you realize what I'm talking about. So this is going to be, in our numerator, we are going to get six x to the third power times two. In our denominator, we're going to have five times three x. And we can see both the numerator and the denominator are divisible by x."}, {"video_title": "Multiplying & dividing rational expressions monomials High School Math Khan Academy.mp3", "Sentence": "So let's work it out together, just so you realize what I'm talking about. So this is going to be, in our numerator, we are going to get six x to the third power times two. In our denominator, we're going to have five times three x. And we can see both the numerator and the denominator are divisible by x. So let's divide the denominator by x, we get one there. Let's divide x to the third by x, we get x squared. And we can also see that both the numerator and denominator are divisible by three."}, {"video_title": "Multiplying & dividing rational expressions monomials High School Math Khan Academy.mp3", "Sentence": "And we can see both the numerator and the denominator are divisible by x. So let's divide the denominator by x, we get one there. Let's divide x to the third by x, we get x squared. And we can also see that both the numerator and denominator are divisible by three. So divide six by three, you get two. Divide three by three, you get one. And we are left with two x squared times two, which is going to be four x squared over five times one times one, over five."}, {"video_title": "Multiplying & dividing rational expressions monomials High School Math Khan Academy.mp3", "Sentence": "And we can also see that both the numerator and denominator are divisible by three. So divide six by three, you get two. Divide three by three, you get one. And we are left with two x squared times two, which is going to be four x squared over five times one times one, over five. And we could also write that as 4 5ths x squared. Now, if someone just presented you on the street with the expression 4 5ths x squared and said, for what x is this defined? You could say, well, I could put any x here."}, {"video_title": "Multiplying & dividing rational expressions monomials High School Math Khan Academy.mp3", "Sentence": "And we are left with two x squared times two, which is going to be four x squared over five times one times one, over five. And we could also write that as 4 5ths x squared. Now, if someone just presented you on the street with the expression 4 5ths x squared and said, for what x is this defined? You could say, well, I could put any x here. x could be zero, because zero squared is zero, times 4 5ths is just going to be zero, so it seems to be defined for zero. And that is true. But if someone says, how would I have to constrain this in order for it to be algebraically equivalent to this first expression?"}, {"video_title": "Multiplying & dividing rational expressions monomials High School Math Khan Academy.mp3", "Sentence": "You could say, well, I could put any x here. x could be zero, because zero squared is zero, times 4 5ths is just going to be zero, so it seems to be defined for zero. And that is true. But if someone says, how would I have to constrain this in order for it to be algebraically equivalent to this first expression? Well, then you'd have to say, well, this first expression is not defined for all x. For example, if x were equal to zero, then you would be dividing by zero right over here, which would make this undefined. So you could explicitly call it out, x cannot be equal to zero."}, {"video_title": "Multiplying & dividing rational expressions monomials High School Math Khan Academy.mp3", "Sentence": "But if someone says, how would I have to constrain this in order for it to be algebraically equivalent to this first expression? Well, then you'd have to say, well, this first expression is not defined for all x. For example, if x were equal to zero, then you would be dividing by zero right over here, which would make this undefined. So you could explicitly call it out, x cannot be equal to zero. And so if you want this one to be algebraically equivalent, you would have to make that same condition. x cannot be equal to zero. Another way to think about it, if you had a function defined this way, if you said f of x is equal to 6 x to the third over five times two over, times two over 3x, and if someone said, well, what is f of zero?"}, {"video_title": "Multiplying & dividing rational expressions monomials High School Math Khan Academy.mp3", "Sentence": "So you could explicitly call it out, x cannot be equal to zero. And so if you want this one to be algebraically equivalent, you would have to make that same condition. x cannot be equal to zero. Another way to think about it, if you had a function defined this way, if you said f of x is equal to 6 x to the third over five times two over, times two over 3x, and if someone said, well, what is f of zero? You would say f of zero is undefined. Undefined. Why is that?"}, {"video_title": "Multiplying & dividing rational expressions monomials High School Math Khan Academy.mp3", "Sentence": "Another way to think about it, if you had a function defined this way, if you said f of x is equal to 6 x to the third over five times two over, times two over 3x, and if someone said, well, what is f of zero? You would say f of zero is undefined. Undefined. Why is that? Because you put x equals zero there, you're gonna get two divided by zero, and it's undefined. But if you said, okay, well, can I simplify this a little bit to get the exact same function? Well, we're saying you can say f of x is equal to 4 5ths times x squared, but if you just left it at that, you would get f of zero is equal to zero."}, {"video_title": "Multiplying & dividing rational expressions monomials High School Math Khan Academy.mp3", "Sentence": "Why is that? Because you put x equals zero there, you're gonna get two divided by zero, and it's undefined. But if you said, okay, well, can I simplify this a little bit to get the exact same function? Well, we're saying you can say f of x is equal to 4 5ths times x squared, but if you just left it at that, you would get f of zero is equal to zero. So now it would be defined at zero, but then this would make it a different function. These are two different functions the way they're written right over here. Instead, to make it clear that this is equivalent to that one, you would have to say x cannot be equal to zero."}, {"video_title": "Multiplying & dividing rational expressions monomials High School Math Khan Academy.mp3", "Sentence": "Well, we're saying you can say f of x is equal to 4 5ths times x squared, but if you just left it at that, you would get f of zero is equal to zero. So now it would be defined at zero, but then this would make it a different function. These are two different functions the way they're written right over here. Instead, to make it clear that this is equivalent to that one, you would have to say x cannot be equal to zero. Now these functions are equivalent, because now if you said f of zero, you'd say, all right, x cannot be equal to zero. This would be the case if x is anything other than zero, and it's not defined for zero, and so you'd say f of zero is undefined. So now these two functions are equivalent, or these two expressions are algebraically equivalent."}, {"video_title": "Multiplying & dividing rational expressions monomials High School Math Khan Academy.mp3", "Sentence": "Instead, to make it clear that this is equivalent to that one, you would have to say x cannot be equal to zero. Now these functions are equivalent, because now if you said f of zero, you'd say, all right, x cannot be equal to zero. This would be the case if x is anything other than zero, and it's not defined for zero, and so you'd say f of zero is undefined. So now these two functions are equivalent, or these two expressions are algebraically equivalent. So thinking about that, let's tackle this division situation here. So immediately when you look at this, you say, well, what are constraints here? Well, x cannot be equal to zero, because if x was zero, this second, this 5x to the fourth over four would be zero, and you'd be dividing."}, {"video_title": "Multiplying & dividing rational expressions monomials High School Math Khan Academy.mp3", "Sentence": "So now these two functions are equivalent, or these two expressions are algebraically equivalent. So thinking about that, let's tackle this division situation here. So immediately when you look at this, you say, well, what are constraints here? Well, x cannot be equal to zero, because if x was zero, this second, this 5x to the fourth over four would be zero, and you'd be dividing. You'd be dividing by zero. So we can explicitly call out that x cannot be equal to zero. And so if x cannot be equal to zero in the original expression, if the result, whatever we get for the resulting expression, in order for it to be algebraically equivalent, we have to give this same constraint."}, {"video_title": "Multiplying & dividing rational expressions monomials High School Math Khan Academy.mp3", "Sentence": "Well, x cannot be equal to zero, because if x was zero, this second, this 5x to the fourth over four would be zero, and you'd be dividing. You'd be dividing by zero. So we can explicitly call out that x cannot be equal to zero. And so if x cannot be equal to zero in the original expression, if the result, whatever we get for the resulting expression, in order for it to be algebraically equivalent, we have to give this same constraint. So let's multiply this, or let's do the division. So this is going to be the same thing as 2x to the fourth power over seven times the reciprocal, times, the reciprocal of this is going to be four over 5x to the fourth, which is going to be equal to, in the numerator, we're gonna have 8x to the fourth. So we're going to have 8x to the fourth, four times 2x to the fourth, over seven times five times x to the fourth is 35x to the fourth."}, {"video_title": "Multiplying & dividing rational expressions monomials High School Math Khan Academy.mp3", "Sentence": "And so if x cannot be equal to zero in the original expression, if the result, whatever we get for the resulting expression, in order for it to be algebraically equivalent, we have to give this same constraint. So let's multiply this, or let's do the division. So this is going to be the same thing as 2x to the fourth power over seven times the reciprocal, times, the reciprocal of this is going to be four over 5x to the fourth, which is going to be equal to, in the numerator, we're gonna have 8x to the fourth. So we're going to have 8x to the fourth, four times 2x to the fourth, over seven times five times x to the fourth is 35x to the fourth. And now there's something, we can do a little bit of simplification here, both the numerator and the denominator are divisible by x to the fourth. So let's divide by x to the fourth, and we get eight over 35. So once again, you just look at 835, you say, well, this is gonna be defined for any x, x isn't even involved in the expression."}, {"video_title": "Multiplying & dividing rational expressions monomials High School Math Khan Academy.mp3", "Sentence": "So we're going to have 8x to the fourth, four times 2x to the fourth, over seven times five times x to the fourth is 35x to the fourth. And now there's something, we can do a little bit of simplification here, both the numerator and the denominator are divisible by x to the fourth. So let's divide by x to the fourth, and we get eight over 35. So once again, you just look at 835, you say, well, this is gonna be defined for any x, x isn't even involved in the expression. But if we want this to be algebraically equivalent to this first expression, then we have to make the same constraint, x does not, cannot be equal to zero. And to see, this even seems a little bit more nonsensical, to say x cannot be equal to zero for an expression that does not even involve x, but one way to think about it is imagine a function that was defined as g of x is equal to all of this business here. Well, g of zero would be undefined."}, {"video_title": "Multiplying & dividing rational expressions monomials High School Math Khan Academy.mp3", "Sentence": "So once again, you just look at 835, you say, well, this is gonna be defined for any x, x isn't even involved in the expression. But if we want this to be algebraically equivalent to this first expression, then we have to make the same constraint, x does not, cannot be equal to zero. And to see, this even seems a little bit more nonsensical, to say x cannot be equal to zero for an expression that does not even involve x, but one way to think about it is imagine a function that was defined as g of x is equal to all of this business here. Well, g of zero would be undefined. But if you said g of x is equal to 835ths, well now g of zero would be defined as 835ths, which would make it a different function. So to make them algebraically equivalent, you could say g of x is equal to 835ths as long as x does not equal zero, and you could say it's undefined if you want, undefined for x equals zero, or you don't even have to include that second row, and that will literally just make it undefined. But now this expression, this algebraic expression is equivalent to our original one, even though we have simplified it."}, {"video_title": "Example 3 Factoring quadratics by taking a common factor and grouping Algebra II Khan Academy.mp3", "Sentence": "But before we even do that, let's see if there's a common factor across all of these terms. And maybe we can get a 1 coefficient out there. If we can't get a 1 coefficient, we'll at least have a lower coefficient here. If we look at all of these numbers, they all look divisible by 5. In fact, their greatest common factor is 5. So let's at least factor out a 5. So this is equal to 5 times 35k squared divided by 5 is 7k squared."}, {"video_title": "Example 3 Factoring quadratics by taking a common factor and grouping Algebra II Khan Academy.mp3", "Sentence": "If we look at all of these numbers, they all look divisible by 5. In fact, their greatest common factor is 5. So let's at least factor out a 5. So this is equal to 5 times 35k squared divided by 5 is 7k squared. 100k divided by 5 is 20k. And then negative 15 divided by 5 is negative 3. So we were able to factor out a 5, but we still don't have a 1 coefficient here."}, {"video_title": "Example 3 Factoring quadratics by taking a common factor and grouping Algebra II Khan Academy.mp3", "Sentence": "So this is equal to 5 times 35k squared divided by 5 is 7k squared. 100k divided by 5 is 20k. And then negative 15 divided by 5 is negative 3. So we were able to factor out a 5, but we still don't have a 1 coefficient here. So we're still going to have to factor by grouping. But at least the numbers here are smaller, so it'll be easier to think about it in terms of finding numbers whose product is equal to 7 times negative 3 and whose sum is equal to 20. So let's think about that."}, {"video_title": "Example 3 Factoring quadratics by taking a common factor and grouping Algebra II Khan Academy.mp3", "Sentence": "So we were able to factor out a 5, but we still don't have a 1 coefficient here. So we're still going to have to factor by grouping. But at least the numbers here are smaller, so it'll be easier to think about it in terms of finding numbers whose product is equal to 7 times negative 3 and whose sum is equal to 20. So let's think about that. Let's figure out two numbers that if I were to add them, or even better, if I were to take the product, I get 7 times negative 3. 7 times negative 3 is equal to 7 times negative 3, which is equal to negative 21. And if I were to take their sum, if I add those two numbers, it needs to be equal to 20."}, {"video_title": "Example 3 Factoring quadratics by taking a common factor and grouping Algebra II Khan Academy.mp3", "Sentence": "So let's think about that. Let's figure out two numbers that if I were to add them, or even better, if I were to take the product, I get 7 times negative 3. 7 times negative 3 is equal to 7 times negative 3, which is equal to negative 21. And if I were to take their sum, if I add those two numbers, it needs to be equal to 20. It needs to be equal to 20. Now, once again, because their product is a negative number, that means they have to be of different signs. So when you add numbers of different signs, you could view it as you're taking the difference of the positive version."}, {"video_title": "Example 3 Factoring quadratics by taking a common factor and grouping Algebra II Khan Academy.mp3", "Sentence": "And if I were to take their sum, if I add those two numbers, it needs to be equal to 20. It needs to be equal to 20. Now, once again, because their product is a negative number, that means they have to be of different signs. So when you add numbers of different signs, you could view it as you're taking the difference of the positive version. So the difference between the positive versions of the number has to be 20. So the number that immediately jumps out is we're probably going to be dealing with 20 and 21. And 1 will be the negative because we want to get to a positive 20."}, {"video_title": "Example 3 Factoring quadratics by taking a common factor and grouping Algebra II Khan Academy.mp3", "Sentence": "So when you add numbers of different signs, you could view it as you're taking the difference of the positive version. So the difference between the positive versions of the number has to be 20. So the number that immediately jumps out is we're probably going to be dealing with 20 and 21. And 1 will be the negative because we want to get to a positive 20. So let's think about it. So if we think of 20 and negative 1, their product is negative 21. Sorry, if we take 21 and negative 1, their product is negative 21."}, {"video_title": "Example 3 Factoring quadratics by taking a common factor and grouping Algebra II Khan Academy.mp3", "Sentence": "And 1 will be the negative because we want to get to a positive 20. So let's think about it. So if we think of 20 and negative 1, their product is negative 21. Sorry, if we take 21 and negative 1, their product is negative 21. 21 times negative 1 is negative 21. And if you take their sum, 21 plus negative 1, that is equal to 20. So these two numbers right there fit the bill."}, {"video_title": "Example 3 Factoring quadratics by taking a common factor and grouping Algebra II Khan Academy.mp3", "Sentence": "Sorry, if we take 21 and negative 1, their product is negative 21. 21 times negative 1 is negative 21. And if you take their sum, 21 plus negative 1, that is equal to 20. So these two numbers right there fit the bill. Now, let's break up this 20k right here into a 21k and a negative 1k. So let's do that. So let's rewrite the whole thing."}, {"video_title": "Example 3 Factoring quadratics by taking a common factor and grouping Algebra II Khan Academy.mp3", "Sentence": "So these two numbers right there fit the bill. Now, let's break up this 20k right here into a 21k and a negative 1k. So let's do that. So let's rewrite the whole thing. We have 5 times 7k squared. And I'm going to break this 20k into a plus 21k minus k. Or you could say minus 1k if you want. I'm using those two factors to break it up."}, {"video_title": "Example 3 Factoring quadratics by taking a common factor and grouping Algebra II Khan Academy.mp3", "Sentence": "So let's rewrite the whole thing. We have 5 times 7k squared. And I'm going to break this 20k into a plus 21k minus k. Or you could say minus 1k if you want. I'm using those two factors to break it up. And then we finally have the minus 3 right there. Now, the whole point of doing that is so that we can now factor each of the two groups. This could be our first group right here."}, {"video_title": "Example 3 Factoring quadratics by taking a common factor and grouping Algebra II Khan Academy.mp3", "Sentence": "I'm using those two factors to break it up. And then we finally have the minus 3 right there. Now, the whole point of doing that is so that we can now factor each of the two groups. This could be our first group right here. And so what can we factor out of that group right there? Well, both of these are divisible by 7k. So we can write this as 7k times 7k squared divided by 7k."}, {"video_title": "Example 3 Factoring quadratics by taking a common factor and grouping Algebra II Khan Academy.mp3", "Sentence": "This could be our first group right here. And so what can we factor out of that group right there? Well, both of these are divisible by 7k. So we can write this as 7k times 7k squared divided by 7k. You're just going to have a k left over. And then plus 21k divided by 7k is just going to be a 3. So that factors into that."}, {"video_title": "Example 3 Factoring quadratics by taking a common factor and grouping Algebra II Khan Academy.mp3", "Sentence": "So we can write this as 7k times 7k squared divided by 7k. You're just going to have a k left over. And then plus 21k divided by 7k is just going to be a 3. So that factors into that. And then we can look at this group right here. They have a common factor. Well, we can factor out a negative 1 if we like."}, {"video_title": "Example 3 Factoring quadratics by taking a common factor and grouping Algebra II Khan Academy.mp3", "Sentence": "So that factors into that. And then we can look at this group right here. They have a common factor. Well, we can factor out a negative 1 if we like. So this is equal to negative 1 times k divided by negative 1 is k. Negative 3 divided by negative 1 is positive 3. And of course, we have this 5 sitting out there. We have this 5 sitting out there the whole time."}, {"video_title": "Example 3 Factoring quadratics by taking a common factor and grouping Algebra II Khan Academy.mp3", "Sentence": "Well, we can factor out a negative 1 if we like. So this is equal to negative 1 times k divided by negative 1 is k. Negative 3 divided by negative 1 is positive 3. And of course, we have this 5 sitting out there. We have this 5 sitting out there the whole time. Now, ignoring that 5 for a second, you see that both of these inside terms have k plus 3 as a factor. This has this k plus 3 as a factor. So we can factor that out."}, {"video_title": "Example 3 Factoring quadratics by taking a common factor and grouping Algebra II Khan Academy.mp3", "Sentence": "We have this 5 sitting out there the whole time. Now, ignoring that 5 for a second, you see that both of these inside terms have k plus 3 as a factor. This has this k plus 3 as a factor. So we can factor that out. So let's ignore this 5 for a second. This inside part right here, the stuff that's inside the parentheses, we can factor k plus 3 out. And it becomes k plus 3 times 7k minus 1."}, {"video_title": "Example 3 Factoring quadratics by taking a common factor and grouping Algebra II Khan Academy.mp3", "Sentence": "So we can factor that out. So let's ignore this 5 for a second. This inside part right here, the stuff that's inside the parentheses, we can factor k plus 3 out. And it becomes k plus 3 times 7k minus 1. And if this seems a little bizarre to you, distribute the k plus 3 onto this. k plus 3 times 7k is that term. k plus 3 times negative 1 is that term."}, {"video_title": "Example 3 Factoring quadratics by taking a common factor and grouping Algebra II Khan Academy.mp3", "Sentence": "And it becomes k plus 3 times 7k minus 1. And if this seems a little bizarre to you, distribute the k plus 3 onto this. k plus 3 times 7k is that term. k plus 3 times negative 1 is that term. And of course, the whole time, you have that 5 sitting outside. You have that 5. We don't have to put parentheses there."}, {"video_title": "Example 3 Factoring quadratics by taking a common factor and grouping Algebra II Khan Academy.mp3", "Sentence": "k plus 3 times negative 1 is that term. And of course, the whole time, you have that 5 sitting outside. You have that 5. We don't have to put parentheses there. 5 times k plus 3 times 7k minus 1. And we factored it. We're done."}, {"video_title": "Function symmetry introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So let's first think about what an even function is. One way to think about an even function is that if you were to flip it over the y-axis, that the function looks the same. So here's a classic example of an even function. It would be this right over here, your classic parabola where your vertex is on the y-axis. This is an even function. So this one is maybe the graph of f of x is equal to x squared. And notice, if you were to flip it over the y-axis, you're going to get the exact same graph."}, {"video_title": "Function symmetry introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "It would be this right over here, your classic parabola where your vertex is on the y-axis. This is an even function. So this one is maybe the graph of f of x is equal to x squared. And notice, if you were to flip it over the y-axis, you're going to get the exact same graph. Now a way that we can talk about that mathematically, and we have talked about this when we introduced the idea of reflection, to say that a function is equal to its reflection over the y-axis, that's just saying that f of x is equal to f of negative x. Because if you were to replace your x's with a negative x, that flips your function over the y-axis. Now what about odd functions?"}, {"video_title": "Function symmetry introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And notice, if you were to flip it over the y-axis, you're going to get the exact same graph. Now a way that we can talk about that mathematically, and we have talked about this when we introduced the idea of reflection, to say that a function is equal to its reflection over the y-axis, that's just saying that f of x is equal to f of negative x. Because if you were to replace your x's with a negative x, that flips your function over the y-axis. Now what about odd functions? So odd functions, you get the same function if you flip over the y and the x-axes. So let me draw a classic example of an odd function. Our classic example would be f of x is equal to x to the third, is equal to x to the third, and it looks something like this."}, {"video_title": "Function symmetry introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Now what about odd functions? So odd functions, you get the same function if you flip over the y and the x-axes. So let me draw a classic example of an odd function. Our classic example would be f of x is equal to x to the third, is equal to x to the third, and it looks something like this. So notice, if you were to flip first over the y-axis, you would get something that looks like this. So I'll do it as a dotted line. If you were to flip just over the y-axis, it would look like this."}, {"video_title": "Function symmetry introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Our classic example would be f of x is equal to x to the third, is equal to x to the third, and it looks something like this. So notice, if you were to flip first over the y-axis, you would get something that looks like this. So I'll do it as a dotted line. If you were to flip just over the y-axis, it would look like this. And then if you were to flip that over the x-axis, well, then you're going to get the same function again. Now how would we write this down mathematically? Well, that means that our function is equivalent to not only flipping it over the y-axis, which would be f of negative x, but then flipping that over the x-axis, which is just taking the negative of that."}, {"video_title": "Function symmetry introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "If you were to flip just over the y-axis, it would look like this. And then if you were to flip that over the x-axis, well, then you're going to get the same function again. Now how would we write this down mathematically? Well, that means that our function is equivalent to not only flipping it over the y-axis, which would be f of negative x, but then flipping that over the x-axis, which is just taking the negative of that. So this is doing two flips. So some of you might be noticing a pattern or think you might be on the verge of seeing a pattern that connects the words even and odd with the notions that we know from earlier in our mathematical lives. I've just shown you an even function where the exponent is an even number."}, {"video_title": "Function symmetry introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Well, that means that our function is equivalent to not only flipping it over the y-axis, which would be f of negative x, but then flipping that over the x-axis, which is just taking the negative of that. So this is doing two flips. So some of you might be noticing a pattern or think you might be on the verge of seeing a pattern that connects the words even and odd with the notions that we know from earlier in our mathematical lives. I've just shown you an even function where the exponent is an even number. And I've just showed you an odd function where the exponent is an odd number. Now I encourage you to try out many, many more polynomials and try out the exponents, but it turns out that if you just have f of x is equal to, if you just have f of x is equal to x to the n, then this is going to be an even function if n is even, and it's going to be an odd function if n is odd. So that's one connection."}, {"video_title": "Function symmetry introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "I've just shown you an even function where the exponent is an even number. And I've just showed you an odd function where the exponent is an odd number. Now I encourage you to try out many, many more polynomials and try out the exponents, but it turns out that if you just have f of x is equal to, if you just have f of x is equal to x to the n, then this is going to be an even function if n is even, and it's going to be an odd function if n is odd. So that's one connection. Now some of you are thinking, wait, but there seem to be a lot of functions that are neither even nor odd, and that is indeed the case. For example, if you just had the graph x squared plus two, this right over here is still going to be even because if you flip it over, you have the symmetry around the y-axis, you're going to get back to itself. But if you had x minus two squared, which looks like this, x minus two, that would shift two to the right, it'll look like that, that is no longer even because notice, if you flip it over the y-axis, you're no longer getting the same function."}, {"video_title": "Function symmetry introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So that's one connection. Now some of you are thinking, wait, but there seem to be a lot of functions that are neither even nor odd, and that is indeed the case. For example, if you just had the graph x squared plus two, this right over here is still going to be even because if you flip it over, you have the symmetry around the y-axis, you're going to get back to itself. But if you had x minus two squared, which looks like this, x minus two, that would shift two to the right, it'll look like that, that is no longer even because notice, if you flip it over the y-axis, you're no longer getting the same function. So it's not just the exponent, it also matters on the structure of the expression itself. If you have something very simple like just x to the n, well then, that could be or that would be even or odd depending on what your n is. Similarly, if we were to shift this f of x, if we were to even shift it up, it's no longer, it is no longer, so if this is x to the third, let's say plus three, this is no longer odd because you flip it over once, you get right over there, but then you flip it again, you're going to get this."}, {"video_title": "Function symmetry introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "But if you had x minus two squared, which looks like this, x minus two, that would shift two to the right, it'll look like that, that is no longer even because notice, if you flip it over the y-axis, you're no longer getting the same function. So it's not just the exponent, it also matters on the structure of the expression itself. If you have something very simple like just x to the n, well then, that could be or that would be even or odd depending on what your n is. Similarly, if we were to shift this f of x, if we were to even shift it up, it's no longer, it is no longer, so if this is x to the third, let's say plus three, this is no longer odd because you flip it over once, you get right over there, but then you flip it again, you're going to get this. You're going to get something like this. So you're no longer back to your original function. Now an interesting thing to think about, can you imagine a function that is both even and odd?"}, {"video_title": "Function symmetry introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Similarly, if we were to shift this f of x, if we were to even shift it up, it's no longer, it is no longer, so if this is x to the third, let's say plus three, this is no longer odd because you flip it over once, you get right over there, but then you flip it again, you're going to get this. You're going to get something like this. So you're no longer back to your original function. Now an interesting thing to think about, can you imagine a function that is both even and odd? So I encourage you to pause that video or pause the video and try to think about it. Is there a function where f of x is equal to f of negative x and f of x is equal to the negative of f of negative x? Well, I'll give you a hint or actually I'll just give you the answer."}, {"video_title": "Function symmetry introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Now an interesting thing to think about, can you imagine a function that is both even and odd? So I encourage you to pause that video or pause the video and try to think about it. Is there a function where f of x is equal to f of negative x and f of x is equal to the negative of f of negative x? Well, I'll give you a hint or actually I'll just give you the answer. Imagine if f of x is just equal to the constant zero. Notice, this thing is just a horizontal line, just like that at y is equal to zero, and if you flip it over the y-axis, you get back to where it was before, then if you flip it over the x-axis again, then you're still back to where you were before. So this over here is both even and odd, very interesting case."}, {"video_title": "Understanding function inverses example Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And it's an interactive one where we can move this line around. And they tell us the graph of h of x is in green. So that's this dotted green line right over here. The graph of h of x is the green dashed line segment shown below, so that's this. Drag the end points of the segment below to graph h inverse of x. So there's a couple of ways to tackle it. Perhaps the simplest one is we say, okay, look, h of x, what does h of x map from and to?"}, {"video_title": "Understanding function inverses example Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "The graph of h of x is the green dashed line segment shown below, so that's this. Drag the end points of the segment below to graph h inverse of x. So there's a couple of ways to tackle it. Perhaps the simplest one is we say, okay, look, h of x, what does h of x map from and to? So h of x, this point shows that h of x, if you input negative eight into h of x, h of negative eight is one. So it's mapping from negative eight to one. Well, the inverse of that then should map from one to negative eight."}, {"video_title": "Understanding function inverses example Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Perhaps the simplest one is we say, okay, look, h of x, what does h of x map from and to? So h of x, this point shows that h of x, if you input negative eight into h of x, h of negative eight is one. So it's mapping from negative eight to one. Well, the inverse of that then should map from one to negative eight. It should map from one to negative eight. So let's put that point on the graph of the inverse. And then let's go on the other end."}, {"video_title": "Understanding function inverses example Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Well, the inverse of that then should map from one to negative eight. It should map from one to negative eight. So let's put that point on the graph of the inverse. And then let's go on the other end. So let me just move this a little bit out of the way. On the other end of h of x, we see that when you input three into h of x, when x is equal to three, h of x is equal to negative four. So this point shows us that it's mapping from three to negative four."}, {"video_title": "Understanding function inverses example Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And then let's go on the other end. So let me just move this a little bit out of the way. On the other end of h of x, we see that when you input three into h of x, when x is equal to three, h of x is equal to negative four. So this point shows us that it's mapping from three to negative four. So the inverse of that would map from negative four to three. So if you input negative four, it should output three. If you input negative four, it should output three."}, {"video_title": "Understanding function inverses example Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So this point shows us that it's mapping from three to negative four. So the inverse of that would map from negative four to three. So if you input negative four, it should output three. If you input negative four, it should output three. And since we took the two endpoints of this line, and we essentially found the inverse mapping of it, what I've just done here is I've graphed the inverse. And another way to think about the inverse is if you were to draw the line y equals x. So I don't have my little pen tool right here because I'm operating on a web page."}, {"video_title": "Understanding function inverses example Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "If you input negative four, it should output three. And since we took the two endpoints of this line, and we essentially found the inverse mapping of it, what I've just done here is I've graphed the inverse. And another way to think about the inverse is if you were to draw the line y equals x. So I don't have my little pen tool right here because I'm operating on a web page. But if you imagine the line y equals x, these things should be reflections around the line y equals x, because one way to think about it is you're swapping the x's for the y's over here. And you see that. If you were to draw the line y equals x, these things, if you flipped it around the line y equals x, the green line, you would actually get the yellow line."}, {"video_title": "Understanding function inverses example Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So I don't have my little pen tool right here because I'm operating on a web page. But if you imagine the line y equals x, these things should be reflections around the line y equals x, because one way to think about it is you're swapping the x's for the y's over here. And you see that. If you were to draw the line y equals x, these things, if you flipped it around the line y equals x, the green line, you would actually get the yellow line. This would flip over there, and this would flip over there. But either way, we're done. We have graphed h inverse of x."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "But I will recommend you memorize it, with the caveat that you also remember how to prove it. Because I don't want you to just remember things and not know where they came from. But with that said, let me show you what I'm talking about. It's the quadratic formula. And as you might guess, it is to solve for the roots, or the zeros, of quadratic equations. So let's speak in very general terms, then I'll show you some examples. So let's say I have an equation of the form ax squared plus bx plus c is equal to 0."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "It's the quadratic formula. And as you might guess, it is to solve for the roots, or the zeros, of quadratic equations. So let's speak in very general terms, then I'll show you some examples. So let's say I have an equation of the form ax squared plus bx plus c is equal to 0. You should recognize this. This is a quadratic equation, where a, b, and c are, well a is the coefficient on the x squared term, or the second degree term. b is the coefficient on the x term."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So let's say I have an equation of the form ax squared plus bx plus c is equal to 0. You should recognize this. This is a quadratic equation, where a, b, and c are, well a is the coefficient on the x squared term, or the second degree term. b is the coefficient on the x term. And then c is, you can imagine, the coefficient on the x to the 0 term, or it's the constant term. Now, given that you have a general quadratic equation like this, the quadratic formula tells us that the solutions to this equation are x is equal to negative b plus or minus the square root of b squared minus 4ac, all of that over 2a. I know it seems crazy and convoluted and hard for you to memorize right now, but as you get a lot more practice, you'll see that it actually is a pretty reasonable formula to stick in your brain someplace."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "b is the coefficient on the x term. And then c is, you can imagine, the coefficient on the x to the 0 term, or it's the constant term. Now, given that you have a general quadratic equation like this, the quadratic formula tells us that the solutions to this equation are x is equal to negative b plus or minus the square root of b squared minus 4ac, all of that over 2a. I know it seems crazy and convoluted and hard for you to memorize right now, but as you get a lot more practice, you'll see that it actually is a pretty reasonable formula to stick in your brain someplace. And you might say, gee, this is a wacky formula. Where did it come from? And in the next video, I'm going to show you where it came from."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "I know it seems crazy and convoluted and hard for you to memorize right now, but as you get a lot more practice, you'll see that it actually is a pretty reasonable formula to stick in your brain someplace. And you might say, gee, this is a wacky formula. Where did it come from? And in the next video, I'm going to show you where it came from. But I want you to get used to using it first. But it really just came from completing the square on this equation right there. If you complete the square here, you're actually going to get this solution."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And in the next video, I'm going to show you where it came from. But I want you to get used to using it first. But it really just came from completing the square on this equation right there. If you complete the square here, you're actually going to get this solution. And that is the quadratic formula right there. So let's apply it to some problem. So let's start off with something that we could have factored, just to verify that it's giving us the same answer."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "If you complete the square here, you're actually going to get this solution. And that is the quadratic formula right there. So let's apply it to some problem. So let's start off with something that we could have factored, just to verify that it's giving us the same answer. So let's say we have x squared plus 4x minus 21 is equal to 0. So in this situation, a is equal to 1. The coefficient on the x squared term is 1. b is equal to 4."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So let's start off with something that we could have factored, just to verify that it's giving us the same answer. So let's say we have x squared plus 4x minus 21 is equal to 0. So in this situation, a is equal to 1. The coefficient on the x squared term is 1. b is equal to 4. The coefficient on the x term. And then c is equal to negative 21, the constant term. And let's just plug it in the formula."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "The coefficient on the x squared term is 1. b is equal to 4. The coefficient on the x term. And then c is equal to negative 21, the constant term. And let's just plug it in the formula. So what do we get? We get x. This tells us that x is going to be equal to negative b."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And let's just plug it in the formula. So what do we get? We get x. This tells us that x is going to be equal to negative b. Negative b is negative 4. I put a negative sign in front of that. Negative b plus or minus the square root of b squared."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "This tells us that x is going to be equal to negative b. Negative b is negative 4. I put a negative sign in front of that. Negative b plus or minus the square root of b squared. b squared is 16. 4 squared is 16. Minus 4 times a, which is 1, times c, which is negative 21."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Negative b plus or minus the square root of b squared. b squared is 16. 4 squared is 16. Minus 4 times a, which is 1, times c, which is negative 21. So we can put a 21 out there. And that negative sign will cancel out just like that. This is the first time we're doing it."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Minus 4 times a, which is 1, times c, which is negative 21. So we can put a 21 out there. And that negative sign will cancel out just like that. This is the first time we're doing it. Let me not skip too many steps. So negative 21, just so you can see how it fit in. And then all of that over 2 times a. a is 1, so all of that over 2."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "This is the first time we're doing it. Let me not skip too many steps. So negative 21, just so you can see how it fit in. And then all of that over 2 times a. a is 1, so all of that over 2. So what does this simplify? Or hopefully it simplifies. So we get x is equal to negative 4 plus or minus the square root of, see we have a negative times a negative."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And then all of that over 2 times a. a is 1, so all of that over 2. So what does this simplify? Or hopefully it simplifies. So we get x is equal to negative 4 plus or minus the square root of, see we have a negative times a negative. That's going to give us a positive. And we have 16 plus, let's see, this is 4 times 1 is 4, times 21 is 84. 16 plus 84 is 100."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So we get x is equal to negative 4 plus or minus the square root of, see we have a negative times a negative. That's going to give us a positive. And we have 16 plus, let's see, this is 4 times 1 is 4, times 21 is 84. 16 plus 84 is 100. That's nice. It's a nice perfect square. All of that over 2."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "16 plus 84 is 100. That's nice. It's a nice perfect square. All of that over 2. And so this is going to be equal to negative 4 plus or minus 10 over 2. We could just divide both of these terms by 2 right now. So this is equal to negative 4 divided by 2 is negative 2."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "All of that over 2. And so this is going to be equal to negative 4 plus or minus 10 over 2. We could just divide both of these terms by 2 right now. So this is equal to negative 4 divided by 2 is negative 2. Plus or minus 10 divided by 2 is 5. So that tells us that x could be equal to negative 2 plus 5, which is 3. Or x could be equal to negative 2 minus 5, which is negative 7."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So this is equal to negative 4 divided by 2 is negative 2. Plus or minus 10 divided by 2 is 5. So that tells us that x could be equal to negative 2 plus 5, which is 3. Or x could be equal to negative 2 minus 5, which is negative 7. So the quadratic formula seems to have given us an answer for this, you can verify just by substituting back in that these do work. Or you could even just try to factor this right here. You say what two numbers, when you take their product, you get negative 21."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Or x could be equal to negative 2 minus 5, which is negative 7. So the quadratic formula seems to have given us an answer for this, you can verify just by substituting back in that these do work. Or you could even just try to factor this right here. You say what two numbers, when you take their product, you get negative 21. And when you take their sum, you get positive 4. So you'd get x plus 7 times x minus 3 is equal to negative 21. Notice 7 times negative 3 is negative 21."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "You say what two numbers, when you take their product, you get negative 21. And when you take their sum, you get positive 4. So you'd get x plus 7 times x minus 3 is equal to negative 21. Notice 7 times negative 3 is negative 21. 7 minus 3 is positive 4. You would get x plus, sorry, it's not negative 21, is equal to 0. There should be a 0 there."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Notice 7 times negative 3 is negative 21. 7 minus 3 is positive 4. You would get x plus, sorry, it's not negative 21, is equal to 0. There should be a 0 there. So you get x plus 7 is equal to 0. Or x minus 3 is equal to 0. x could be equal to negative 7. Or x could be equal to 3."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "There should be a 0 there. So you get x plus 7 is equal to 0. Or x minus 3 is equal to 0. x could be equal to negative 7. Or x could be equal to 3. So it definitely gives us the same answer as factoring. So you might say, hey, why bother with this crazy mess? And the reason we want to bother with this crazy mess is it'll also work for problems that are hard to factor."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Or x could be equal to 3. So it definitely gives us the same answer as factoring. So you might say, hey, why bother with this crazy mess? And the reason we want to bother with this crazy mess is it'll also work for problems that are hard to factor. And let's do a couple of those. Let's do some hard to factor problems right now. So let's scroll down, get some fresh real estate."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And the reason we want to bother with this crazy mess is it'll also work for problems that are hard to factor. And let's do a couple of those. Let's do some hard to factor problems right now. So let's scroll down, get some fresh real estate. Let's rewrite the formula again, just in case we haven't had it memorized yet. x is going to be equal to negative b plus or minus the square root of b squared minus 4ac. All of that over 2a."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So let's scroll down, get some fresh real estate. Let's rewrite the formula again, just in case we haven't had it memorized yet. x is going to be equal to negative b plus or minus the square root of b squared minus 4ac. All of that over 2a. Now let's apply this to another problem. Let's say we have the equation 3x squared plus 6x is equal to negative 10. Well, the first thing we want to do is get it in the form where all of our terms are on the left-hand side."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "All of that over 2a. Now let's apply this to another problem. Let's say we have the equation 3x squared plus 6x is equal to negative 10. Well, the first thing we want to do is get it in the form where all of our terms are on the left-hand side. So let's add 10 to both sides of this equation. We get 3x squared plus 6x plus 10 is equal to 0. And now we can use the quadratic formula."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Well, the first thing we want to do is get it in the form where all of our terms are on the left-hand side. So let's add 10 to both sides of this equation. We get 3x squared plus 6x plus 10 is equal to 0. And now we can use the quadratic formula. So let's apply it here. So a is equal to 3. That is a."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And now we can use the quadratic formula. So let's apply it here. So a is equal to 3. That is a. This is b. And this right there is c. So the quadratic formula tells us the solutions to this equation, the roots of this quadratic function, I guess we could call it. x is going to be equal to negative b. b is 6."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "That is a. This is b. And this right there is c. So the quadratic formula tells us the solutions to this equation, the roots of this quadratic function, I guess we could call it. x is going to be equal to negative b. b is 6. So negative 6 plus or minus the square root of b squared. b is 6. So we get 6 squared minus 4 times a, which is 3, times c, which is 10."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "x is going to be equal to negative b. b is 6. So negative 6 plus or minus the square root of b squared. b is 6. So we get 6 squared minus 4 times a, which is 3, times c, which is 10. Stretch out the radical a little bit. All of that over 2 times a. 2 times 3."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So we get 6 squared minus 4 times a, which is 3, times c, which is 10. Stretch out the radical a little bit. All of that over 2 times a. 2 times 3. So we get x is equal to negative 6 plus or minus the square root of 36 minus 4 times 3 times 10. So this is minus 120. All of that over 6."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "2 times 3. So we get x is equal to negative 6 plus or minus the square root of 36 minus 4 times 3 times 10. So this is minus 120. All of that over 6. So this is interesting. You might already realize why it's interesting. What is this going to simplify to?"}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "All of that over 6. So this is interesting. You might already realize why it's interesting. What is this going to simplify to? 36 minus 120 is what? That's 84. If I'm doing my 120 minus 36, we make this into a 10."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "What is this going to simplify to? 36 minus 120 is what? That's 84. If I'm doing my 120 minus 36, we make this into a 10. Then this will become an 11. This is a 4. It is 84."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "If I'm doing my 120 minus 36, we make this into a 10. Then this will become an 11. This is a 4. It is 84. So this is going to be equal to negative 6 plus or minus the square root of. But not positive 84. That's if it's 120 minus 36."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "It is 84. So this is going to be equal to negative 6 plus or minus the square root of. But not positive 84. That's if it's 120 minus 36. We have 36 minus 120. It's going to be negative 84. All of that over 6."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "That's if it's 120 minus 36. We have 36 minus 120. It's going to be negative 84. All of that over 6. So you might say, gee, this is crazy. What is this silly quadratic formula you're introducing me to, Sal? It's worthless."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "All of that over 6. So you might say, gee, this is crazy. What is this silly quadratic formula you're introducing me to, Sal? It's worthless. It just gave me a square root of a negative number. It's not giving me an answer. And the reason why it's not giving you an answer, or at least an answer that you might want, is because this will have no real solutions."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "It's worthless. It just gave me a square root of a negative number. It's not giving me an answer. And the reason why it's not giving you an answer, or at least an answer that you might want, is because this will have no real solutions. In the future, we're going to introduce something called an imaginary number, which is a square root of a negative number, and then we can actually express this in terms of those numbers. So this actually does have solutions, but they involve imaginary numbers. So this actually has no real solution."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And the reason why it's not giving you an answer, or at least an answer that you might want, is because this will have no real solutions. In the future, we're going to introduce something called an imaginary number, which is a square root of a negative number, and then we can actually express this in terms of those numbers. So this actually does have solutions, but they involve imaginary numbers. So this actually has no real solution. We're taking the square root of a negative number. So the b squared with the b squared minus 4ac, if this term right here is negative, then you're not going to have any real solutions. And let's verify that for ourselves."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So this actually has no real solution. We're taking the square root of a negative number. So the b squared with the b squared minus 4ac, if this term right here is negative, then you're not going to have any real solutions. And let's verify that for ourselves. Let's get our graphing calculator out. Let's graph this equation right here. So let's get the graph."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And let's verify that for ourselves. Let's get our graphing calculator out. Let's graph this equation right here. So let's get the graph. The y is equal to, that's what I had there before. Let's see, I have 3x. Let me clear this, right."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So let's get the graph. The y is equal to, that's what I had there before. Let's see, I have 3x. Let me clear this, right. So I get 3x squared plus 6x plus 10. So that's the equation. We're going to see where it intersects the x-axis."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Let me clear this, right. So I get 3x squared plus 6x plus 10. So that's the equation. We're going to see where it intersects the x-axis. Where does it equal 0? So let me graph it. Let's graph it."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "We're going to see where it intersects the x-axis. Where does it equal 0? So let me graph it. Let's graph it. Notice, this thing just comes down and then goes back up. Its vertex is sitting here above the x-axis, and it's upward opening. It never intersects the x-axis."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Let's graph it. Notice, this thing just comes down and then goes back up. Its vertex is sitting here above the x-axis, and it's upward opening. It never intersects the x-axis. So at no point will this expression, will this function equal 0. At no point will y equal 0 on this graph. So once again, the quadratic formula seems to be working."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "It never intersects the x-axis. So at no point will this expression, will this function equal 0. At no point will y equal 0 on this graph. So once again, the quadratic formula seems to be working. Let's do one more example. You can never see enough examples here. And I want to do ones that are maybe not so obvious to factor."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So once again, the quadratic formula seems to be working. Let's do one more example. You can never see enough examples here. And I want to do ones that are maybe not so obvious to factor. So let's say we get negative 3x squared plus 12x plus 1 is equal to 0. Now let's try to do it just having the quadratic formula in our brain. So the x's that satisfy this equation are going to be negative b."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And I want to do ones that are maybe not so obvious to factor. So let's say we get negative 3x squared plus 12x plus 1 is equal to 0. Now let's try to do it just having the quadratic formula in our brain. So the x's that satisfy this equation are going to be negative b. This is b. So negative b is negative 12 plus or minus the square root of b squared of 144. That's b squared minus 4 times a, which is negative 3, times c, which is 1."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So the x's that satisfy this equation are going to be negative b. This is b. So negative b is negative 12 plus or minus the square root of b squared of 144. That's b squared minus 4 times a, which is negative 3, times c, which is 1. All of that over 2 times a. Over 2 times negative 3. So all of that over negative 6."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "That's b squared minus 4 times a, which is negative 3, times c, which is 1. All of that over 2 times a. Over 2 times negative 3. So all of that over negative 6. This is going to be equal to negative 12 plus or minus the square root of, what is this? The negative times the negative. They cancel out."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So all of that over negative 6. This is going to be equal to negative 12 plus or minus the square root of, what is this? The negative times the negative. They cancel out. So I have 144 plus 12. So that is 156. 144 plus 12."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "They cancel out. So I have 144 plus 12. So that is 156. 144 plus 12. All of that over negative 6. Now, I suspect we can simplify this 156. We can maybe bring some things out of the radical sign."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "144 plus 12. All of that over negative 6. Now, I suspect we can simplify this 156. We can maybe bring some things out of the radical sign. So let's attempt to do that. So let's do a prime factorization of 156. Sometimes this is the hardest part, simplifying the radical."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "We can maybe bring some things out of the radical sign. So let's attempt to do that. So let's do a prime factorization of 156. Sometimes this is the hardest part, simplifying the radical. So 156 is the same thing as 2 times 78. 78 is the same thing as 2 times what? That's 2 times 39."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Sometimes this is the hardest part, simplifying the radical. So 156 is the same thing as 2 times 78. 78 is the same thing as 2 times what? That's 2 times 39. So the square root of 156 is equal to the square root of 2 times 2 times 39. Or we could say that's the square root of 2 times 2 times the square root of 39. And this obviously is just going to be the square root of 4, or this is the square root of 2 times 2."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "That's 2 times 39. So the square root of 156 is equal to the square root of 2 times 2 times 39. Or we could say that's the square root of 2 times 2 times the square root of 39. And this obviously is just going to be the square root of 4, or this is the square root of 2 times 2. It's just 2. 2 square roots of 39, if I did that properly. Let's see, 4 times 39."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And this obviously is just going to be the square root of 4, or this is the square root of 2 times 2. It's just 2. 2 square roots of 39, if I did that properly. Let's see, 4 times 39. Yeah, it looks like it's right. 120, yep. So this up here will simplify to negative 12 plus or minus 2 times the square root of 39."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Let's see, 4 times 39. Yeah, it looks like it's right. 120, yep. So this up here will simplify to negative 12 plus or minus 2 times the square root of 39. All of that over negative 6. Now we can divide the numerator and the denominator by 2. So this will be equal to negative 6 plus or minus the square root of 39 over negative 3."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So this up here will simplify to negative 12 plus or minus 2 times the square root of 39. All of that over negative 6. Now we can divide the numerator and the denominator by 2. So this will be equal to negative 6 plus or minus the square root of 39 over negative 3. Or we could separate these two terms out. We could say this is equal to negative 6 over negative 3 plus or minus the square root of 39 over negative 3. Now this is just a 2 right here, right?"}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So this will be equal to negative 6 plus or minus the square root of 39 over negative 3. Or we could separate these two terms out. We could say this is equal to negative 6 over negative 3 plus or minus the square root of 39 over negative 3. Now this is just a 2 right here, right? These cancel out. 6 divided by 3 is 2. So we get 2."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Now this is just a 2 right here, right? These cancel out. 6 divided by 3 is 2. So we get 2. And now notice, if this is plus and we use this minus sign, the plus will become negative and the negative will become positive. But it still doesn't matter, right? We could say minus or plus, or that's the same thing as plus or minus the square root of 39 over 3."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So we get 2. And now notice, if this is plus and we use this minus sign, the plus will become negative and the negative will become positive. But it still doesn't matter, right? We could say minus or plus, or that's the same thing as plus or minus the square root of 39 over 3. I think that's about as simple as we can get this answer. Now I want to make a very clear point of what I did at that last step. I did not forget about this negative sign."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "We could say minus or plus, or that's the same thing as plus or minus the square root of 39 over 3. I think that's about as simple as we can get this answer. Now I want to make a very clear point of what I did at that last step. I did not forget about this negative sign. I just said it doesn't matter. It's going to turn the positive into the negative. It's going to turn the negative into the positive."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "I did not forget about this negative sign. I just said it doesn't matter. It's going to turn the positive into the negative. It's going to turn the negative into the positive. Let me rewrite this. So this right here can be rewritten as 2 plus the square root of 39 over negative 3. Or 2 minus the square root of 39 over negative 3."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "It's going to turn the negative into the positive. Let me rewrite this. So this right here can be rewritten as 2 plus the square root of 39 over negative 3. Or 2 minus the square root of 39 over negative 3. That's what the plus or minus means. It could be this or that, or both of them really. Now in this situation, this negative 3 will turn into 2 minus the square root of 39 over 3."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Or 2 minus the square root of 39 over negative 3. That's what the plus or minus means. It could be this or that, or both of them really. Now in this situation, this negative 3 will turn into 2 minus the square root of 39 over 3. I'm just taking this negative out. Here the negative and the negative will become a positive, and you get 2 plus the square root of 39 over 3. Negative times a negative is a positive."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Now in this situation, this negative 3 will turn into 2 minus the square root of 39 over 3. I'm just taking this negative out. Here the negative and the negative will become a positive, and you get 2 plus the square root of 39 over 3. Negative times a negative is a positive. So once again, you have 2 plus or minus the square root of 39 over 3. 2 plus or minus the square root of 39 over 3 are solutions to this equation right there. Let's verify."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Negative times a negative is a positive. So once again, you have 2 plus or minus the square root of 39 over 3. 2 plus or minus the square root of 39 over 3 are solutions to this equation right there. Let's verify. I'm just curious what the graph looks like. So let's just look at it. So let me clear this."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Let's verify. I'm just curious what the graph looks like. So let's just look at it. So let me clear this. Where's the clear button? So we have negative 3x squared plus 12x plus 1. And let's graph it."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So let me clear this. Where's the clear button? So we have negative 3x squared plus 12x plus 1. And let's graph it. Let's see where it intersects the x axis. It goes up there and then back down again. So 2 plus or minus the square root of 39 is going to be a little bit more than 6, right?"}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And let's graph it. Let's see where it intersects the x axis. It goes up there and then back down again. So 2 plus or minus the square root of 39 is going to be a little bit more than 6, right? Because 36 is 6 squared. So it's going to be a little bit more than 6. So this is going to be a little bit more than 2."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So 2 plus or minus the square root of 39 is going to be a little bit more than 6, right? Because 36 is 6 squared. So it's going to be a little bit more than 6. So this is going to be a little bit more than 2. A little bit more than 6 divided by 3 is a little bit more than 2. So you're going to get one value that's a little bit more than 4, and then another value that should be a little bit less than 1. And that looks like the case."}, {"video_title": "How to use the quadratic formula Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So this is going to be a little bit more than 2. A little bit more than 6 divided by 3 is a little bit more than 2. So you're going to get one value that's a little bit more than 4, and then another value that should be a little bit less than 1. And that looks like the case. You have 1, 2, 3, 4. You have a value that's pretty close to 4. And then you have another value that is a little bit, maybe it looks close to 0, but a little bit less than that."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "And what you're going to see, it's really just an application of the Pythagorean theorem. So let's start with an example. Let's say I have the point. I'll do it in a darker color so we can see it on the graph paper, let's say I have the point 3, negative 4. So if I were to graph it, I'd go 1, 2, 3, and then I'd go down 4. 1, 2, 3, 4 right there is 3, negative 4. And let's say I also have the point 6, 0."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "I'll do it in a darker color so we can see it on the graph paper, let's say I have the point 3, negative 4. So if I were to graph it, I'd go 1, 2, 3, and then I'd go down 4. 1, 2, 3, 4 right there is 3, negative 4. And let's say I also have the point 6, 0. So 1, 2, 3, 4, 5, 6. And then there's no movement in the y direction. We're just sitting on the x-axis."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "And let's say I also have the point 6, 0. So 1, 2, 3, 4, 5, 6. And then there's no movement in the y direction. We're just sitting on the x-axis. The y-coordinate is 0, so that's 6, 0. And what I want to figure out is the distance between these two points. How far is this blue point away from this orange point?"}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "We're just sitting on the x-axis. The y-coordinate is 0, so that's 6, 0. And what I want to figure out is the distance between these two points. How far is this blue point away from this orange point? And at first you're like, gee, Sal, I don't think I've ever seen anything about how to solve for a distance like this, and what are you even talking about, the Pythagorean theorem? I don't see a triangle there. And if you don't see a triangle, let me draw it for you."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "How far is this blue point away from this orange point? And at first you're like, gee, Sal, I don't think I've ever seen anything about how to solve for a distance like this, and what are you even talking about, the Pythagorean theorem? I don't see a triangle there. And if you don't see a triangle, let me draw it for you. Let me draw the triangle. Let me draw this triangle right there, just like that. Let me actually do several colors here just to really hit the point home."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "And if you don't see a triangle, let me draw it for you. Let me draw the triangle. Let me draw this triangle right there, just like that. Let me actually do several colors here just to really hit the point home. So there's our triangle. And you might immediately recognize this is a right triangle. This is a right angle right there."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Let me actually do several colors here just to really hit the point home. So there's our triangle. And you might immediately recognize this is a right triangle. This is a right angle right there. The base goes straight left to right. The right side goes straight up and down. So we're dealing with a right triangle."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "This is a right angle right there. The base goes straight left to right. The right side goes straight up and down. So we're dealing with a right triangle. So if we could just figure out what the base length is and what this height is, we could use the Pythagorean theorem to figure out this long side, the side that is opposite the right angle, the hypotenuse. This right here, the distance is the hypotenuse of this right triangle. Let me write that down."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So we're dealing with a right triangle. So if we could just figure out what the base length is and what this height is, we could use the Pythagorean theorem to figure out this long side, the side that is opposite the right angle, the hypotenuse. This right here, the distance is the hypotenuse of this right triangle. Let me write that down. The distance is equal to the hypotenuse of this right triangle. So let me draw it a little bit bigger. So this is the hypotenuse right there."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Let me write that down. The distance is equal to the hypotenuse of this right triangle. So let me draw it a little bit bigger. So this is the hypotenuse right there. And then we have the side on the right, the side that goes straight up and down. And then we have our base. Now, how do we figure out?"}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So this is the hypotenuse right there. And then we have the side on the right, the side that goes straight up and down. And then we have our base. Now, how do we figure out? Let's call this d for distance. That's the length of our hypotenuse. How do we figure out the lengths of this up and down side and the base side right here?"}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Now, how do we figure out? Let's call this d for distance. That's the length of our hypotenuse. How do we figure out the lengths of this up and down side and the base side right here? So let's look at the base first. What is this distance? You could even count it on this graph paper."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "How do we figure out the lengths of this up and down side and the base side right here? So let's look at the base first. What is this distance? You could even count it on this graph paper. But here, we're at x is equal to 3. And here, we're at x is equal to 6. We're just moving straight right."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "You could even count it on this graph paper. But here, we're at x is equal to 3. And here, we're at x is equal to 6. We're just moving straight right. This is the same distance as that distance right there. So to figure out that distance, it's literally the end x point. And you can actually go either way, because you're going to square everything."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "We're just moving straight right. This is the same distance as that distance right there. So to figure out that distance, it's literally the end x point. And you can actually go either way, because you're going to square everything. So it doesn't matter if you get negative numbers. So it's going to be 6. The distance here is going to be 6 minus 3."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "And you can actually go either way, because you're going to square everything. So it doesn't matter if you get negative numbers. So it's going to be 6. The distance here is going to be 6 minus 3. That's this distance right here, which is equal to 3. So we figured out the base. And just to remind ourselves, that is equal to the change in x."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "The distance here is going to be 6 minus 3. That's this distance right here, which is equal to 3. So we figured out the base. And just to remind ourselves, that is equal to the change in x. That was equal to your finishing x minus your starting x. 6 minus 3. This is our delta x."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "And just to remind ourselves, that is equal to the change in x. That was equal to your finishing x minus your starting x. 6 minus 3. This is our delta x. Now, by the same exact line of reasoning, this height right here is going to be your change in y. Up here, you're at y is equal to 0. That's kind of where you finish."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "This is our delta x. Now, by the same exact line of reasoning, this height right here is going to be your change in y. Up here, you're at y is equal to 0. That's kind of where you finish. That's your higher y point. And over here, you're at y is equal to negative 4. So change in y is equal to 0 minus negative 4."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "That's kind of where you finish. That's your higher y point. And over here, you're at y is equal to negative 4. So change in y is equal to 0 minus negative 4. I'm just taking the larger y value minus the smaller y value, the larger x value minus the smaller x value. But you're going to see, we're going to square it in a second. So even if you did it the other way around, you get a negative number, but you still get the same answer."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So change in y is equal to 0 minus negative 4. I'm just taking the larger y value minus the smaller y value, the larger x value minus the smaller x value. But you're going to see, we're going to square it in a second. So even if you did it the other way around, you get a negative number, but you still get the same answer. So this is equal to 4. So this side is equal to 4. You could even count it on the graph paper if you like."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So even if you did it the other way around, you get a negative number, but you still get the same answer. So this is equal to 4. So this side is equal to 4. You could even count it on the graph paper if you like. And this side is equal to 3. And now we can do the Pythagorean theorem. This distance is the distance squared."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "You could even count it on the graph paper if you like. And this side is equal to 3. And now we can do the Pythagorean theorem. This distance is the distance squared. Be careful. The distance squared is going to be equal to this delta x squared, the change in x squared plus the change in y squared. This is nothing fancy."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "This distance is the distance squared. Be careful. The distance squared is going to be equal to this delta x squared, the change in x squared plus the change in y squared. This is nothing fancy. Sometimes people will call this the distance formula. It's just the Pythagorean theorem. This side squared plus that side squared is equal to hypotenuse squared, because this is a right triangle."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "This is nothing fancy. Sometimes people will call this the distance formula. It's just the Pythagorean theorem. This side squared plus that side squared is equal to hypotenuse squared, because this is a right triangle. So let's apply it with these numbers, the numbers that we have at hand. So the distance squared is going to be equal to delta x squared is 3 squared plus delta y squared plus 4 squared, which is equal to 9 plus 16, which is equal to 25. So the distance is equal to, let me write that, d squared is equal to 25. d, our distance, is equal to, you don't want to take the negative square root, because you can't have a negative distance, so only the principal root, the positive square root of 25, which is equal to 5."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "This side squared plus that side squared is equal to hypotenuse squared, because this is a right triangle. So let's apply it with these numbers, the numbers that we have at hand. So the distance squared is going to be equal to delta x squared is 3 squared plus delta y squared plus 4 squared, which is equal to 9 plus 16, which is equal to 25. So the distance is equal to, let me write that, d squared is equal to 25. d, our distance, is equal to, you don't want to take the negative square root, because you can't have a negative distance, so only the principal root, the positive square root of 25, which is equal to 5. So this distance right here is 5. Or if we look at this distance right here, that was the original problem. How far is this point from that point?"}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So the distance is equal to, let me write that, d squared is equal to 25. d, our distance, is equal to, you don't want to take the negative square root, because you can't have a negative distance, so only the principal root, the positive square root of 25, which is equal to 5. So this distance right here is 5. Or if we look at this distance right here, that was the original problem. How far is this point from that point? It is 5 units away. So what you'll see here, they call it the distance formula, but it's just the Pythagorean theorem. And just so you're exposed to all of the ways that you'll see the distance formula, sometimes people will say, oh, if I have two points, if I have one point, let's call it x1 and y1."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "How far is this point from that point? It is 5 units away. So what you'll see here, they call it the distance formula, but it's just the Pythagorean theorem. And just so you're exposed to all of the ways that you'll see the distance formula, sometimes people will say, oh, if I have two points, if I have one point, let's call it x1 and y1. So that's just a particular point. And let's say I have another point, that is x2, y2. Sometimes you'll see this formula, that the distance, you'll see it in different ways, but you'll see that the distance is equal to, and it looks like this really complicated formula, but I want you to see that this is really just the Pythagorean theorem."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "And just so you're exposed to all of the ways that you'll see the distance formula, sometimes people will say, oh, if I have two points, if I have one point, let's call it x1 and y1. So that's just a particular point. And let's say I have another point, that is x2, y2. Sometimes you'll see this formula, that the distance, you'll see it in different ways, but you'll see that the distance is equal to, and it looks like this really complicated formula, but I want you to see that this is really just the Pythagorean theorem. You'll see that the distance is equal to x2 minus x1 squared plus y2 minus y1 squared. You'll see this written in a lot of textbooks as the distance formula. It's a complete waste of your time to memorize it, because it's really just the Pythagorean theorem."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Sometimes you'll see this formula, that the distance, you'll see it in different ways, but you'll see that the distance is equal to, and it looks like this really complicated formula, but I want you to see that this is really just the Pythagorean theorem. You'll see that the distance is equal to x2 minus x1 squared plus y2 minus y1 squared. You'll see this written in a lot of textbooks as the distance formula. It's a complete waste of your time to memorize it, because it's really just the Pythagorean theorem. This is your change in x. And it really doesn't matter which x you pick to be first or second, because even if you get the negative of this value, when you square it, the negative disappears. This right here is your change in y."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "It's a complete waste of your time to memorize it, because it's really just the Pythagorean theorem. This is your change in x. And it really doesn't matter which x you pick to be first or second, because even if you get the negative of this value, when you square it, the negative disappears. This right here is your change in y. So it's just saying that the distance squared, remember if you square both sides of this equation, the radical will disappear, and this will be the distance squared is equal to this expression squared, delta x squared, change in x. Delta means change in. Delta x squared plus delta y squared. Don't want to confuse you."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "This right here is your change in y. So it's just saying that the distance squared, remember if you square both sides of this equation, the radical will disappear, and this will be the distance squared is equal to this expression squared, delta x squared, change in x. Delta means change in. Delta x squared plus delta y squared. Don't want to confuse you. Delta y just means change in y. I should have probably said that earlier in the video. But let's apply it to a couple of more. And I'll just pick some points at random."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Don't want to confuse you. Delta y just means change in y. I should have probably said that earlier in the video. But let's apply it to a couple of more. And I'll just pick some points at random. Let's say I have the point 1, 2, 3, 4, 5, 6. Negative 6 comma negative 4. And let's say I want to find the distance between that and 1 comma 1, 2, 3, 4, 5, 6, 7."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "And I'll just pick some points at random. Let's say I have the point 1, 2, 3, 4, 5, 6. Negative 6 comma negative 4. And let's say I want to find the distance between that and 1 comma 1, 2, 3, 4, 5, 6, 7. And the point 1 comma 7. So I want to find this distance right here. So it's the exact same idea."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "And let's say I want to find the distance between that and 1 comma 1, 2, 3, 4, 5, 6, 7. And the point 1 comma 7. So I want to find this distance right here. So it's the exact same idea. We just use the Pythagorean theorem. You figure out this distance, which is our change in x. This distance, which is our change in y."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So it's the exact same idea. We just use the Pythagorean theorem. You figure out this distance, which is our change in x. This distance, which is our change in y. This distance squared plus this distance squared is going to equal that distance squared. So let's do it. So our change in x, you just take, it doesn't matter."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "This distance, which is our change in y. This distance squared plus this distance squared is going to equal that distance squared. So let's do it. So our change in x, you just take, it doesn't matter. I mean, in general, you want to take the larger x value minus the smaller x value, but you could do it either way. So we could write the distance squared is equal to, what's our change in x? So let's take the larger x minus the smaller x."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So our change in x, you just take, it doesn't matter. I mean, in general, you want to take the larger x value minus the smaller x value, but you could do it either way. So we could write the distance squared is equal to, what's our change in x? So let's take the larger x minus the smaller x. 1 minus negative 6 squared plus the change in y. The larger y is here. It's 7 minus negative 4 squared."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So let's take the larger x minus the smaller x. 1 minus negative 6 squared plus the change in y. The larger y is here. It's 7 minus negative 4 squared. And I just picked these numbers at random, so they're probably not going to come out too cleanly. So we get that the distance squared is equal to 1 minus negative 6, that is 7 squared. And you'll even see it over here if you count it."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "It's 7 minus negative 4 squared. And I just picked these numbers at random, so they're probably not going to come out too cleanly. So we get that the distance squared is equal to 1 minus negative 6, that is 7 squared. And you'll even see it over here if you count it. You go 1, 2, 3, 4, 5, 6, 7. That's that number right here. That's what your change in x is."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "And you'll even see it over here if you count it. You go 1, 2, 3, 4, 5, 6, 7. That's that number right here. That's what your change in x is. Plus 7 minus negative 4. That's 11. This is this distance right here."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "That's what your change in x is. Plus 7 minus negative 4. That's 11. This is this distance right here. And you can count it on the blocks. We're going up 11. We're just taking 7 minus negative 4 to get a distance of 11."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "This is this distance right here. And you can count it on the blocks. We're going up 11. We're just taking 7 minus negative 4 to get a distance of 11. So plus 11 squared is equal to d squared. So let me just take the calculator out. So the distance, let's just take, if we get 7 squared plus 11 squared is equal to 170."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "We're just taking 7 minus negative 4 to get a distance of 11. So plus 11 squared is equal to d squared. So let me just take the calculator out. So the distance, let's just take, if we get 7 squared plus 11 squared is equal to 170. That distance is going to be the square root of that, right? d squared is equal to 170. So let's take the square root of 170 and we get roughly 13.04."}, {"video_title": "Factoring difference of squares shared factors High School Math Khan Academy.mp3", "Sentence": "We're told that the quadratic expressions, m squared minus four m minus 45, and six m squared minus 150, share a common binomial factor. What binomial factor do they share? And like always, pause the video and see if you can work through this. All right, now let's work through this together. And the way I'm gonna do it is I'm just gonna try to factor both of them into the product of binomials and maybe some other things, and see if we have any common binomial factors. So first, let's focus on m squared minus four m minus 45. So let me write it over here."}, {"video_title": "Factoring difference of squares shared factors High School Math Khan Academy.mp3", "Sentence": "All right, now let's work through this together. And the way I'm gonna do it is I'm just gonna try to factor both of them into the product of binomials and maybe some other things, and see if we have any common binomial factors. So first, let's focus on m squared minus four m minus 45. So let me write it over here. M squared minus four m minus 45. So when you're factoring a quadratic expression like this, where the coefficient on the, in this case, m squared term, on the second degree term is one, we could factor it as being equal to m plus a times m plus b, where a plus b is going to be equal to this coefficient right over here, and a times b is going to be equal to this coefficient right over here. So let's be clear."}, {"video_title": "Factoring difference of squares shared factors High School Math Khan Academy.mp3", "Sentence": "So let me write it over here. M squared minus four m minus 45. So when you're factoring a quadratic expression like this, where the coefficient on the, in this case, m squared term, on the second degree term is one, we could factor it as being equal to m plus a times m plus b, where a plus b is going to be equal to this coefficient right over here, and a times b is going to be equal to this coefficient right over here. So let's be clear. So a, let me just, another color. So a plus b needs to be equal to negative four. A plus b needs to be equal to negative four."}, {"video_title": "Factoring difference of squares shared factors High School Math Khan Academy.mp3", "Sentence": "So let's be clear. So a, let me just, another color. So a plus b needs to be equal to negative four. A plus b needs to be equal to negative four. And then a times b needs to be equal to negative 45. A times b is equal to negative 45. Now I like to focus on the a times b and think about, well, what could a and b be to get to negative 45?"}, {"video_title": "Factoring difference of squares shared factors High School Math Khan Academy.mp3", "Sentence": "A plus b needs to be equal to negative four. And then a times b needs to be equal to negative 45. A times b is equal to negative 45. Now I like to focus on the a times b and think about, well, what could a and b be to get to negative 45? Well, if I'm taking the product of two things, and if the product is negative, that means that they're going to have different signs. And if when we add them, we get a negative number, that means that the negative one has a larger magnitude. So let's think about this a little bit."}, {"video_title": "Factoring difference of squares shared factors High School Math Khan Academy.mp3", "Sentence": "Now I like to focus on the a times b and think about, well, what could a and b be to get to negative 45? Well, if I'm taking the product of two things, and if the product is negative, that means that they're going to have different signs. And if when we add them, we get a negative number, that means that the negative one has a larger magnitude. So let's think about this a little bit. So a times b is equal to negative 45. So this could be, let's try some values out. So one and 45, those are too far apart."}, {"video_title": "Factoring difference of squares shared factors High School Math Khan Academy.mp3", "Sentence": "So let's think about this a little bit. So a times b is equal to negative 45. So this could be, let's try some values out. So one and 45, those are too far apart. Let's see, three and 15, those still seem pretty far apart. Let's see, it looks like five and nine seem interesting. So if we say five times, if we were to say five times negative nine, that indeed is equal to negative 45."}, {"video_title": "Factoring difference of squares shared factors High School Math Khan Academy.mp3", "Sentence": "So one and 45, those are too far apart. Let's see, three and 15, those still seem pretty far apart. Let's see, it looks like five and nine seem interesting. So if we say five times, if we were to say five times negative nine, that indeed is equal to negative 45. And five plus negative nine is indeed equal to negative four. So a could be equal to five, and b could be equal to negative nine. And so if we were to factor this, this is going to be m plus five times m, I could say m plus negative nine, but I'll just write m minus nine."}, {"video_title": "Factoring difference of squares shared factors High School Math Khan Academy.mp3", "Sentence": "So if we say five times, if we were to say five times negative nine, that indeed is equal to negative 45. And five plus negative nine is indeed equal to negative four. So a could be equal to five, and b could be equal to negative nine. And so if we were to factor this, this is going to be m plus five times m, I could say m plus negative nine, but I'll just write m minus nine. So just like that, I've been able to factor, I've been able to factor this first quadratic expression right over there as a product of two binomials. So now let's try to factor the other quadratic expression. Let's try to factor six m squared minus 150."}, {"video_title": "Factoring difference of squares shared factors High School Math Khan Academy.mp3", "Sentence": "And so if we were to factor this, this is going to be m plus five times m, I could say m plus negative nine, but I'll just write m minus nine. So just like that, I've been able to factor, I've been able to factor this first quadratic expression right over there as a product of two binomials. So now let's try to factor the other quadratic expression. Let's try to factor six m squared minus 150. And let's see, the first thing I might wanna do is both six, both six m squared and 150, they're both divisible by six. So let me write it this way. I could write it as six, actually I'll just write six m squared minus six times, let's see, six goes into 150 25 times."}, {"video_title": "Factoring difference of squares shared factors High School Math Khan Academy.mp3", "Sentence": "Let's try to factor six m squared minus 150. And let's see, the first thing I might wanna do is both six, both six m squared and 150, they're both divisible by six. So let me write it this way. I could write it as six, actually I'll just write six m squared minus six times, let's see, six goes into 150 25 times. So all I did is I rewrote this, and really I just wrote 150 as six times 25. And now you can clearly see that we can factor out a six. You can view this as undistributing the six."}, {"video_title": "Factoring difference of squares shared factors High School Math Khan Academy.mp3", "Sentence": "I could write it as six, actually I'll just write six m squared minus six times, let's see, six goes into 150 25 times. So all I did is I rewrote this, and really I just wrote 150 as six times 25. And now you can clearly see that we can factor out a six. You can view this as undistributing the six. So this is the same thing as six times m squared minus 25, which we recognize this is a difference of squares, so it's all gonna be six times m plus five times m minus five. And so we've factored this out as a product of binomials and a constant factor here, six. And so what is their shared common, or what is their common binomial factor that they share?"}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Dominique from Dominique's Pizza bakes the same amount of pizza every day. She used to spend $8 each day on using the oven and $1.50 on ingredients for each pizza. So $8 each day on using the oven and $1.50 on ingredients for each pizza. One day the price for the ingredients increased from $1.50 to $2 per pizza. Dominique made some calculations and found that she should bake eight pizzas more each day. So the expenses for a single pizza would remain the same. And I assume they're saying the total expenses for a single pizza because clearly the ingredients cost is not the same."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "One day the price for the ingredients increased from $1.50 to $2 per pizza. Dominique made some calculations and found that she should bake eight pizzas more each day. So the expenses for a single pizza would remain the same. And I assume they're saying the total expenses for a single pizza because clearly the ingredients cost is not the same. We're talking about the total ingredients. So if we were to spread the cost of the oven across all of the pizzas, write an equation to find, or the total cost for the oven per day to spread that across all the pizzas. Write an equation to find out how many pizzas Dominique baked each day before the change in price."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And I assume they're saying the total expenses for a single pizza because clearly the ingredients cost is not the same. We're talking about the total ingredients. So if we were to spread the cost of the oven across all of the pizzas, write an equation to find, or the total cost for the oven per day to spread that across all the pizzas. Write an equation to find out how many pizzas Dominique baked each day before the change in price. Use P to represent the number of pizzas. So let's just think about her total cost per pizza before and then her total cost per pizza after if she bakes eight more pizzas. So before, we're gonna use P to say that's the number of pizzas she baked per day before the change in price."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Write an equation to find out how many pizzas Dominique baked each day before the change in price. Use P to represent the number of pizzas. So let's just think about her total cost per pizza before and then her total cost per pizza after if she bakes eight more pizzas. So before, we're gonna use P to say that's the number of pizzas she baked per day before the change in price. So before the change in price, on a given day, she would spend $8 on the oven and then $1.50 on ingredients for each pizza. So 1.5 or $1.50 times the number of pizzas. This would be her total cost on all the pizzas in that day."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So before, we're gonna use P to say that's the number of pizzas she baked per day before the change in price. So before the change in price, on a given day, she would spend $8 on the oven and then $1.50 on ingredients for each pizza. So 1.5 or $1.50 times the number of pizzas. This would be her total cost on all the pizzas in that day. It's the oven cost plus it's the ingredients cost. So if you wanted this on a per pizza basis, you would just divide by the number of pizzas. Now let's think about what happens after the change in price."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "This would be her total cost on all the pizzas in that day. It's the oven cost plus it's the ingredients cost. So if you wanted this on a per pizza basis, you would just divide by the number of pizzas. Now let's think about what happens after the change in price. After the change in price, her cost per day for the oven is still $8. But now she has to spend $2 per pizza on ingredients. So $2 per pizza."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Now let's think about what happens after the change in price. After the change in price, her cost per day for the oven is still $8. But now she has to spend $2 per pizza on ingredients. So $2 per pizza. And instead of saying that she's baking P pizzas, let's say that she's now baking eight more pizzas each day. So it's going to be P plus eight. And so this is going to be her total cost for all of the pizzas she's now baking."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So $2 per pizza. And instead of saying that she's baking P pizzas, let's say that she's now baking eight more pizzas each day. So it's going to be P plus eight. And so this is going to be her total cost for all of the pizzas she's now baking. And so if you want it on a per pizza basis, well, she's now making P plus eight pizzas, you would divide by P plus eight. And the problem tells us that these two things are equivalent. Here you had a higher cost of ingredients per pizzas, but since you are now baking more pizzas, you're spreading the oven cost amongst more and more pizzas."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And so this is going to be her total cost for all of the pizzas she's now baking. And so if you want it on a per pizza basis, well, she's now making P plus eight pizzas, you would divide by P plus eight. And the problem tells us that these two things are equivalent. Here you had a higher cost of ingredients per pizzas, but since you are now baking more pizzas, you're spreading the oven cost amongst more and more pizzas. So let's think about what P has to be. P has to be some number, some number of pizzas, so that these two expressions are equal. Her total cost per pizza before, when she only made P, is going to be the same as her total cost per pizza when she's making P plus eight pizzas."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Here you had a higher cost of ingredients per pizzas, but since you are now baking more pizzas, you're spreading the oven cost amongst more and more pizzas. So let's think about what P has to be. P has to be some number, some number of pizzas, so that these two expressions are equal. Her total cost per pizza before, when she only made P, is going to be the same as her total cost per pizza when she's making P plus eight pizzas. So these two things need to be equal. So we did that first part, or we did what they asked us. We wrote an equation to find out how many pizzas Dominique baked each day before the change in price, and we used P to represent the number of pizzas."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Her total cost per pizza before, when she only made P, is going to be the same as her total cost per pizza when she's making P plus eight pizzas. So these two things need to be equal. So we did that first part, or we did what they asked us. We wrote an equation to find out how many pizzas Dominique baked each day before the change in price, and we used P to represent the number of pizzas. But now for fun, let's actually just solve for P. So let's just simplify things a little bit. And actually, so this part right over here, actually, let's just cross multiply this on both sides, or another way of thinking is, multiply both sides times P plus eight, and multiply both sides times P. So if we multiply by P plus eight, and we multiply by P, we multiply by P plus eight, and we multiply by P, that cancels with that, that cancels with that. On the left-hand side, so let's see, we have to just do the distributive property twice right over here."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "We wrote an equation to find out how many pizzas Dominique baked each day before the change in price, and we used P to represent the number of pizzas. But now for fun, let's actually just solve for P. So let's just simplify things a little bit. And actually, so this part right over here, actually, let's just cross multiply this on both sides, or another way of thinking is, multiply both sides times P plus eight, and multiply both sides times P. So if we multiply by P plus eight, and we multiply by P, we multiply by P plus eight, and we multiply by P, that cancels with that, that cancels with that. On the left-hand side, so let's see, we have to just do the distributive property twice right over here. What is P times eight plus 1.5P? Well, that's going to be eight P, I'm just multiplying the P times this stuff first, plus 1.5P squared, and now let's multiply, now let's multiply the eight times both of these terms. So plus 64, plus eight times 1.5, that is 12, plus 12P, and that's going to be equal to, let's see, let's multiply P times all of this business."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "On the left-hand side, so let's see, we have to just do the distributive property twice right over here. What is P times eight plus 1.5P? Well, that's going to be eight P, I'm just multiplying the P times this stuff first, plus 1.5P squared, and now let's multiply, now let's multiply the eight times both of these terms. So plus 64, plus eight times 1.5, that is 12, plus 12P, and that's going to be equal to, let's see, let's multiply P times all of this business. So that's going to be equal to eight P, eight times P is eight P, and let's see, I could distribute these terms and then multiply by P. So two times P is two P, times P is two P squared, plus two P squared, and then two times eight is 16, times P is 16P. So now we have, well, we essentially end up with a quadratic equation, but let's simplify it a little bit so that we can either factor it or apply the quadratic formula. So let's see, let's throw, let's subtract two P squared from, actually, let's subtract 1.5P squared from both sides."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So plus 64, plus eight times 1.5, that is 12, plus 12P, and that's going to be equal to, let's see, let's multiply P times all of this business. So that's going to be equal to eight P, eight times P is eight P, and let's see, I could distribute these terms and then multiply by P. So two times P is two P, times P is two P squared, plus two P squared, and then two times eight is 16, times P is 16P. So now we have, well, we essentially end up with a quadratic equation, but let's simplify it a little bit so that we can either factor it or apply the quadratic formula. So let's see, let's throw, let's subtract two P squared from, actually, let's subtract 1.5P squared from both sides. So subtract 1.5P squared, subtract 1.5P squared. Actually, let me just put everything on the left-hand side just because that's, might be a little bit more intuitive. So let's subtract two P squared from both sides, subtract two P squared from both sides."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So let's see, let's throw, let's subtract two P squared from, actually, let's subtract 1.5P squared from both sides. So subtract 1.5P squared, subtract 1.5P squared. Actually, let me just put everything on the left-hand side just because that's, might be a little bit more intuitive. So let's subtract two P squared from both sides, subtract two P squared from both sides. Let's subtract 16P from both sides. We have an eight P and a 12P, and then we're gonna subtract a 16P from both sides. And then, oh, actually, let's subtract an eight P as well from both sides."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So let's subtract two P squared from both sides, subtract two P squared from both sides. Let's subtract 16P from both sides. We have an eight P and a 12P, and then we're gonna subtract a 16P from both sides. And then, oh, actually, let's subtract an eight P as well from both sides. We have a 16P and an eight P. So that actually works out quite well. So now we've subtracted eight P from both sides, 16P from both sides. We've essentially subtracted all of this stuff from both sides, and we are left with, let's see, 1.5, I'll do it in degree order."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And then, oh, actually, let's subtract an eight P as well from both sides. We have a 16P and an eight P. So that actually works out quite well. So now we've subtracted eight P from both sides, 16P from both sides. We've essentially subtracted all of this stuff from both sides, and we are left with, let's see, 1.5, I'll do it in degree order. 1.5P squared minus two P squared is negative, negative 0.5P squared. Now let's see, these cancel out. 12P minus 16P is minus four P. And then we have plus 64, plus 64, and then that is going to be equal to zero."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "We've essentially subtracted all of this stuff from both sides, and we are left with, let's see, 1.5, I'll do it in degree order. 1.5P squared minus two P squared is negative, negative 0.5P squared. Now let's see, these cancel out. 12P minus 16P is minus four P. And then we have plus 64, plus 64, and then that is going to be equal to zero. That is going to be equal to zero. And just to simplify this a little bit, or just to make this a little bit cleaner, let's multiply both sides of this equation by negative two. By negative two."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "12P minus 16P is minus four P. And then we have plus 64, plus 64, and then that is going to be equal to zero. That is going to be equal to zero. And just to simplify this a little bit, or just to make this a little bit cleaner, let's multiply both sides of this equation by negative two. By negative two. I want the coefficient over here to be one. So then we get P squared plus eight P, P squared plus eight P is going to be equal to, is going to be equal to, let's see, negative, oh, negative times negative two. So minus 128 is going to be equal to zero."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "By negative two. I want the coefficient over here to be one. So then we get P squared plus eight P, P squared plus eight P is going to be equal to, is going to be equal to, let's see, negative, oh, negative times negative two. So minus 128 is going to be equal to zero. So let's see if we can factor this. Can we think of two numbers where if we take their product, we get negative 128, and if we were to, and if we were to add them together, we get positive eight. So they're gonna have different signs right over here."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So minus 128 is going to be equal to zero. So let's see if we can factor this. Can we think of two numbers where if we take their product, we get negative 128, and if we were to, and if we were to add them together, we get positive eight. So they're gonna have different signs right over here. So let's see, if we say 12 times, let's see, 12 times, well, let's see, what numbers could this be? So if we were to think about 128 is the same thing as, let's see, 16, let's see, 16 goes into 128. Let me work through this."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So they're gonna have different signs right over here. So let's see, if we say 12 times, let's see, 12 times, well, let's see, what numbers could this be? So if we were to think about 128 is the same thing as, let's see, 16, let's see, 16 goes into 128. Let me work through this. 16 goes into 128. Does it go eight times? Eight times six is, eight times six is 48."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Let me work through this. 16 goes into 128. Does it go eight times? Eight times six is, eight times six is 48. Eight times 10 is 80 plus 40 is 128. Yep, it goes eight times. So 16 and eight seem to work."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Eight times six is, eight times six is 48. Eight times 10 is 80 plus 40 is 128. Yep, it goes eight times. So 16 and eight seem to work. So if you have positive 16 and negative eight, their product would be negative 128. So we can factor this out as P plus 16 times P minus eight is equal to zero. Now this is going to be equal to zero if one of these two, at least one of these is going to be equal to zero."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So 16 and eight seem to work. So if you have positive 16 and negative eight, their product would be negative 128. So we can factor this out as P plus 16 times P minus eight is equal to zero. Now this is going to be equal to zero if one of these two, at least one of these is going to be equal to zero. So we have two solutions. Either P plus 16 is going to be equal to zero or P minus eight is equal to zero. This one right over here, subtract 16 from both sides, you get P is equal to negative 16."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Now this is going to be equal to zero if one of these two, at least one of these is going to be equal to zero. So we have two solutions. Either P plus 16 is going to be equal to zero or P minus eight is equal to zero. This one right over here, subtract 16 from both sides, you get P is equal to negative 16. Here you get P is equal to eight if you add eight to both sides. Now we're talking about a number of pizzas made. So we're not, this one doesn't apply."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "This one right over here, subtract 16 from both sides, you get P is equal to negative 16. Here you get P is equal to eight if you add eight to both sides. Now we're talking about a number of pizzas made. So we're not, this one doesn't apply. This would be like her, Dominique eating 16 pizzas or somehow destroying 16 pizzas a day. We're not interested in that solution. So if we want the solution to the original question, the number of pizzas she made before the increase in price, she made eight pizzas, she made eight pizzas per day."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So we're not, this one doesn't apply. This would be like her, Dominique eating 16 pizzas or somehow destroying 16 pizzas a day. We're not interested in that solution. So if we want the solution to the original question, the number of pizzas she made before the increase in price, she made eight pizzas, she made eight pizzas per day. So P right over here needs to be equal to eight. So before the change in price, she made eight pizzas a day. After the change in price, she made eight more pizzas a day or 16 pizzas per day."}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "So let's just review real quick what a logarithm even is. So if I write, let's say I write log base x of a is equal to, I don't know, make up a letter, n. What does this mean? Well, this just means that x to the n equals a. I think we already know that. We've learned that in the logarithm video. And so it's very important to realize that when you evaluate a logarithm expression like log base x of a, the answer, or when you evaluate, what you get is an exponent, right? This n is really just an exponent. This is equal to this thing, right?"}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "We've learned that in the logarithm video. And so it's very important to realize that when you evaluate a logarithm expression like log base x of a, the answer, or when you evaluate, what you get is an exponent, right? This n is really just an exponent. This is equal to this thing, right? You could have written it just like this. You could have, because this n is equal to this, you could just write x, it's going to get a little messy, to the log base x of a is equal to a. All I did is I took this n and I replaced it with this term."}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "This is equal to this thing, right? You could have written it just like this. You could have, because this n is equal to this, you could just write x, it's going to get a little messy, to the log base x of a is equal to a. All I did is I took this n and I replaced it with this term. And I wanted to write it this way because I want you to really get an intuitive understanding of the notion that a logarithm, when you evaluate it, really is an exponent. And we're going to take that notion and that's where really all of the logarithm properties come from. So let me just do, what I actually want to do is I want to stumble upon the logarithm properties by playing around."}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "All I did is I took this n and I replaced it with this term. And I wanted to write it this way because I want you to really get an intuitive understanding of the notion that a logarithm, when you evaluate it, really is an exponent. And we're going to take that notion and that's where really all of the logarithm properties come from. So let me just do, what I actually want to do is I want to stumble upon the logarithm properties by playing around. And then later on I'll summarize them and clean it all up. But I wanted to show you maybe how people originally discovered this stuff. So let's say that, I don't know, let's say that x, let me switch colors, I think that keeps things interesting."}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "So let me just do, what I actually want to do is I want to stumble upon the logarithm properties by playing around. And then later on I'll summarize them and clean it all up. But I wanted to show you maybe how people originally discovered this stuff. So let's say that, I don't know, let's say that x, let me switch colors, I think that keeps things interesting. So let's say that x to the, I don't know, l is equal to a. Well if we write that as a logarithm, that same relationship as a logarithm, we could write that log base x of a is equal to l, right? I just rewrote what I wrote on the top line."}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "So let's say that, I don't know, let's say that x, let me switch colors, I think that keeps things interesting. So let's say that x to the, I don't know, l is equal to a. Well if we write that as a logarithm, that same relationship as a logarithm, we could write that log base x of a is equal to l, right? I just rewrote what I wrote on the top line. Now let me switch colors. If I were to say that x to the m is equal to b, same thing, I just switch letters. But that just means that log base x of b is equal to m. I just did the same thing that I did in this line, I just switched letters."}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "I just rewrote what I wrote on the top line. Now let me switch colors. If I were to say that x to the m is equal to b, same thing, I just switch letters. But that just means that log base x of b is equal to m. I just did the same thing that I did in this line, I just switched letters. So let's just keep going and see what happens. So let's say, let me get another color, I might have to run out of color, I have infinite color, I will never run out. So let's say I have x to the n, you're probably saying, Sal, where are you going with this?"}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "But that just means that log base x of b is equal to m. I just did the same thing that I did in this line, I just switched letters. So let's just keep going and see what happens. So let's say, let me get another color, I might have to run out of color, I have infinite color, I will never run out. So let's say I have x to the n, you're probably saying, Sal, where are you going with this? But you'll see, it's pretty neat. x to the n is equal to a times b. x to the n is equal to a times b. And that's just like saying that log base x is equal to a times b."}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "So let's say I have x to the n, you're probably saying, Sal, where are you going with this? But you'll see, it's pretty neat. x to the n is equal to a times b. x to the n is equal to a times b. And that's just like saying that log base x is equal to a times b. So what can we do with all of this? Let's start with this right here. x to the n is equal to a times b."}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "And that's just like saying that log base x is equal to a times b. So what can we do with all of this? Let's start with this right here. x to the n is equal to a times b. So how can we rewrite this? Well, a is this, and b is this, right? So let's rewrite that."}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "x to the n is equal to a times b. So how can we rewrite this? Well, a is this, and b is this, right? So let's rewrite that. So we know that x to the n is equal to a, a is this, x to the l, and what's b? Times b. Well, b is x to the m, right?"}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "So let's rewrite that. So we know that x to the n is equal to a, a is this, x to the l, and what's b? Times b. Well, b is x to the m, right? Not doing anything fancy right now. Well, what's x to the l times x to the m? Well, we know from the exponents, when you multiply two expressions that have the same base and different exponents, you just add the exponents."}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, b is x to the m, right? Not doing anything fancy right now. Well, what's x to the l times x to the m? Well, we know from the exponents, when you multiply two expressions that have the same base and different exponents, you just add the exponents. So this is equal to, let me take a neutral color. I don't know if I said that verbally correct, but you get the point. When you have the same base and you're multiplying, you can just add the exponents."}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, we know from the exponents, when you multiply two expressions that have the same base and different exponents, you just add the exponents. So this is equal to, let me take a neutral color. I don't know if I said that verbally correct, but you get the point. When you have the same base and you're multiplying, you can just add the exponents. That equals x to the, I want to keep switching colors because I think that's useful. L plus m. It's kind of onerous to keep switching colors, but you get what I'm saying. So x to the n is equal to x to the l plus m. Let me put the x here."}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "When you have the same base and you're multiplying, you can just add the exponents. That equals x to the, I want to keep switching colors because I think that's useful. L plus m. It's kind of onerous to keep switching colors, but you get what I'm saying. So x to the n is equal to x to the l plus m. Let me put the x here. Oh, I wanted that to be green. x to the l plus m. So what do we know? We know x to the n is equal to x to the l plus m, right?"}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "So x to the n is equal to x to the l plus m. Let me put the x here. Oh, I wanted that to be green. x to the l plus m. So what do we know? We know x to the n is equal to x to the l plus m, right? Well, we have the same base. These exponents must equal each other, right? So we know that n is equal to l plus m. Well, what does that do for us?"}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "We know x to the n is equal to x to the l plus m, right? Well, we have the same base. These exponents must equal each other, right? So we know that n is equal to l plus m. Well, what does that do for us? I've kind of just been playing around with logarithms. Am I getting anywhere? I think you'll see that I am."}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "So we know that n is equal to l plus m. Well, what does that do for us? I've kind of just been playing around with logarithms. Am I getting anywhere? I think you'll see that I am. Well, what's another way of writing n? So we said x to the n is equal to a times b. Oh, I actually skipped a step here. So that means, so going back here, x to the n is equal to a times b."}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "I think you'll see that I am. Well, what's another way of writing n? So we said x to the n is equal to a times b. Oh, I actually skipped a step here. So that means, so going back here, x to the n is equal to a times b. That means that log base x of a times b is equal to n. You knew that. I hope you don't realize I'm not backtracking or anything. I just forgot to write this down when I first did it."}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "So that means, so going back here, x to the n is equal to a times b. That means that log base x of a times b is equal to n. You knew that. I hope you don't realize I'm not backtracking or anything. I just forgot to write this down when I first did it. But anyway, so what's n? What's another way of writing n? Well, another way of writing n is right here."}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "I just forgot to write this down when I first did it. But anyway, so what's n? What's another way of writing n? Well, another way of writing n is right here. Log base x of a times b. So now we know that if we just substitute n for that, we get log base x of a times b. And what does that equal?"}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, another way of writing n is right here. Log base x of a times b. So now we know that if we just substitute n for that, we get log base x of a times b. And what does that equal? Well, that equals l. Another way to write l is right up here. It equals log base x of a plus m. And what's m? m is right here."}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "And what does that equal? Well, that equals l. Another way to write l is right up here. It equals log base x of a plus m. And what's m? m is right here. It's log base x of b. And there we have our first logarithm property. The log base x of a times b, well, that just equals the log of base x of a plus the log base x of b."}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "m is right here. It's log base x of b. And there we have our first logarithm property. The log base x of a times b, well, that just equals the log of base x of a plus the log base x of b. And this is, hopefully, this proves that to you. And if you want the intuition of why this works out, it falls from the fact that logarithms are nothing but exponents. So with that, I'll leave you in this video."}, {"video_title": "Finding inverses of rational functions Equations Algebra 2 Khan Academy.mp3", "Sentence": "All right, let's say that we have the function f of x, and it's equal to two x plus five over four minus three x. And what we wanna do is figure out what is the inverse of our function. Pause this video and try to figure that out before we work on that together. All right, now let's work on it together. And just as a reminder of what a function and an inverse even does, if this is the domain of a function, and that's the set of all values that you could input into the function for x and get a valid output, and so let's say you have an x here, it's a member of the domain, and if I were to apply the function to it, if I were to input that x into that function, then the function is going to output a value in the range of the function, and we call that value f of x. Now an inverse, that goes the other way. If you were to input the f of x value into the function, that's going to take us back to x."}, {"video_title": "Finding inverses of rational functions Equations Algebra 2 Khan Academy.mp3", "Sentence": "All right, now let's work on it together. And just as a reminder of what a function and an inverse even does, if this is the domain of a function, and that's the set of all values that you could input into the function for x and get a valid output, and so let's say you have an x here, it's a member of the domain, and if I were to apply the function to it, if I were to input that x into that function, then the function is going to output a value in the range of the function, and we call that value f of x. Now an inverse, that goes the other way. If you were to input the f of x value into the function, that's going to take us back to x. So that's exactly what f inverse does. Now how do we actually figure out the inverse of a function, especially a function that's defined with a rational expression like this? Well, the way that I think about it is, let's say that y is equal to our function of x, or y is a function of x."}, {"video_title": "Finding inverses of rational functions Equations Algebra 2 Khan Academy.mp3", "Sentence": "If you were to input the f of x value into the function, that's going to take us back to x. So that's exactly what f inverse does. Now how do we actually figure out the inverse of a function, especially a function that's defined with a rational expression like this? Well, the way that I think about it is, let's say that y is equal to our function of x, or y is a function of x. So we could say that y is equal to two x plus five over four minus three x. For our inverse, the relationship between x and y is going to be swapped, and so in our inverse, it's going to be true that x is going to be equal to two y plus five over four minus three y. And then to be able to express this as a function of x, to say that what is y is a function of x for our inverse, we now have to solve for y."}, {"video_title": "Finding inverses of rational functions Equations Algebra 2 Khan Academy.mp3", "Sentence": "Well, the way that I think about it is, let's say that y is equal to our function of x, or y is a function of x. So we could say that y is equal to two x plus five over four minus three x. For our inverse, the relationship between x and y is going to be swapped, and so in our inverse, it's going to be true that x is going to be equal to two y plus five over four minus three y. And then to be able to express this as a function of x, to say that what is y is a function of x for our inverse, we now have to solve for y. So it's just a little bit of algebra here. So let's see if we can do that. So the first thing that I would do is multiply both sides of this equation by four minus three y."}, {"video_title": "Finding inverses of rational functions Equations Algebra 2 Khan Academy.mp3", "Sentence": "And then to be able to express this as a function of x, to say that what is y is a function of x for our inverse, we now have to solve for y. So it's just a little bit of algebra here. So let's see if we can do that. So the first thing that I would do is multiply both sides of this equation by four minus three y. If we do that, on the left-hand side, we are going to get x times each of these terms, so we're going to get four x minus three y x, and then that's going to be equal to, on the right-hand side, since we multiplied by the denominator here, we're just going to be left with the numerator, it's going to be equal to two y plus five. And this could be a little bit intimidating because we're seeing an x's and y's, what are we trying to do? Remember, we're trying to solve for y."}, {"video_title": "Finding inverses of rational functions Equations Algebra 2 Khan Academy.mp3", "Sentence": "So the first thing that I would do is multiply both sides of this equation by four minus three y. If we do that, on the left-hand side, we are going to get x times each of these terms, so we're going to get four x minus three y x, and then that's going to be equal to, on the right-hand side, since we multiplied by the denominator here, we're just going to be left with the numerator, it's going to be equal to two y plus five. And this could be a little bit intimidating because we're seeing an x's and y's, what are we trying to do? Remember, we're trying to solve for y. So let's gather all the y terms on one side and all the non-y terms on the other side. So let's get rid of this two y here. Actually, well, I could go either way."}, {"video_title": "Finding inverses of rational functions Equations Algebra 2 Khan Academy.mp3", "Sentence": "Remember, we're trying to solve for y. So let's gather all the y terms on one side and all the non-y terms on the other side. So let's get rid of this two y here. Actually, well, I could go either way. Let's get rid of this two y here. So let's subtract two y from both sides. And let's get rid of this four x from the left-hand side."}, {"video_title": "Finding inverses of rational functions Equations Algebra 2 Khan Academy.mp3", "Sentence": "Actually, well, I could go either way. Let's get rid of this two y here. So let's subtract two y from both sides. And let's get rid of this four x from the left-hand side. So let's subtract four x from both sides. And then what are we going to be left with? On the left-hand side, we're left with minus, or negative five, or actually, it would be this way."}, {"video_title": "Finding inverses of rational functions Equations Algebra 2 Khan Academy.mp3", "Sentence": "And let's get rid of this four x from the left-hand side. So let's subtract four x from both sides. And then what are we going to be left with? On the left-hand side, we're left with minus, or negative five, or actually, it would be this way. It would be negative three y x minus two y, and you might say, hey, where is this going? But I'll show you in a second. Is equal to, those cancel out, and we're going to have five minus four x."}, {"video_title": "Finding inverses of rational functions Equations Algebra 2 Khan Academy.mp3", "Sentence": "On the left-hand side, we're left with minus, or negative five, or actually, it would be this way. It would be negative three y x minus two y, and you might say, hey, where is this going? But I'll show you in a second. Is equal to, those cancel out, and we're going to have five minus four x. Now, once again, we are trying to solve for y. So let's factor out a y here, and then we are going to have y times negative three x minus two is equal to five minus four x. And now this is the home stretch."}, {"video_title": "Finding inverses of rational functions Equations Algebra 2 Khan Academy.mp3", "Sentence": "Is equal to, those cancel out, and we're going to have five minus four x. Now, once again, we are trying to solve for y. So let's factor out a y here, and then we are going to have y times negative three x minus two is equal to five minus four x. And now this is the home stretch. We can just divide both sides of this equation by negative three x minus two, and we're going to get y is equal to five minus four x over negative three x minus two. Now, another way that you could express this, you could multiply both the numerator and the denominator by a negative one. That won't change the value."}, {"video_title": "Finding inverses of rational functions Equations Algebra 2 Khan Academy.mp3", "Sentence": "And now this is the home stretch. We can just divide both sides of this equation by negative three x minus two, and we're going to get y is equal to five minus four x over negative three x minus two. Now, another way that you could express this, you could multiply both the numerator and the denominator by a negative one. That won't change the value. And then you would get, you would get in the numerator four x minus five, and in the denominator, you would get a three x plus two. So there you have it. Our f inverse as a function of x, which you could say is equal to this y, is equal to this right over there."}, {"video_title": "Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "And we can see the area right over here, and they broke it up. This green area is 12x to the fourth, this purple area is 6x to the third, this blue area is 15x squared. You add them all together, you get this entire rectangle, which would be the combined areas, 12x to the fourth plus 6x to the third plus 15x squared. The length of the rectangle in meters, so this is the length right over here that we're talking about, we're talking about this distance. The length of the rectangle in meters is equal to the greatest common monomial factor of 12x to the fourth, 6x to the third, and 15x squared. What is the length and width of the rectangle? I encourage you to pause the video and try to work through it on your own."}, {"video_title": "Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "The length of the rectangle in meters, so this is the length right over here that we're talking about, we're talking about this distance. The length of the rectangle in meters is equal to the greatest common monomial factor of 12x to the fourth, 6x to the third, and 15x squared. What is the length and width of the rectangle? I encourage you to pause the video and try to work through it on your own. Well, the key realization here is that the length times the width, the length times the width, is going to be equal to this area. If the length is the greatest common monomial factor of these terms, of 12x to the fourth, 6x to the third, and 15x squared, well then we can factor that out, and then what we have left over is going to be the width. So let's figure out what is the greatest common monomial factor of these three terms."}, {"video_title": "Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "I encourage you to pause the video and try to work through it on your own. Well, the key realization here is that the length times the width, the length times the width, is going to be equal to this area. If the length is the greatest common monomial factor of these terms, of 12x to the fourth, 6x to the third, and 15x squared, well then we can factor that out, and then what we have left over is going to be the width. So let's figure out what is the greatest common monomial factor of these three terms. The first thing we can look at is let's look at the coefficients. Let's figure out what's the greatest common factor of 12, 6, and 15. And there's a couple of ways you could do it."}, {"video_title": "Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "So let's figure out what is the greatest common monomial factor of these three terms. The first thing we can look at is let's look at the coefficients. Let's figure out what's the greatest common factor of 12, 6, and 15. And there's a couple of ways you could do it. You could do it by looking at a prime factorization. You could say, all right, well 12 is two times six, which is two times three. That's the prime factorization of 12."}, {"video_title": "Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "And there's a couple of ways you could do it. You could do it by looking at a prime factorization. You could say, all right, well 12 is two times six, which is two times three. That's the prime factorization of 12. Prime factorization of six is just two times three. Prime factorization of 15 is three times five. And so the greatest common factor, the largest factor that's divisible into all of them, so let's see, we can throw a three in there."}, {"video_title": "Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "That's the prime factorization of 12. Prime factorization of six is just two times three. Prime factorization of 15 is three times five. And so the greatest common factor, the largest factor that's divisible into all of them, so let's see, we can throw a three in there. Three is divisible into all of them. And that's it, because we can't say a three and a two. A three and a two would be divisible into 12 and six, but there's no two that's divisible into 15."}, {"video_title": "Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "And so the greatest common factor, the largest factor that's divisible into all of them, so let's see, we can throw a three in there. Three is divisible into all of them. And that's it, because we can't say a three and a two. A three and a two would be divisible into 12 and six, but there's no two that's divisible into 15. We can't say a three and a five, because five isn't divisible into 12 or six. So the greatest common factor is going to be three. Another way we could have done this is we could have said, what are the non-prime factors of each of these numbers?"}, {"video_title": "Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "A three and a two would be divisible into 12 and six, but there's no two that's divisible into 15. We can't say a three and a five, because five isn't divisible into 12 or six. So the greatest common factor is going to be three. Another way we could have done this is we could have said, what are the non-prime factors of each of these numbers? 12, you could have said, okay, I can get 12 by saying one times 12, or two times six, or three times four. Six, you could have said, let's see, that could be one times six, or two times three. So those are the factors of six."}, {"video_title": "Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "Another way we could have done this is we could have said, what are the non-prime factors of each of these numbers? 12, you could have said, okay, I can get 12 by saying one times 12, or two times six, or three times four. Six, you could have said, let's see, that could be one times six, or two times three. So those are the factors of six. And then 15, you could have said, well, one times 15, or three times five. And so you say the greatest common factor, well, three is the largest number that I've listed here that is common to all three of these factors. So once again, the greatest common factor of 12, six, and 15 is three."}, {"video_title": "Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "So those are the factors of six. And then 15, you could have said, well, one times 15, or three times five. And so you say the greatest common factor, well, three is the largest number that I've listed here that is common to all three of these factors. So once again, the greatest common factor of 12, six, and 15 is three. So when we're looking at the greatest common monomial factor, the coefficient is going to be three. And then we look at these powers of x. We have x to the fourth, we have x to the, let me do this in a different color."}, {"video_title": "Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "So once again, the greatest common factor of 12, six, and 15 is three. So when we're looking at the greatest common monomial factor, the coefficient is going to be three. And then we look at these powers of x. We have x to the fourth, we have x to the, let me do this in a different color. We have x to the fourth, x to the third, and x squared. Well, what's the largest power of x that's divisible into all of those? Well, it's going to be x squared."}, {"video_title": "Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "We have x to the fourth, we have x to the, let me do this in a different color. We have x to the fourth, x to the third, and x squared. Well, what's the largest power of x that's divisible into all of those? Well, it's going to be x squared. X squared is divisible into x to the fourth and x to the third, and of course, x squared itself. So the greatest common monomial factor is three x squared. This length right over here, this is three x squared."}, {"video_title": "Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "Well, it's going to be x squared. X squared is divisible into x to the fourth and x to the third, and of course, x squared itself. So the greatest common monomial factor is three x squared. This length right over here, this is three x squared. So if this is three x squared, we can then figure out what the width is. So what's, if we were to divide 12 x to the fourth by three x squared, what do we get? Well, 12 divided by three is four, and x to the fourth divided by x squared is x squared."}, {"video_title": "Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "This length right over here, this is three x squared. So if this is three x squared, we can then figure out what the width is. So what's, if we were to divide 12 x to the fourth by three x squared, what do we get? Well, 12 divided by three is four, and x to the fourth divided by x squared is x squared. Notice, three x squared times four x squared is 12 x to the fourth. And then we move over to this purple section. If we take six x to the third divided by three x squared, six divided by three is two, and then x to the third divided by x squared is just going to be x."}, {"video_title": "Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "Well, 12 divided by three is four, and x to the fourth divided by x squared is x squared. Notice, three x squared times four x squared is 12 x to the fourth. And then we move over to this purple section. If we take six x to the third divided by three x squared, six divided by three is two, and then x to the third divided by x squared is just going to be x. And then last but not least, we have 15 divided by three is going to be five. X squared divided by x squared is just one, so it's just going to be five. So the width is going to be four x squared plus two x plus five."}, {"video_title": "Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "If we take six x to the third divided by three x squared, six divided by three is two, and then x to the third divided by x squared is just going to be x. And then last but not least, we have 15 divided by three is going to be five. X squared divided by x squared is just one, so it's just going to be five. So the width is going to be four x squared plus two x plus five. So once again, the length, we figured that out, it was the greatest common monomial factor of these terms. It's three x squared, and the width is four x squared plus two x plus five. And one way to think about it is we just factored, we just factored this expression over here."}, {"video_title": "Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "So the width is going to be four x squared plus two x plus five. So once again, the length, we figured that out, it was the greatest common monomial factor of these terms. It's three x squared, and the width is four x squared plus two x plus five. And one way to think about it is we just factored, we just factored this expression over here. We could write, we could write that, excuse me, I wanna see the original thing. We could write that three x squared times four x squared plus two x plus five, which is the entire width, well that's going to be equal to the area. That's going to be equal to our original expression, 12x to the fourth power plus six x to the third plus 15x squared."}, {"video_title": "Recognize functions from verbal descriptions (example 2) Algebra II Khan Academy.mp3", "Sentence": "For orders less than $20, shipping costs $4. For orders $20 or more, shipping is $7. Can the dollar amount of the order be represented as a function of shipping costs? So they're saying, can the dollar amount of the order, so can the amount of order be represented as a function of shipping? Let me do that in that blue color. Can it be represented as a function of shipping costs? So if we have shipping costs as an input, will that map to exactly, for a given input, will we get exactly one output for the amount of order?"}, {"video_title": "Recognize functions from verbal descriptions (example 2) Algebra II Khan Academy.mp3", "Sentence": "So they're saying, can the dollar amount of the order, so can the amount of order be represented as a function of shipping? Let me do that in that blue color. Can it be represented as a function of shipping costs? So if we have shipping costs as an input, will that map to exactly, for a given input, will we get exactly one output for the amount of order? In order for this to be represented as a function, in order for this to be represented as a function, we have to input a shipping cost, a shipping cost where this relationship is defined. We need to input a shipping cost, put it into our relationship, and get exactly $1 amount of the order in order for this to be a function. If we get multiple dollar amounts of the order, then the relationship, well, it's still a relationship, but it's not going to be a function."}, {"video_title": "Recognize functions from verbal descriptions (example 2) Algebra II Khan Academy.mp3", "Sentence": "So if we have shipping costs as an input, will that map to exactly, for a given input, will we get exactly one output for the amount of order? In order for this to be represented as a function, in order for this to be represented as a function, we have to input a shipping cost, a shipping cost where this relationship is defined. We need to input a shipping cost, put it into our relationship, and get exactly $1 amount of the order in order for this to be a function. If we get multiple dollar amounts of the order, then the relationship, well, it's still a relationship, but it's not going to be a function. So let's think about it. What are the possible inputs here? Well, there's only two possible shipping costs."}, {"video_title": "Recognize functions from verbal descriptions (example 2) Algebra II Khan Academy.mp3", "Sentence": "If we get multiple dollar amounts of the order, then the relationship, well, it's still a relationship, but it's not going to be a function. So let's think about it. What are the possible inputs here? Well, there's only two possible shipping costs. Shipping costs are either going to be $4 or they're going to be $7. So let's think about what happens when we input $4 in as a shipping cost. So if we input $4 into our relationship, so we input $4 into our little potential function box, so $4 into it, what is the output?"}, {"video_title": "Recognize functions from verbal descriptions (example 2) Algebra II Khan Academy.mp3", "Sentence": "Well, there's only two possible shipping costs. Shipping costs are either going to be $4 or they're going to be $7. So let's think about what happens when we input $4 in as a shipping cost. So if we input $4 into our relationship, so we input $4 into our little potential function box, so $4 into it, what is the output? What is going to be the amount of the order? Well, if the shipping cost is $4, the amount of the order just has to be anything less than $20. So it could have been $1."}, {"video_title": "Recognize functions from verbal descriptions (example 2) Algebra II Khan Academy.mp3", "Sentence": "So if we input $4 into our relationship, so we input $4 into our little potential function box, so $4 into it, what is the output? What is going to be the amount of the order? Well, if the shipping cost is $4, the amount of the order just has to be anything less than $20. So it could have been $1. It could have been $1.50. It could have been $7. It literally can take on any value between up to $20."}, {"video_title": "Recognize functions from verbal descriptions (example 2) Algebra II Khan Academy.mp3", "Sentence": "So it could have been $1. It could have been $1.50. It could have been $7. It literally can take on any value between up to $20. So it could even be $19.99. We could do a similar thing if we inputted $7 into this relationship. If $7 was, then I could put literally an infinite number of numbers."}, {"video_title": "Recognize functions from verbal descriptions (example 2) Algebra II Khan Academy.mp3", "Sentence": "It literally can take on any value between up to $20. So it could even be $19.99. We could do a similar thing if we inputted $7 into this relationship. If $7 was, then I could put literally an infinite number of numbers. It could be a million dollars. So if I input $7 into this relationship that we're trying to test whether it's a function, if $7 is the shipping cost, then the order we just know is over $20, $20 or more. So it could be $20."}, {"video_title": "Recognize functions from verbal descriptions (example 2) Algebra II Khan Academy.mp3", "Sentence": "If $7 was, then I could put literally an infinite number of numbers. It could be a million dollars. So if I input $7 into this relationship that we're trying to test whether it's a function, if $7 is the shipping cost, then the order we just know is over $20, $20 or more. So it could be $20. It could be $800. It could be $1 million. There's actually an infinite number of values that it could take on right over here."}, {"video_title": "Recognize functions from verbal descriptions (example 2) Algebra II Khan Academy.mp3", "Sentence": "So it could be $20. It could be $800. It could be $1 million. There's actually an infinite number of values that it could take on right over here. So because for a valid shipping cost, for each of these valid shipping costs, I can get many, many, many potential outputs. I don't know which output it necessarily will output to. If someone tells you the shipping cost, you don't necessarily know what the order size was."}, {"video_title": "Recognize functions from verbal descriptions (example 2) Algebra II Khan Academy.mp3", "Sentence": "There's actually an infinite number of values that it could take on right over here. So because for a valid shipping cost, for each of these valid shipping costs, I can get many, many, many potential outputs. I don't know which output it necessarily will output to. If someone tells you the shipping cost, you don't necessarily know what the order size was. This is not a function. You cannot represent the amount of order as a function of the shipping cost. So no."}, {"video_title": "Writing equations of perpendicular lines Mathematics I High School Math Khan Academy.mp3", "Sentence": "And they say that line B contains the point 6, negative 7. And they tell us lines A and B are perpendicular. So that means that slope of B must be negative inverse of slope of A. So what we'll do is we'll figure out the slope of A, then take the negative inverse of it, then we'll know the slope of B, then we can use this point right here to fill in the gaps and figure out B's y-intercept. So what's the slope of A? This is already in slope-intercept form. The slope of A is right there."}, {"video_title": "Writing equations of perpendicular lines Mathematics I High School Math Khan Academy.mp3", "Sentence": "So what we'll do is we'll figure out the slope of A, then take the negative inverse of it, then we'll know the slope of B, then we can use this point right here to fill in the gaps and figure out B's y-intercept. So what's the slope of A? This is already in slope-intercept form. The slope of A is right there. It's the 2, mx plus b. So the slope here is equal to 2. So the slope of A is 2."}, {"video_title": "Writing equations of perpendicular lines Mathematics I High School Math Khan Academy.mp3", "Sentence": "The slope of A is right there. It's the 2, mx plus b. So the slope here is equal to 2. So the slope of A is 2. What is the slope of B? So what is B's slope going to have to be? Well, it's perpendicular to A, so it's going to be the negative inverse of this."}, {"video_title": "Writing equations of perpendicular lines Mathematics I High School Math Khan Academy.mp3", "Sentence": "So the slope of A is 2. What is the slope of B? So what is B's slope going to have to be? Well, it's perpendicular to A, so it's going to be the negative inverse of this. The inverse of 2 is 1 half. The negative inverse of that is negative 1 half. So B's slope is negative 1 half."}, {"video_title": "Writing equations of perpendicular lines Mathematics I High School Math Khan Academy.mp3", "Sentence": "Well, it's perpendicular to A, so it's going to be the negative inverse of this. The inverse of 2 is 1 half. The negative inverse of that is negative 1 half. So B's slope is negative 1 half. So we know that B's equation has to be y is equal to its slope m times x plus some y-intercept. We still don't know what the y-intercept of B is, but we can use this information to figure it out. We know that y is equal to negative 7 when x is equal to 6, negative 1 half times 6 plus b. I just know that this is on the point."}, {"video_title": "Writing equations of perpendicular lines Mathematics I High School Math Khan Academy.mp3", "Sentence": "So B's slope is negative 1 half. So we know that B's equation has to be y is equal to its slope m times x plus some y-intercept. We still don't know what the y-intercept of B is, but we can use this information to figure it out. We know that y is equal to negative 7 when x is equal to 6, negative 1 half times 6 plus b. I just know that this is on the point. So this point must satisfy the equation of line B. So let's work out what B must be. So in this, or what the B, the y-intercept, this is the lowercase b, not the line b."}, {"video_title": "Writing equations of perpendicular lines Mathematics I High School Math Khan Academy.mp3", "Sentence": "We know that y is equal to negative 7 when x is equal to 6, negative 1 half times 6 plus b. I just know that this is on the point. So this point must satisfy the equation of line B. So let's work out what B must be. So in this, or what the B, the y-intercept, this is the lowercase b, not the line b. So we have negative 7 is equal to what's negative 1 half times 6? That's not a b there, that's a 6. What's negative 1 half times 6?"}, {"video_title": "Writing equations of perpendicular lines Mathematics I High School Math Khan Academy.mp3", "Sentence": "So in this, or what the B, the y-intercept, this is the lowercase b, not the line b. So we have negative 7 is equal to what's negative 1 half times 6? That's not a b there, that's a 6. What's negative 1 half times 6? It's negative 3 is equal to negative 3 plus our y-intercept. Let's add 3 to both sides of this equation. So if we add 3 to both sides, I just want to get rid of this 3 right here, what do we get?"}, {"video_title": "Writing equations of perpendicular lines Mathematics I High School Math Khan Academy.mp3", "Sentence": "What's negative 1 half times 6? It's negative 3 is equal to negative 3 plus our y-intercept. Let's add 3 to both sides of this equation. So if we add 3 to both sides, I just want to get rid of this 3 right here, what do we get? The left-hand side, negative 7 plus 3 is negative 4. And that's going to be equal to, these guys cancel out, that's equal to b, our y-intercept. So this right here is a negative 4."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So what you see here, this is a screenshot of the Desmos online graphing calculator. You can use it at desmos.com, and I encourage you to use this after this video or even while I'm doing this video. But the goal here is to think about reflection of functions. So let's just start with some examples. Let's say that I had a function f of x, and it is equal to the square root of x. So that's what it looks like, fairly reasonable. Now let's make another function, g of x, and I'll start off by also making that the square root of x."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So let's just start with some examples. Let's say that I had a function f of x, and it is equal to the square root of x. So that's what it looks like, fairly reasonable. Now let's make another function, g of x, and I'll start off by also making that the square root of x. So no surprise there, g of x was graphed right on top of f of x. But what would happen if, instead of it just being the square root of x, what would happen if we put a negative out front, right over there? What do you think is going to happen when I do that?"}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Now let's make another function, g of x, and I'll start off by also making that the square root of x. So no surprise there, g of x was graphed right on top of f of x. But what would happen if, instead of it just being the square root of x, what would happen if we put a negative out front, right over there? What do you think is going to happen when I do that? Well, let's just try it out. When I put the negative, it looks like it flipped it over the x-axis. It looks like it reflected it over the x-axis."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "What do you think is going to happen when I do that? Well, let's just try it out. When I put the negative, it looks like it flipped it over the x-axis. It looks like it reflected it over the x-axis. Now, instead of doing it that way, what if we had another function, h of x, and I'll start off by making it identical to f of x. So once again, it's right over there. It traces out f of x."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "It looks like it reflected it over the x-axis. Now, instead of doing it that way, what if we had another function, h of x, and I'll start off by making it identical to f of x. So once again, it's right over there. It traces out f of x. Instead of putting the negative out in front of the radical sign, what if we put it under the radical sign? What if we replaced x with a negative x? What do you think is going to happen there?"}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "It traces out f of x. Instead of putting the negative out in front of the radical sign, what if we put it under the radical sign? What if we replaced x with a negative x? What do you think is going to happen there? Well, let's try it out. If we replace it, that shifted it over the y-axis. And then, pause this video and think about how would you shift it over both axes?"}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "What do you think is going to happen there? Well, let's try it out. If we replace it, that shifted it over the y-axis. And then, pause this video and think about how would you shift it over both axes? Well, we could do a, well, I'm running out of letters. Maybe I will do a, I don't know, k of x is equal to, so I'm gonna put the negative outside the radical sign, and then I'm gonna take the square root, and I'm gonna put a negative inside the radical sign. And notice, it flipped it over both."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And then, pause this video and think about how would you shift it over both axes? Well, we could do a, well, I'm running out of letters. Maybe I will do a, I don't know, k of x is equal to, so I'm gonna put the negative outside the radical sign, and then I'm gonna take the square root, and I'm gonna put a negative inside the radical sign. And notice, it flipped it over both. It flipped it over both the x-axis and the y-axis to go over here. Now, why does this happen? Well, let's just start with the g of x."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And notice, it flipped it over both. It flipped it over both the x-axis and the y-axis to go over here. Now, why does this happen? Well, let's just start with the g of x. So, when you put the negative out in front, when you negate everything that's in the expression that defines a function, whatever value you would have gotten of the function before, you're now going to get the opposite of it. So, when x is zero, we got zero. When x is one, instead of one now, you're taking the negative of it, so you're gonna get negative one."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Well, let's just start with the g of x. So, when you put the negative out in front, when you negate everything that's in the expression that defines a function, whatever value you would have gotten of the function before, you're now going to get the opposite of it. So, when x is zero, we got zero. When x is one, instead of one now, you're taking the negative of it, so you're gonna get negative one. When x is four, instead of getting positive two, you're now going to get negative two. When x is equal to nine, instead of getting positive three, you now get negative three. So, hopefully that makes sense why putting a negative out front of an entire expression is going to flip it over, flip its graph over the x-axis."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "When x is one, instead of one now, you're taking the negative of it, so you're gonna get negative one. When x is four, instead of getting positive two, you're now going to get negative two. When x is equal to nine, instead of getting positive three, you now get negative three. So, hopefully that makes sense why putting a negative out front of an entire expression is going to flip it over, flip its graph over the x-axis. Now, what about replacing an x with a negative x? Well, one way to think about it now is whenever you inputted one before, that would now be a negative one that you're trying to evaluate the principal root of. And we know that the principal root function is not defined for negative one."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So, hopefully that makes sense why putting a negative out front of an entire expression is going to flip it over, flip its graph over the x-axis. Now, what about replacing an x with a negative x? Well, one way to think about it now is whenever you inputted one before, that would now be a negative one that you're trying to evaluate the principal root of. And we know that the principal root function is not defined for negative one. But when x is equal to negative one, our original function wasn't defined there when x is equal to negative one. But if you take the negative of that, well, now you're taking the principal root of one. And so, that's why it is now defined."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And we know that the principal root function is not defined for negative one. But when x is equal to negative one, our original function wasn't defined there when x is equal to negative one. But if you take the negative of that, well, now you're taking the principal root of one. And so, that's why it is now defined. So, whatever value the function would have taken on at a given value of x, it now takes that value on the corresponding opposite value of x, on the negative value of that x. And so, that's why it flips it over the y-axis. And this is true with many types of functions."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And so, that's why it is now defined. So, whatever value the function would have taken on at a given value of x, it now takes that value on the corresponding opposite value of x, on the negative value of that x. And so, that's why it flips it over the y-axis. And this is true with many types of functions. We don't have to do this just with a square root function. Let's try another function. Let's say we tried this for e to the x power."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And this is true with many types of functions. We don't have to do this just with a square root function. Let's try another function. Let's say we tried this for e to the x power. So, there you go. We have a very classic exponential there. Now, let's say that g of x is equal to negative e to the x."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Let's say we tried this for e to the x power. So, there you go. We have a very classic exponential there. Now, let's say that g of x is equal to negative e to the x. And if what we expect to happen happens, this will flip it over the x-axis. So, negative e to the x power. And indeed, that is what happens."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Now, let's say that g of x is equal to negative e to the x. And if what we expect to happen happens, this will flip it over the x-axis. So, negative e to the x power. And indeed, that is what happens. And then, how would we flip it over the y-axis? Well, let's do an h of x. That's going to be equal to e to the, instead of putting an x there, we will put a negative x, negative x."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And indeed, that is what happens. And then, how would we flip it over the y-axis? Well, let's do an h of x. That's going to be equal to e to the, instead of putting an x there, we will put a negative x, negative x. And there you have it. Notice, it flipped it over the y-axis. Now, both examples that I just did, these are very simple expressions."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "That's going to be equal to e to the, instead of putting an x there, we will put a negative x, negative x. And there you have it. Notice, it flipped it over the y-axis. Now, both examples that I just did, these are very simple expressions. Let's imagine something that's a little bit more complex. Let's say that f of x, let's give it a nice higher degree polynomial. So, let's say it's x to the third minus two x squared."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Now, both examples that I just did, these are very simple expressions. Let's imagine something that's a little bit more complex. Let's say that f of x, let's give it a nice higher degree polynomial. So, let's say it's x to the third minus two x squared. That's a nice one. And actually, let's just add another term here. So, plus two x. Oh, no, I wanna make it minus two x. I wanna accentuate some of those curves."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So, let's say it's x to the third minus two x squared. That's a nice one. And actually, let's just add another term here. So, plus two x. Oh, no, I wanna make it minus two x. I wanna accentuate some of those curves. All right, so that's a pretty interesting graph. Now, how would I flip it over the x-axis? Well, the way that I would do that is I could define a g of x. I could do it two ways."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So, plus two x. Oh, no, I wanna make it minus two x. I wanna accentuate some of those curves. All right, so that's a pretty interesting graph. Now, how would I flip it over the x-axis? Well, the way that I would do that is I could define a g of x. I could do it two ways. I could say g of x is equal to the negative of f of x, and we get that. So, that's essentially just taking this entire expression and multiplying it by negative one. And notice, it's multiplying it, it's flipping it over the x-axis."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Well, the way that I would do that is I could define a g of x. I could do it two ways. I could say g of x is equal to the negative of f of x, and we get that. So, that's essentially just taking this entire expression and multiplying it by negative one. And notice, it's multiplying it, it's flipping it over the x-axis. Another way we could have done it is, instead of that, we could have said the negative of x to the third minus two x squared, and then minus two x, and then we close those parentheses, and we get the same effect. Now, what if we wanted to flip it over the y-axis? Well, then, instead of putting a negative on the entire expression, what we wanna do is replace our x's with a negative x."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And notice, it's multiplying it, it's flipping it over the x-axis. Another way we could have done it is, instead of that, we could have said the negative of x to the third minus two x squared, and then minus two x, and then we close those parentheses, and we get the same effect. Now, what if we wanted to flip it over the y-axis? Well, then, instead of putting a negative on the entire expression, what we wanna do is replace our x's with a negative x. So, you could do it like this. You could say that that's going to be f of negative x, and that has the effect of, everywhere you saw an x before, you replace it with a negative x. And notice, it did exactly what we expect."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Well, then, instead of putting a negative on the entire expression, what we wanna do is replace our x's with a negative x. So, you could do it like this. You could say that that's going to be f of negative x, and that has the effect of, everywhere you saw an x before, you replace it with a negative x. And notice, it did exactly what we expect. It flipped it over, over the y-axis. Now, the other way we could have done that, just to make it clear, that's the same thing as negative x to the third power minus two times negative x squared minus two times negative x. And of course, we could simplify that expression, but notice, it has the exact same idea."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And notice, it did exactly what we expect. It flipped it over, over the y-axis. Now, the other way we could have done that, just to make it clear, that's the same thing as negative x to the third power minus two times negative x squared minus two times negative x. And of course, we could simplify that expression, but notice, it has the exact same idea. And if we wanted to flip it over both the x and y-axes, well, we've already flipped it over the y-axis. To flip it over the x-axis, whoops, I just deleted it. To flip it over the, I'm having issues here."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And of course, we could simplify that expression, but notice, it has the exact same idea. And if we wanted to flip it over both the x and y-axes, well, we've already flipped it over the y-axis. To flip it over the x-axis, whoops, I just deleted it. To flip it over the, I'm having issues here. To flip it over the x-axis as well, we would, oh, it gave me a parentheses already. I would just put a negative out front. So, I put a negative out front, and there you have it."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "To flip it over the, I'm having issues here. To flip it over the x-axis as well, we would, oh, it gave me a parentheses already. I would just put a negative out front. So, I put a negative out front, and there you have it. This flipped it over both the x and y-axis. You can do them in either order, and you will get to this green curve. Now, an easier way of writing that would have been just the negative of f of negative x, and you would have gotten to that same place."}, {"video_title": "Writing equations in all forms Algebra I Khan Academy.mp3", "Sentence": "A line passes through the points negative 3, 6, and 6, 0. Find the equation of this line in point-slope form, slope-intercept form, standard form. And the way to think about these, these are just three different ways of writing the same equation. So if you give me one of them, we can manipulate it to get any of the other ones. But just so you know what these are, point-slope form, let's say that x, let's say the point x1, y1, let's say that that is a point on the line. And when someone puts this little subscript here, so if they just write an x, that means that we're talking about a variable that can take on any value. If someone writes x with a subscript 1 and a y with a subscript 1, that's like saying a particular value of x and a particular value of y or a particular coordinate."}, {"video_title": "Writing equations in all forms Algebra I Khan Academy.mp3", "Sentence": "So if you give me one of them, we can manipulate it to get any of the other ones. But just so you know what these are, point-slope form, let's say that x, let's say the point x1, y1, let's say that that is a point on the line. And when someone puts this little subscript here, so if they just write an x, that means that we're talking about a variable that can take on any value. If someone writes x with a subscript 1 and a y with a subscript 1, that's like saying a particular value of x and a particular value of y or a particular coordinate. And you'll see that when we do the example. But point-slope form says that, look, if I know a particular point and if I know the slope of the line, then putting that line in point-slope form would be y minus y1 is equal to m times x minus x1. So for example, and we'll see do that in this video, if the point negative 3, 6 is on the line, then we'd say y minus 6 is equal to m times x minus negative 3."}, {"video_title": "Writing equations in all forms Algebra I Khan Academy.mp3", "Sentence": "If someone writes x with a subscript 1 and a y with a subscript 1, that's like saying a particular value of x and a particular value of y or a particular coordinate. And you'll see that when we do the example. But point-slope form says that, look, if I know a particular point and if I know the slope of the line, then putting that line in point-slope form would be y minus y1 is equal to m times x minus x1. So for example, and we'll see do that in this video, if the point negative 3, 6 is on the line, then we'd say y minus 6 is equal to m times x minus negative 3. So it'll end up becoming x plus 3. So this is a particular x and a particular y. It could be a negative 3 and 6."}, {"video_title": "Writing equations in all forms Algebra I Khan Academy.mp3", "Sentence": "So for example, and we'll see do that in this video, if the point negative 3, 6 is on the line, then we'd say y minus 6 is equal to m times x minus negative 3. So it'll end up becoming x plus 3. So this is a particular x and a particular y. It could be a negative 3 and 6. So that's point-slope form. Slope-intercept form is y is equal to mx plus b, where once again, m is the slope, b is the y-intercept. Where does the line intersect the y-axis?"}, {"video_title": "Writing equations in all forms Algebra I Khan Academy.mp3", "Sentence": "It could be a negative 3 and 6. So that's point-slope form. Slope-intercept form is y is equal to mx plus b, where once again, m is the slope, b is the y-intercept. Where does the line intersect the y-axis? What value does y take on when x is 0? And then standard form is the form ax plus by is equal to c, where these are just two numbers, essentially. They really don't have any interpretation directly on the graph."}, {"video_title": "Writing equations in all forms Algebra I Khan Academy.mp3", "Sentence": "Where does the line intersect the y-axis? What value does y take on when x is 0? And then standard form is the form ax plus by is equal to c, where these are just two numbers, essentially. They really don't have any interpretation directly on the graph. So let's do this. Let's figure out all of these forms. So the first thing we want to do is figure out the slope."}, {"video_title": "Writing equations in all forms Algebra I Khan Academy.mp3", "Sentence": "They really don't have any interpretation directly on the graph. So let's do this. Let's figure out all of these forms. So the first thing we want to do is figure out the slope. Once we figure out the slope, then point-slope form is actually very, very, very straightforward to calculate. So just to remind ourselves, slope, which is equal to m, which is going to be equal to the change in y over the change in x. Now, what is the change in y?"}, {"video_title": "Writing equations in all forms Algebra I Khan Academy.mp3", "Sentence": "So the first thing we want to do is figure out the slope. Once we figure out the slope, then point-slope form is actually very, very, very straightforward to calculate. So just to remind ourselves, slope, which is equal to m, which is going to be equal to the change in y over the change in x. Now, what is the change in y? If we view this as our endpoint, if we imagine that we're going from here to that point, what is the change in y? Well, we have our endpoint, which is 0. y ends up at 0, and y was at 6. So our finishing y point is 0."}, {"video_title": "Writing equations in all forms Algebra I Khan Academy.mp3", "Sentence": "Now, what is the change in y? If we view this as our endpoint, if we imagine that we're going from here to that point, what is the change in y? Well, we have our endpoint, which is 0. y ends up at 0, and y was at 6. So our finishing y point is 0. Our starting y point is 6. What was our finishing x point, or x coordinate? Our finishing x coordinate was 6."}, {"video_title": "Writing equations in all forms Algebra I Khan Academy.mp3", "Sentence": "So our finishing y point is 0. Our starting y point is 6. What was our finishing x point, or x coordinate? Our finishing x coordinate was 6. Let me make this very clear. I don't want to confuse you. So this 0, we have that 0."}, {"video_title": "Writing equations in all forms Algebra I Khan Academy.mp3", "Sentence": "Our finishing x coordinate was 6. Let me make this very clear. I don't want to confuse you. So this 0, we have that 0. That is that 0 right there. And then we have this 6, which was our starting y point. That is that 6 right there."}, {"video_title": "Writing equations in all forms Algebra I Khan Academy.mp3", "Sentence": "So this 0, we have that 0. That is that 0 right there. And then we have this 6, which was our starting y point. That is that 6 right there. And then we want our finishing x value. That is that 6 right there. And we want to subtract from that our starting x value."}, {"video_title": "Writing equations in all forms Algebra I Khan Academy.mp3", "Sentence": "That is that 6 right there. And then we want our finishing x value. That is that 6 right there. And we want to subtract from that our starting x value. Well, our starting x value is that right over there. That's that negative 3. And just to make sure we know what we're doing, this negative 3 is that negative 3 right there."}, {"video_title": "Writing equations in all forms Algebra I Khan Academy.mp3", "Sentence": "And we want to subtract from that our starting x value. Well, our starting x value is that right over there. That's that negative 3. And just to make sure we know what we're doing, this negative 3 is that negative 3 right there. I'm just saying, if we go from that point to that point, our y went down by 6. We went from 6 to 0. Our y went down by 6."}, {"video_title": "Writing equations in all forms Algebra I Khan Academy.mp3", "Sentence": "And just to make sure we know what we're doing, this negative 3 is that negative 3 right there. I'm just saying, if we go from that point to that point, our y went down by 6. We went from 6 to 0. Our y went down by 6. So we get 0 minus 6 is negative 6. That makes sense. y went down by 6."}, {"video_title": "Writing equations in all forms Algebra I Khan Academy.mp3", "Sentence": "Our y went down by 6. So we get 0 minus 6 is negative 6. That makes sense. y went down by 6. And if we went from that point to that point, what happened to x? We went from negative 3 to 6. It should go up by 9."}, {"video_title": "Writing equations in all forms Algebra I Khan Academy.mp3", "Sentence": "y went down by 6. And if we went from that point to that point, what happened to x? We went from negative 3 to 6. It should go up by 9. And if you calculate this, take your 6 minus negative 3. That's the same thing as 6 plus 3. That is 9."}, {"video_title": "Writing equations in all forms Algebra I Khan Academy.mp3", "Sentence": "It should go up by 9. And if you calculate this, take your 6 minus negative 3. That's the same thing as 6 plus 3. That is 9. And what is negative 6 9? Well, if you simplify it, it is negative 2 3rds. You divide the numerator and the denominator by 3."}, {"video_title": "Writing equations in all forms Algebra I Khan Academy.mp3", "Sentence": "That is 9. And what is negative 6 9? Well, if you simplify it, it is negative 2 3rds. You divide the numerator and the denominator by 3. So that is our slope, negative 2 3rds. So we're pretty much ready to use point-slope form. We have a point."}, {"video_title": "Writing equations in all forms Algebra I Khan Academy.mp3", "Sentence": "You divide the numerator and the denominator by 3. So that is our slope, negative 2 3rds. So we're pretty much ready to use point-slope form. We have a point. We could pick one of these points. I'll just go with the negative 3 6. And we have our slope."}, {"video_title": "Writing equations in all forms Algebra I Khan Academy.mp3", "Sentence": "We have a point. We could pick one of these points. I'll just go with the negative 3 6. And we have our slope. So let's put it in point-slope form. So point-slope form. All we have to do is we say y minus."}, {"video_title": "Writing equations in all forms Algebra I Khan Academy.mp3", "Sentence": "And we have our slope. So let's put it in point-slope form. So point-slope form. All we have to do is we say y minus. Now, we could have taken either of these points. I'll take this one. So y minus the y value over here."}, {"video_title": "Writing equations in all forms Algebra I Khan Academy.mp3", "Sentence": "All we have to do is we say y minus. Now, we could have taken either of these points. I'll take this one. So y minus the y value over here. So y minus 6 is equal to our slope, which is negative 2 3rds times x minus our x coordinate. Well, our x coordinate. So x minus our x coordinate is negative 3."}, {"video_title": "Writing equations in all forms Algebra I Khan Academy.mp3", "Sentence": "So y minus the y value over here. So y minus 6 is equal to our slope, which is negative 2 3rds times x minus our x coordinate. Well, our x coordinate. So x minus our x coordinate is negative 3. And we're done. We can simplify it a little bit. This becomes y minus 6 is equal to negative 2 3rds times x. x minus negative 3 is the same thing as x plus 3."}, {"video_title": "Writing equations in all forms Algebra I Khan Academy.mp3", "Sentence": "So x minus our x coordinate is negative 3. And we're done. We can simplify it a little bit. This becomes y minus 6 is equal to negative 2 3rds times x. x minus negative 3 is the same thing as x plus 3. This is our point-slope form. Now, we can literally just algebraically manipulate this guy right here to put it into our slope-intercept form. Let's do that."}, {"video_title": "Writing equations in all forms Algebra I Khan Academy.mp3", "Sentence": "This becomes y minus 6 is equal to negative 2 3rds times x. x minus negative 3 is the same thing as x plus 3. This is our point-slope form. Now, we can literally just algebraically manipulate this guy right here to put it into our slope-intercept form. Let's do that. So let's do slope-intercept in orange. So we have slope-intercept. So what can we do here to simplify this?"}, {"video_title": "Writing equations in all forms Algebra I Khan Academy.mp3", "Sentence": "Let's do that. So let's do slope-intercept in orange. So we have slope-intercept. So what can we do here to simplify this? Well, we can multiply out the negative 2 3rds. So you get y minus 6 is equal to. I'm just distributing the negative 2 3rds."}, {"video_title": "Writing equations in all forms Algebra I Khan Academy.mp3", "Sentence": "So what can we do here to simplify this? Well, we can multiply out the negative 2 3rds. So you get y minus 6 is equal to. I'm just distributing the negative 2 3rds. So negative 2 3rds times x is negative 2 3rds x. And then negative 2 3rds times 3 is negative 2. And now to get it in slope-intercept form, we just have to add the 6 to both sides so we get rid of it on the left-hand side."}, {"video_title": "Writing equations in all forms Algebra I Khan Academy.mp3", "Sentence": "I'm just distributing the negative 2 3rds. So negative 2 3rds times x is negative 2 3rds x. And then negative 2 3rds times 3 is negative 2. And now to get it in slope-intercept form, we just have to add the 6 to both sides so we get rid of it on the left-hand side. So let's add 6 to both sides of this equation. Left-hand side of the equation, we're just left with a y. These guys cancel out."}, {"video_title": "Writing equations in all forms Algebra I Khan Academy.mp3", "Sentence": "And now to get it in slope-intercept form, we just have to add the 6 to both sides so we get rid of it on the left-hand side. So let's add 6 to both sides of this equation. Left-hand side of the equation, we're just left with a y. These guys cancel out. You get a y is equal to negative 2 3rds x. Negative 2 plus 6 is plus 4. So there you have it."}, {"video_title": "Writing equations in all forms Algebra I Khan Academy.mp3", "Sentence": "These guys cancel out. You get a y is equal to negative 2 3rds x. Negative 2 plus 6 is plus 4. So there you have it. That is our slope-intercept form, mx plus b. That's our y-intercept. Now the last thing we need to do is get it into the standard form."}, {"video_title": "Writing equations in all forms Algebra I Khan Academy.mp3", "Sentence": "So there you have it. That is our slope-intercept form, mx plus b. That's our y-intercept. Now the last thing we need to do is get it into the standard form. So once again, we just have to algebraically manipulate it so that the x's and the y's are both on this side of the equation. So let's just add 2 3rds x to both sides of this equation. So let me start here."}, {"video_title": "Writing equations in all forms Algebra I Khan Academy.mp3", "Sentence": "Now the last thing we need to do is get it into the standard form. So once again, we just have to algebraically manipulate it so that the x's and the y's are both on this side of the equation. So let's just add 2 3rds x to both sides of this equation. So let me start here. So we have y is equal to negative 2 3rds x plus 4. That's slope-intercept form. Let's add 2 3rds x."}, {"video_title": "Writing equations in all forms Algebra I Khan Academy.mp3", "Sentence": "So let me start here. So we have y is equal to negative 2 3rds x plus 4. That's slope-intercept form. Let's add 2 3rds x. So plus 2 3rds x to both sides of this equation. I'm doing that so I don't have this 2 3rds x on the right-hand side, this negative 2 3rds x. So the left-hand side of the equation, I squenched it up a little bit, maybe more than I should have."}, {"video_title": "Writing equations in all forms Algebra I Khan Academy.mp3", "Sentence": "Let's add 2 3rds x. So plus 2 3rds x to both sides of this equation. I'm doing that so I don't have this 2 3rds x on the right-hand side, this negative 2 3rds x. So the left-hand side of the equation, I squenched it up a little bit, maybe more than I should have. The left-hand side of this equation is what? It is 2 3rds x, because 2 over 3x plus this y, that's my left-hand side, is equal to, these guys cancel out, is equal to 4. So this by itself, we are in standard form."}, {"video_title": "Writing equations in all forms Algebra I Khan Academy.mp3", "Sentence": "So the left-hand side of the equation, I squenched it up a little bit, maybe more than I should have. The left-hand side of this equation is what? It is 2 3rds x, because 2 over 3x plus this y, that's my left-hand side, is equal to, these guys cancel out, is equal to 4. So this by itself, we are in standard form. This is the standard form of the equation. If we wanted to make it look extra clean, have no fractions here, we could multiply both sides of this equation by 3. If we do that, what do we get?"}, {"video_title": "Writing equations in all forms Algebra I Khan Academy.mp3", "Sentence": "So this by itself, we are in standard form. This is the standard form of the equation. If we wanted to make it look extra clean, have no fractions here, we could multiply both sides of this equation by 3. If we do that, what do we get? 2 3rds x times 3 is just 2x. y times 3 is 3y. And then 4 times 3 is 12."}, {"video_title": "Writing equations in all forms Algebra I Khan Academy.mp3", "Sentence": "If we do that, what do we get? 2 3rds x times 3 is just 2x. y times 3 is 3y. And then 4 times 3 is 12. These are the same equations, I just multiplied every term by 3. If you do it to the left-hand side, you can do it to the right-hand side, or you have to do it to the right-hand side. And we are in standard form."}, {"video_title": "Writing standard equation of a circle Mathematics II High School Math Khan Academy.mp3", "Sentence": "And so with that information, I want you to pause the video and see if you can figure out the equation for this circle. All right, let's work through this together. So let's first think about the center of the circle, and the center of the circle is just going to be the coordinates of that point. So the x-coordinate is negative one, and then the y-coordinate is one. So center is negative one, comma, one. And now let's think about what the radius of the circle is. Well, the radius is going to be the distance between the center and any point on the circle."}, {"video_title": "Writing standard equation of a circle Mathematics II High School Math Khan Academy.mp3", "Sentence": "So the x-coordinate is negative one, and then the y-coordinate is one. So center is negative one, comma, one. And now let's think about what the radius of the circle is. Well, the radius is going to be the distance between the center and any point on the circle. So for example, this distance, the distance of that line, let's see, I can do a thicker version of that. This line right over there, something strange about my, something strange about my pen tool is making that very thin. Let me do it one more time."}, {"video_title": "Writing standard equation of a circle Mathematics II High School Math Khan Academy.mp3", "Sentence": "Well, the radius is going to be the distance between the center and any point on the circle. So for example, this distance, the distance of that line, let's see, I can do a thicker version of that. This line right over there, something strange about my, something strange about my pen tool is making that very thin. Let me do it one more time. Okay, that's better. The distance of that line right over there, that is going to be the radius. So how can we figure that out?"}, {"video_title": "Writing standard equation of a circle Mathematics II High School Math Khan Academy.mp3", "Sentence": "Let me do it one more time. Okay, that's better. The distance of that line right over there, that is going to be the radius. So how can we figure that out? Well, we can set up a right triangle and essentially use the distance formula which comes from the Pythagorean theorem. To figure out the length of that line, so this is the radius, we could figure out a change in x. So if we look at our change in x right over here, our change in x as we go from the center to this point, so this is our change in x, and then we could say that this is our change in y."}, {"video_title": "Writing standard equation of a circle Mathematics II High School Math Khan Academy.mp3", "Sentence": "So how can we figure that out? Well, we can set up a right triangle and essentially use the distance formula which comes from the Pythagorean theorem. To figure out the length of that line, so this is the radius, we could figure out a change in x. So if we look at our change in x right over here, our change in x as we go from the center to this point, so this is our change in x, and then we could say that this is our change in y. That right over there is our change in y. And so our change in x squared plus our change in y squared is going to be our radius squared. That comes straight out of the Pythagorean theorem."}, {"video_title": "Writing standard equation of a circle Mathematics II High School Math Khan Academy.mp3", "Sentence": "So if we look at our change in x right over here, our change in x as we go from the center to this point, so this is our change in x, and then we could say that this is our change in y. That right over there is our change in y. And so our change in x squared plus our change in y squared is going to be our radius squared. That comes straight out of the Pythagorean theorem. This is a right triangle. And so we can say that r squared is going to be equal to our change in x squared plus our change in y squared. Plus our change in y squared."}, {"video_title": "Writing standard equation of a circle Mathematics II High School Math Khan Academy.mp3", "Sentence": "That comes straight out of the Pythagorean theorem. This is a right triangle. And so we can say that r squared is going to be equal to our change in x squared plus our change in y squared. Plus our change in y squared. Now what is our change in x going to be? Our change in x is going to be equal to, well when we go from the radius to this point over here, our x goes from negative one to six. So you could view it as our ending x minus our starting x."}, {"video_title": "Writing standard equation of a circle Mathematics II High School Math Khan Academy.mp3", "Sentence": "Plus our change in y squared. Now what is our change in x going to be? Our change in x is going to be equal to, well when we go from the radius to this point over here, our x goes from negative one to six. So you could view it as our ending x minus our starting x. So negative one minus, sorry, six minus negative one is equal to seven. So let me, so we have our change in x, this right over here, is equal to seven. If we viewed this as the start point and this as the end point, it would be negative seven, but we really care about the absolute value of the change in x, and once you square it, it all becomes a positive anyway."}, {"video_title": "Writing standard equation of a circle Mathematics II High School Math Khan Academy.mp3", "Sentence": "So you could view it as our ending x minus our starting x. So negative one minus, sorry, six minus negative one is equal to seven. So let me, so we have our change in x, this right over here, is equal to seven. If we viewed this as the start point and this as the end point, it would be negative seven, but we really care about the absolute value of the change in x, and once you square it, it all becomes a positive anyway. So our change in x right over here is going to be positive seven. And our change in y, well, we are starting at, we are starting at y is equal to one, and we are going to y is equal to negative four, so it would be negative four minus one, which is equal to negative five. And so our change in y is negative five."}, {"video_title": "Writing standard equation of a circle Mathematics II High School Math Khan Academy.mp3", "Sentence": "If we viewed this as the start point and this as the end point, it would be negative seven, but we really care about the absolute value of the change in x, and once you square it, it all becomes a positive anyway. So our change in x right over here is going to be positive seven. And our change in y, well, we are starting at, we are starting at y is equal to one, and we are going to y is equal to negative four, so it would be negative four minus one, which is equal to negative five. And so our change in y is negative five. You could view this distance right over here as the absolute value of our change in y, which of course would be the absolute value of five. But once you square it, it doesn't matter, the negative sign goes away. And so this is going to simplify to seven squared, change in x squared is 49."}, {"video_title": "Writing standard equation of a circle Mathematics II High School Math Khan Academy.mp3", "Sentence": "And so our change in y is negative five. You could view this distance right over here as the absolute value of our change in y, which of course would be the absolute value of five. But once you square it, it doesn't matter, the negative sign goes away. And so this is going to simplify to seven squared, change in x squared is 49. Change in y squared, negative five squared is 25. So we get r squared, we get r squared is equal to 49 plus 25. So what's 49 plus 25?"}, {"video_title": "Writing standard equation of a circle Mathematics II High School Math Khan Academy.mp3", "Sentence": "And so this is going to simplify to seven squared, change in x squared is 49. Change in y squared, negative five squared is 25. So we get r squared, we get r squared is equal to 49 plus 25. So what's 49 plus 25? Let's see, that's going to be 54, was it 74? R squared is equal to 74. Did I do that right?"}, {"video_title": "Writing standard equation of a circle Mathematics II High School Math Khan Academy.mp3", "Sentence": "So what's 49 plus 25? Let's see, that's going to be 54, was it 74? R squared is equal to 74. Did I do that right? Yep, 74. And so now we can write the equation for the circle. The circle is going to be all of the points that are, well, let me write, so if r squared is equal to 74, r is equal to the square root of 74."}, {"video_title": "Writing standard equation of a circle Mathematics II High School Math Khan Academy.mp3", "Sentence": "Did I do that right? Yep, 74. And so now we can write the equation for the circle. The circle is going to be all of the points that are, well, let me write, so if r squared is equal to 74, r is equal to the square root of 74. And so the equation of the circle is going to be all points x comma y that are this far away from the center. And so what are those points going to be? Well, the distance is going to be x minus the x coordinate of the center, x minus negative one squared."}, {"video_title": "Writing standard equation of a circle Mathematics II High School Math Khan Academy.mp3", "Sentence": "The circle is going to be all of the points that are, well, let me write, so if r squared is equal to 74, r is equal to the square root of 74. And so the equation of the circle is going to be all points x comma y that are this far away from the center. And so what are those points going to be? Well, the distance is going to be x minus the x coordinate of the center, x minus negative one squared. Let me do that in blue color. Minus negative one squared plus y minus, y minus the y coordinate of the center, y minus one squared, is equal, is going to be equal to r squared, is going to be equal to the length of the radius squared. Well, r squared we already know is going to be 74."}, {"video_title": "Writing standard equation of a circle Mathematics II High School Math Khan Academy.mp3", "Sentence": "Well, the distance is going to be x minus the x coordinate of the center, x minus negative one squared. Let me do that in blue color. Minus negative one squared plus y minus, y minus the y coordinate of the center, y minus one squared, is equal, is going to be equal to r squared, is going to be equal to the length of the radius squared. Well, r squared we already know is going to be 74. 74. And then if we want to simplify a little bit, you subtract a negative, this becomes a positive. So it simplifies to x plus one squared plus y minus one squared is equal to 74."}, {"video_title": "Worked example Solving equations by completing the square High School Math Khan Academy.mp3", "Sentence": "So let's see if we can solve this quadratic equation right over here. X squared minus two x minus eight is equal to zero. And actually they're cutting down some trees outside, so my apologies if you hear some chopping of trees. Well, I'll try to ignore it myself. Alright, so back to the problem at hand. And there's actually several ways that you could attack this problem. You could just try to factor the left-hand side and go that way."}, {"video_title": "Worked example Solving equations by completing the square High School Math Khan Academy.mp3", "Sentence": "Well, I'll try to ignore it myself. Alright, so back to the problem at hand. And there's actually several ways that you could attack this problem. You could just try to factor the left-hand side and go that way. But the way we're going to tackle it is by completing the square. And what does that mean? Well, that means that I want to write, I want to write the left-hand side of this equation into the form x plus a squared plus b."}, {"video_title": "Worked example Solving equations by completing the square High School Math Khan Academy.mp3", "Sentence": "You could just try to factor the left-hand side and go that way. But the way we're going to tackle it is by completing the square. And what does that mean? Well, that means that I want to write, I want to write the left-hand side of this equation into the form x plus a squared plus b. And as we'll see, if we can write the left-hand in this form, then we can actually solve it in a pretty straightforward way. So let's see if we could do that. Well, let's just remind ourselves how we need to rearrange the left-hand side in order to get to this form."}, {"video_title": "Worked example Solving equations by completing the square High School Math Khan Academy.mp3", "Sentence": "Well, that means that I want to write, I want to write the left-hand side of this equation into the form x plus a squared plus b. And as we'll see, if we can write the left-hand in this form, then we can actually solve it in a pretty straightforward way. So let's see if we could do that. Well, let's just remind ourselves how we need to rearrange the left-hand side in order to get to this form. If I were to expand out x plus a squared, let me do that in a different color. So if I were to expand out x plus a squared, that is x squared plus two a x, make sure that plus sign you can see, plus two a x plus a squared, and of course you still have that plus b there, plus b. So let's see if we can write this in that form."}, {"video_title": "Worked example Solving equations by completing the square High School Math Khan Academy.mp3", "Sentence": "Well, let's just remind ourselves how we need to rearrange the left-hand side in order to get to this form. If I were to expand out x plus a squared, let me do that in a different color. So if I were to expand out x plus a squared, that is x squared plus two a x, make sure that plus sign you can see, plus two a x plus a squared, and of course you still have that plus b there, plus b. So let's see if we can write this in that form. So what I'm going to do, and this is what you typically do when you try to complete the square, I'll write the x squared minus two x. Now I'm gonna have a little bit of a gap and I'm going to have minus eight, and I have another little bit of a gap, and I'm gonna say equals zero. So I just rewrote this equation, but I gave myself some space so I can add or subtract some things that might make it a little bit easier to get into this form."}, {"video_title": "Worked example Solving equations by completing the square High School Math Khan Academy.mp3", "Sentence": "So let's see if we can write this in that form. So what I'm going to do, and this is what you typically do when you try to complete the square, I'll write the x squared minus two x. Now I'm gonna have a little bit of a gap and I'm going to have minus eight, and I have another little bit of a gap, and I'm gonna say equals zero. So I just rewrote this equation, but I gave myself some space so I can add or subtract some things that might make it a little bit easier to get into this form. So if we just match our terms, x squared, x squared, two a x negative two x. So if this is two a x, that means that two a is negative two. Two a is equal to negative two, or a is equal to negative one."}, {"video_title": "Worked example Solving equations by completing the square High School Math Khan Academy.mp3", "Sentence": "So I just rewrote this equation, but I gave myself some space so I can add or subtract some things that might make it a little bit easier to get into this form. So if we just match our terms, x squared, x squared, two a x negative two x. So if this is two a x, that means that two a is negative two. Two a is equal to negative two, or a is equal to negative one. Another way to think about it, your a is going to be half of your first degree coefficient, so the coefficient on the x term. So the coefficient on the x term is a negative two, half of that is a negative one. And then we want to have, and then we want to have an a squared."}, {"video_title": "Worked example Solving equations by completing the square High School Math Khan Academy.mp3", "Sentence": "Two a is equal to negative two, or a is equal to negative one. Another way to think about it, your a is going to be half of your first degree coefficient, so the coefficient on the x term. So the coefficient on the x term is a negative two, half of that is a negative one. And then we want to have, and then we want to have an a squared. So if a is negative one, a squared would be plus one. So let's throw a plus one there. But like we've done said before, we can't just willy nilly add something on one side of the equation without adding it to the other, or without subtracting it again on the same side."}, {"video_title": "Worked example Solving equations by completing the square High School Math Khan Academy.mp3", "Sentence": "And then we want to have, and then we want to have an a squared. So if a is negative one, a squared would be plus one. So let's throw a plus one there. But like we've done said before, we can't just willy nilly add something on one side of the equation without adding it to the other, or without subtracting it again on the same side. Otherwise you're fundamentally changing the truth of the equation. So if I add one on that side, I either have to add one on the, if I add one on the left side, I either have to add one on the right side to make the equation still hold true, or I could add one and subtract one from the left hand side, so I'm not really changing the value of the left hand side. All I've done is added one and subtracted one from the left hand side."}, {"video_title": "Worked example Solving equations by completing the square High School Math Khan Academy.mp3", "Sentence": "But like we've done said before, we can't just willy nilly add something on one side of the equation without adding it to the other, or without subtracting it again on the same side. Otherwise you're fundamentally changing the truth of the equation. So if I add one on that side, I either have to add one on the, if I add one on the left side, I either have to add one on the right side to make the equation still hold true, or I could add one and subtract one from the left hand side, so I'm not really changing the value of the left hand side. All I've done is added one and subtracted one from the left hand side. Now why did I do this again? Well now I've been able, I haven't changed its value, I just added and subtracted the same thing, but this part of the left hand side now matches this pattern right over here. It's x squared plus two a x, where a is negative one, so it's minus two x, plus a squared, plus negative one squared, and then this, this part right over here, is the plus b."}, {"video_title": "Worked example Solving equations by completing the square High School Math Khan Academy.mp3", "Sentence": "All I've done is added one and subtracted one from the left hand side. Now why did I do this again? Well now I've been able, I haven't changed its value, I just added and subtracted the same thing, but this part of the left hand side now matches this pattern right over here. It's x squared plus two a x, where a is negative one, so it's minus two x, plus a squared, plus negative one squared, and then this, this part right over here, is the plus b. So we already know that b is equal to negative nine. Negative eight minus one is negative nine, and so that's going to be our b right over there. And so we can rewrite this as, what I squared off in green, that's going to be x plus a squared."}, {"video_title": "Worked example Solving equations by completing the square High School Math Khan Academy.mp3", "Sentence": "It's x squared plus two a x, where a is negative one, so it's minus two x, plus a squared, plus negative one squared, and then this, this part right over here, is the plus b. So we already know that b is equal to negative nine. Negative eight minus one is negative nine, and so that's going to be our b right over there. And so we can rewrite this as, what I squared off in green, that's going to be x plus a squared. So we could write it as x plus, and I could write a as negative one, actually let me, I could write it like that first. X plus a squared, or x plus negative one, well that's just x minus one. So I'm just going to write it as x minus negative one, squared, and then we have minus nine, minus nine is equal to zero."}, {"video_title": "Worked example Solving equations by completing the square High School Math Khan Academy.mp3", "Sentence": "And so we can rewrite this as, what I squared off in green, that's going to be x plus a squared. So we could write it as x plus, and I could write a as negative one, actually let me, I could write it like that first. X plus a squared, or x plus negative one, well that's just x minus one. So I'm just going to write it as x minus negative one, squared, and then we have minus nine, minus nine is equal to zero. Is equal to zero. And then I can add nine to both sides, so I just have this squared expression on the left hand side. So let's do that."}, {"video_title": "Worked example Solving equations by completing the square High School Math Khan Academy.mp3", "Sentence": "So I'm just going to write it as x minus negative one, squared, and then we have minus nine, minus nine is equal to zero. Is equal to zero. And then I can add nine to both sides, so I just have this squared expression on the left hand side. So let's do that. Let me add nine to both sides, and what I am going to be left with, so let me just, on the left hand side, those cancel out, that's why I added the nine. I'm just going to be left with the x minus one squared, it's going to be equal to, it's going to be equal to on this side, it's going to, zero plus nine is nine. So if x minus one, let me do that blue color."}, {"video_title": "Worked example Solving equations by completing the square High School Math Khan Academy.mp3", "Sentence": "So let's do that. Let me add nine to both sides, and what I am going to be left with, so let me just, on the left hand side, those cancel out, that's why I added the nine. I'm just going to be left with the x minus one squared, it's going to be equal to, it's going to be equal to on this side, it's going to, zero plus nine is nine. So if x minus one, let me do that blue color. So, it's going to be nine. And so x minus one squared, is nine. If I have something squared is equal to nine that means that that something is either going to be the positive or the negative square root of nine."}, {"video_title": "Worked example Solving equations by completing the square High School Math Khan Academy.mp3", "Sentence": "So if x minus one, let me do that blue color. So, it's going to be nine. And so x minus one squared, is nine. If I have something squared is equal to nine that means that that something is either going to be the positive or the negative square root of nine. So it's either going to be a positive or negative three. So we can say x minus one is equal to positive three, or x minus one is equal to negative three. And you could see that here, if x minus one is three, three squared is nine."}, {"video_title": "Worked example Solving equations by completing the square High School Math Khan Academy.mp3", "Sentence": "If I have something squared is equal to nine that means that that something is either going to be the positive or the negative square root of nine. So it's either going to be a positive or negative three. So we can say x minus one is equal to positive three, or x minus one is equal to negative three. And you could see that here, if x minus one is three, three squared is nine. If x minus one is negative three, negative three squared is nine. And so here, we can just add one to both sides of this equation. Add one to both sides of this equation, and you get x is equal to four, or x is, if we add one to both sides of this equation, we get, my digital ink is acting up, I don't know."}, {"video_title": "Worked example Solving equations by completing the square High School Math Khan Academy.mp3", "Sentence": "And you could see that here, if x minus one is three, three squared is nine. If x minus one is negative three, negative three squared is nine. And so here, we can just add one to both sides of this equation. Add one to both sides of this equation, and you get x is equal to four, or x is, if we add one to both sides of this equation, we get, my digital ink is acting up, I don't know. All right, then we get x is equal to negative three plus one is negative two. So x could be equal to four, or x could be equal to negative two. And we're done."}, {"video_title": "Worked example Solving equations by completing the square High School Math Khan Academy.mp3", "Sentence": "Add one to both sides of this equation, and you get x is equal to four, or x is, if we add one to both sides of this equation, we get, my digital ink is acting up, I don't know. All right, then we get x is equal to negative three plus one is negative two. So x could be equal to four, or x could be equal to negative two. And we're done. Now, some of you might be saying, well, why did we go through the trouble of completing the square? I might have been able to just factor this and then solve it that way. And you could have, actually, for this particular problem."}, {"video_title": "Worked example Solving equations by completing the square High School Math Khan Academy.mp3", "Sentence": "And we're done. Now, some of you might be saying, well, why did we go through the trouble of completing the square? I might have been able to just factor this and then solve it that way. And you could have, actually, for this particular problem. But completing the square is very powerful because you can actually always apply this. And what you will, in the future, you will learn the quadratic formula, and the quadratic formula actually comes directly out of completing the square. In fact, when you're applying the quadratic formula, you're essentially applying the result of completing the square."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So that would be 1, 2, 3, and then down 4. 1, 2, 3, 4. So that's 3, negative 4. And I also have the point 6, 1. So 1, 2, 3, 4, 5, 6, 1. So just like that. 6, 1."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "And I also have the point 6, 1. So 1, 2, 3, 4, 5, 6, 1. So just like that. 6, 1. In the last video, we figured out that we could just use the Pythagorean theorem if we wanted to figure out the distance between these two points. We just drew a triangle there and realized that this was the hypotenuse. In this video, we're going to try to figure out what is the coordinate of the point that is exactly halfway between this point and that point."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "6, 1. In the last video, we figured out that we could just use the Pythagorean theorem if we wanted to figure out the distance between these two points. We just drew a triangle there and realized that this was the hypotenuse. In this video, we're going to try to figure out what is the coordinate of the point that is exactly halfway between this point and that point. So this right here is kind of the distance, the line that connects them. Now what is the coordinate of the point that is exactly halfway in between the two? What is this coordinate right here?"}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "In this video, we're going to try to figure out what is the coordinate of the point that is exactly halfway between this point and that point. So this right here is kind of the distance, the line that connects them. Now what is the coordinate of the point that is exactly halfway in between the two? What is this coordinate right here? It's something comma something. And to do that, let me draw it really big here. Because I think you're going to find out that it's actually pretty straightforward."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "What is this coordinate right here? It's something comma something. And to do that, let me draw it really big here. Because I think you're going to find out that it's actually pretty straightforward. At first it seems like a really tough problem. Gee, let me use the distance formula with some variables. But you're going to see it's actually one of the simplest things you'll learn in algebra and geometry."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Because I think you're going to find out that it's actually pretty straightforward. At first it seems like a really tough problem. Gee, let me use the distance formula with some variables. But you're going to see it's actually one of the simplest things you'll learn in algebra and geometry. So let's say that this is my triangle right there. That is my triangle right there. This right here is the point 6, 1."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "But you're going to see it's actually one of the simplest things you'll learn in algebra and geometry. So let's say that this is my triangle right there. That is my triangle right there. This right here is the point 6, 1. This down here is the point 3 comma negative 4. And we're looking for the point that is smack dab in between those two points. What are its coordinates?"}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "This right here is the point 6, 1. This down here is the point 3 comma negative 4. And we're looking for the point that is smack dab in between those two points. What are its coordinates? It seems very hard at first. But it's easy when you think about it in terms of just the x and the y coordinates. What's this guy's x coordinate going to be?"}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "What are its coordinates? It seems very hard at first. But it's easy when you think about it in terms of just the x and the y coordinates. What's this guy's x coordinate going to be? This line out here represents x is equal to 6. This over here, let me do it in a little darker color, this over here represents x is equal to 6. This over here represents x is equal to 3."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "What's this guy's x coordinate going to be? This line out here represents x is equal to 6. This over here, let me do it in a little darker color, this over here represents x is equal to 6. This over here represents x is equal to 3. What will this guy's x coordinate be? Well, his x coordinate is going to be smack dab in between the two x coordinates. This is x is equal to 3."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "This over here represents x is equal to 3. What will this guy's x coordinate be? Well, his x coordinate is going to be smack dab in between the two x coordinates. This is x is equal to 3. This is x equal to 6. He's going to be right in between. This distance is going to be equal to that distance."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "This is x is equal to 3. This is x equal to 6. He's going to be right in between. This distance is going to be equal to that distance. His x coordinate is going to be right in between the 3 and the 6. So what do we call the number that's right in between the 3 and the 6? Well, we could even call that the midpoint."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "This distance is going to be equal to that distance. His x coordinate is going to be right in between the 3 and the 6. So what do we call the number that's right in between the 3 and the 6? Well, we could even call that the midpoint. Or we could call it the mean or the average, or however you want to talk about it. We just want to know what's the average of 3 and 6. So to figure out this point, the point halfway between 3 and 6, you literally just figure out 3 plus 6 over 2, which is equal to 4.5."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Well, we could even call that the midpoint. Or we could call it the mean or the average, or however you want to talk about it. We just want to know what's the average of 3 and 6. So to figure out this point, the point halfway between 3 and 6, you literally just figure out 3 plus 6 over 2, which is equal to 4.5. So this x coordinate is going to be 4.5. Let me draw that on this graph. 1, 2, 3, 4.5."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So to figure out this point, the point halfway between 3 and 6, you literally just figure out 3 plus 6 over 2, which is equal to 4.5. So this x coordinate is going to be 4.5. Let me draw that on this graph. 1, 2, 3, 4.5. And you see it's smack dab in between. That's its x coordinate. Now, by the exact same logic, this guy's y coordinate is going to be smack dab between y is equal to negative 4 and y is equal to 1."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "1, 2, 3, 4.5. And you see it's smack dab in between. That's its x coordinate. Now, by the exact same logic, this guy's y coordinate is going to be smack dab between y is equal to negative 4 and y is equal to 1. So it's going to be right in between those. So this is the x right there. The y coordinate is just going to be right in between y is equal to negative 4 and y is equal to 1."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Now, by the exact same logic, this guy's y coordinate is going to be smack dab between y is equal to negative 4 and y is equal to 1. So it's going to be right in between those. So this is the x right there. The y coordinate is just going to be right in between y is equal to negative 4 and y is equal to 1. So you just take the average. 1 plus negative 4 over 2, that's equal to negative 3 over 2, or you could say negative 1.5. So you go down 1.5."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "The y coordinate is just going to be right in between y is equal to negative 4 and y is equal to 1. So you just take the average. 1 plus negative 4 over 2, that's equal to negative 3 over 2, or you could say negative 1.5. So you go down 1.5. It is literally right there. So just like that, you literally take the average of the x's, take the average of the y's, or maybe I should say the mean to be a little bit more specific, a mean of only two points, and you will get the midpoint of those two points, the point that's equidistant from both of them. It's the midpoint of the line that connects them."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So you go down 1.5. It is literally right there. So just like that, you literally take the average of the x's, take the average of the y's, or maybe I should say the mean to be a little bit more specific, a mean of only two points, and you will get the midpoint of those two points, the point that's equidistant from both of them. It's the midpoint of the line that connects them. So the coordinates are 4.5 comma negative 1.5. Let's do a couple more of these. These actually, you're going to find are very, very straightforward."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "It's the midpoint of the line that connects them. So the coordinates are 4.5 comma negative 1.5. Let's do a couple more of these. These actually, you're going to find are very, very straightforward. But just to visualize it, let me graph it. Let's say I have the point 4, negative 5. So 1, 2, 3, 4, and then go down 5."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "These actually, you're going to find are very, very straightforward. But just to visualize it, let me graph it. Let's say I have the point 4, negative 5. So 1, 2, 3, 4, and then go down 5. 1, 2, 3, 4, 5. So that's 4, negative 5. And I have the point 8 comma 2."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So 1, 2, 3, 4, and then go down 5. 1, 2, 3, 4, 5. So that's 4, negative 5. And I have the point 8 comma 2. So 1, 2, 3, 4, 5, 6, 7, 8 comma 2. So what is the coordinate of the midpoint of these two points, the point that is smack dab in between them? Well, we just average the x's, average the y's."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "And I have the point 8 comma 2. So 1, 2, 3, 4, 5, 6, 7, 8 comma 2. So what is the coordinate of the midpoint of these two points, the point that is smack dab in between them? Well, we just average the x's, average the y's. So the midpoint is going to be the x values are 8 and 4. So it's going to be 8 plus 4 over 2. And the y value is going to be, well, we have a 2 and a negative 5, so you get 2 plus negative 5 over 2."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Well, we just average the x's, average the y's. So the midpoint is going to be the x values are 8 and 4. So it's going to be 8 plus 4 over 2. And the y value is going to be, well, we have a 2 and a negative 5, so you get 2 plus negative 5 over 2. And what is this equal to? This is 12 over 2, which is 6 comma 2 minus 5 is negative 3. Negative 3 over 2 is negative 1.5."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "And the y value is going to be, well, we have a 2 and a negative 5, so you get 2 plus negative 5 over 2. And what is this equal to? This is 12 over 2, which is 6 comma 2 minus 5 is negative 3. Negative 3 over 2 is negative 1.5. So that right there is the midpoint. You literally just average the x's and average the y's or find their mean. So let's graph it just to make sure it looks like the midpoint."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Negative 3 over 2 is negative 1.5. So that right there is the midpoint. You literally just average the x's and average the y's or find their mean. So let's graph it just to make sure it looks like the midpoint. 6, negative 5. 1, 2, 3, 4, 5, 6. Negative 1.5."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So let's graph it just to make sure it looks like the midpoint. 6, negative 5. 1, 2, 3, 4, 5, 6. Negative 1.5. Negative 1, negative 1.5. Yep, looks pretty good. It looks like it's equidistant from this point and that point up there."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Negative 1.5. Negative 1, negative 1.5. Yep, looks pretty good. It looks like it's equidistant from this point and that point up there. Now, that's all you have to remember. Average the x or take the mean of the x or find the x that's right in between the two. Average the y's."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "It looks like it's equidistant from this point and that point up there. Now, that's all you have to remember. Average the x or take the mean of the x or find the x that's right in between the two. Average the y's. You've got the midpoint. What I'm going to show you now is what's in many textbooks. They'll write, oh, if I have the point x1, y1, and then I have the point, actually I'm just sticking yellow."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Average the y's. You've got the midpoint. What I'm going to show you now is what's in many textbooks. They'll write, oh, if I have the point x1, y1, and then I have the point, actually I'm just sticking yellow. It's kind of painful to switch colors all the time. And then I have the point x2, y2. Many books will give you something called the midpoint formula, which once again, I think is kind of silly to memorize."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "They'll write, oh, if I have the point x1, y1, and then I have the point, actually I'm just sticking yellow. It's kind of painful to switch colors all the time. And then I have the point x2, y2. Many books will give you something called the midpoint formula, which once again, I think is kind of silly to memorize. Just remember, you just average. Find the x in between, find the y in between. So midpoint formula."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Many books will give you something called the midpoint formula, which once again, I think is kind of silly to memorize. Just remember, you just average. Find the x in between, find the y in between. So midpoint formula. What they'll really say is the midpoint. So maybe we'll say the midpoint x, or maybe I'll call it this way. I'm just making up notation."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So midpoint formula. What they'll really say is the midpoint. So maybe we'll say the midpoint x, or maybe I'll call it this way. I'm just making up notation. The x midpoint and the y midpoint is going to be equal to, and they'll give you this formula, x1 plus x2 over 2. And then y1 plus y2 over 2. And it looks like something you have to memorize."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "I'm just making up notation. The x midpoint and the y midpoint is going to be equal to, and they'll give you this formula, x1 plus x2 over 2. And then y1 plus y2 over 2. And it looks like something you have to memorize. But all you have to say is, look, that's just the average or the mean of these two numbers. This is just the average or the mean of these two numbers. I'm just saying I'm adding the two together, dividing by 2."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "And it looks like something you have to memorize. But all you have to say is, look, that's just the average or the mean of these two numbers. This is just the average or the mean of these two numbers. I'm just saying I'm adding the two together, dividing by 2. Adding these two together, dividing by 2. And then I get the midpoint. That's all the midpoint formula is."}, {"video_title": "Simplifying square roots comment response Algebra I Khan Academy.mp3", "Sentence": "So just as a quick review of what we did in the last video, we said that this is the same thing as 3 times the principal root of 500. And I'm going to do it a little bit different than I did in the last video, just to make it interesting. This is 3 times the principal root of 500 times the principal root of x to the third. And 500, we can rewrite it, because 500 is not a perfect square. We can rewrite 500 as 100 times 5, or even better, we can rewrite that as 10 squared times 5. 10 squared is the same thing as 100. So we can rewrite this first part over here as 3 times the principal root of 10 squared times 5 times the principal root of x squared times x."}, {"video_title": "Simplifying square roots comment response Algebra I Khan Academy.mp3", "Sentence": "And 500, we can rewrite it, because 500 is not a perfect square. We can rewrite 500 as 100 times 5, or even better, we can rewrite that as 10 squared times 5. 10 squared is the same thing as 100. So we can rewrite this first part over here as 3 times the principal root of 10 squared times 5 times the principal root of x squared times x. That's the same thing as x to the third. Now, the one thing I'm going to do here, actually I won't talk about it just yet, of how we're going to do it differently than we did in the last video. This radical right here can be rewritten as, so this is going to be 3 times the square root, or the principal root I should say, of 10 squared times the square root of 5."}, {"video_title": "Simplifying square roots comment response Algebra I Khan Academy.mp3", "Sentence": "So we can rewrite this first part over here as 3 times the principal root of 10 squared times 5 times the principal root of x squared times x. That's the same thing as x to the third. Now, the one thing I'm going to do here, actually I won't talk about it just yet, of how we're going to do it differently than we did in the last video. This radical right here can be rewritten as, so this is going to be 3 times the square root, or the principal root I should say, of 10 squared times the square root of 5. If we take the square root of the product of two things, it's the same thing as taking the square root of each of them and taking the product. And so this over here is going to be times the square root of, or the principal root of x squared times the principal root of x. And the principal root of 10 squared is 10."}, {"video_title": "Simplifying square roots comment response Algebra I Khan Academy.mp3", "Sentence": "This radical right here can be rewritten as, so this is going to be 3 times the square root, or the principal root I should say, of 10 squared times the square root of 5. If we take the square root of the product of two things, it's the same thing as taking the square root of each of them and taking the product. And so this over here is going to be times the square root of, or the principal root of x squared times the principal root of x. And the principal root of 10 squared is 10. And then what I said in the last video is that the principal root of x squared is going to be the absolute value of x, just in case x itself is a negative number. And so then if you simplify all of this, you get 3 times 10, which is 30, times, and I'm just going to switch the order here, times the absolute value of x. And then you have the square root of 5, or the principal root of 5, times the principal root of x."}, {"video_title": "Simplifying square roots comment response Algebra I Khan Academy.mp3", "Sentence": "And the principal root of 10 squared is 10. And then what I said in the last video is that the principal root of x squared is going to be the absolute value of x, just in case x itself is a negative number. And so then if you simplify all of this, you get 3 times 10, which is 30, times, and I'm just going to switch the order here, times the absolute value of x. And then you have the square root of 5, or the principal root of 5, times the principal root of x. And this is just going to be equal to the principal root of 5x. Taking the square root of something and multiplying that times the square root of something else is the same thing as just taking the square root of 5x. So all of this simplified down to 30 times the absolute value of x times the principal root of 5x."}, {"video_title": "Simplifying square roots comment response Algebra I Khan Academy.mp3", "Sentence": "And then you have the square root of 5, or the principal root of 5, times the principal root of x. And this is just going to be equal to the principal root of 5x. Taking the square root of something and multiplying that times the square root of something else is the same thing as just taking the square root of 5x. So all of this simplified down to 30 times the absolute value of x times the principal root of 5x. And this is what we got in the last video. And the interesting thing here is if we assume we're only dealing with real numbers, the domain of x right over here, the x's that will make this expression defined in the real numbers, then x has to be greater than or equal to 0. So let me, so maybe I could write it this way."}, {"video_title": "Simplifying square roots comment response Algebra I Khan Academy.mp3", "Sentence": "So all of this simplified down to 30 times the absolute value of x times the principal root of 5x. And this is what we got in the last video. And the interesting thing here is if we assume we're only dealing with real numbers, the domain of x right over here, the x's that will make this expression defined in the real numbers, then x has to be greater than or equal to 0. So let me, so maybe I could write it this way. The domain here is that x is, x, any real number greater than or equal to 0. And the reason why I say that is if you put a negative number in here, if you put a negative number in here and you cube it, you're going to get another negative number. And then it doesn't make, at least in the real numbers, you won't get an actual value."}, {"video_title": "Simplifying square roots comment response Algebra I Khan Academy.mp3", "Sentence": "So let me, so maybe I could write it this way. The domain here is that x is, x, any real number greater than or equal to 0. And the reason why I say that is if you put a negative number in here, if you put a negative number in here and you cube it, you're going to get another negative number. And then it doesn't make, at least in the real numbers, you won't get an actual value. You'll get a square root of a negative number here. So if you make this, if you assume this right here, we're dealing with the real numbers. We're not dealing with any complex numbers."}, {"video_title": "Simplifying square roots comment response Algebra I Khan Academy.mp3", "Sentence": "And then it doesn't make, at least in the real numbers, you won't get an actual value. You'll get a square root of a negative number here. So if you make this, if you assume this right here, we're dealing with the real numbers. We're not dealing with any complex numbers. When you open up the complex numbers, then you can expand the domain more broadly. But if you're dealing with real numbers, you can say that x is going to be greater than or equal to 0. And then the absolute value of x is just going to be x because it's not going to be a negative number."}, {"video_title": "Simplifying square roots comment response Algebra I Khan Academy.mp3", "Sentence": "We're not dealing with any complex numbers. When you open up the complex numbers, then you can expand the domain more broadly. But if you're dealing with real numbers, you can say that x is going to be greater than or equal to 0. And then the absolute value of x is just going to be x because it's not going to be a negative number. And so if we're assuming that the domain of x is, or if this expression is going to be a valuable, or it's going to have a positive number, then this can be written as 30x times the square root of 5x. If you had the situation where we were dealing with complex numbers, then you would. So numbers that were, and if you don't know what a complex number is or an imaginary number, don't worry too much about it."}, {"video_title": "Example 2 Factoring a difference of squares with leading coefficient other than 1 Khan Academy.mp3", "Sentence": "45x squared minus 125. So whenever I see something like this, I have a second degree term here, I have a subtraction sign, my temptation is to look at this as a difference of squares. We've already seen this multiple times. We've already seen that if we have something of the form a squared minus b squared, that this can be factored as a plus b times a minus b. So let's look over here. Well, over here it's not obvious that this right over here is a perfect square. Neither is it obvious that this right over here is a perfect square."}, {"video_title": "Example 2 Factoring a difference of squares with leading coefficient other than 1 Khan Academy.mp3", "Sentence": "We've already seen that if we have something of the form a squared minus b squared, that this can be factored as a plus b times a minus b. So let's look over here. Well, over here it's not obvious that this right over here is a perfect square. Neither is it obvious that this right over here is a perfect square. So it's not clear to me that this is a difference of squares. But what is interesting is that both 45 and 125 have some factors in common. And the one that jumps out at me is 5."}, {"video_title": "Example 2 Factoring a difference of squares with leading coefficient other than 1 Khan Academy.mp3", "Sentence": "Neither is it obvious that this right over here is a perfect square. So it's not clear to me that this is a difference of squares. But what is interesting is that both 45 and 125 have some factors in common. And the one that jumps out at me is 5. So let's see if we can factor out a 5. And by doing that, whether we can get something that's a little bit closer to this pattern right over here. So if we factor out a 5, this becomes 5 times 45x squared divided by 5 is going to be 9x squared."}, {"video_title": "Example 2 Factoring a difference of squares with leading coefficient other than 1 Khan Academy.mp3", "Sentence": "And the one that jumps out at me is 5. So let's see if we can factor out a 5. And by doing that, whether we can get something that's a little bit closer to this pattern right over here. So if we factor out a 5, this becomes 5 times 45x squared divided by 5 is going to be 9x squared. And then 125 divided by 5 is 25. Now this is interesting. 9x squared, that's a perfect square."}, {"video_title": "Example 2 Factoring a difference of squares with leading coefficient other than 1 Khan Academy.mp3", "Sentence": "So if we factor out a 5, this becomes 5 times 45x squared divided by 5 is going to be 9x squared. And then 125 divided by 5 is 25. Now this is interesting. 9x squared, that's a perfect square. If we call this a squared, then that tells us that a would be equal to 3x. 3x, the whole thing squared is 9x squared. Similarly, I can never say similarly correctly, 25 is clearly just 5 squared."}, {"video_title": "Example 2 Factoring a difference of squares with leading coefficient other than 1 Khan Academy.mp3", "Sentence": "9x squared, that's a perfect square. If we call this a squared, then that tells us that a would be equal to 3x. 3x, the whole thing squared is 9x squared. Similarly, I can never say similarly correctly, 25 is clearly just 5 squared. So in this case, if we're looking at this template, b would be equal to 5. So now this is a difference of squares. And we can factor it completely."}, {"video_title": "Example 2 Factoring a difference of squares with leading coefficient other than 1 Khan Academy.mp3", "Sentence": "Similarly, I can never say similarly correctly, 25 is clearly just 5 squared. So in this case, if we're looking at this template, b would be equal to 5. So now this is a difference of squares. And we can factor it completely. So we can't forget our 5 out front that we factored out. So it's going to be 5 times a plus b. So let me write this."}, {"video_title": "Example 2 Factoring a difference of squares with leading coefficient other than 1 Khan Academy.mp3", "Sentence": "And we can factor it completely. So we can't forget our 5 out front that we factored out. So it's going to be 5 times a plus b. So let me write this. So it's going to be 5 times a plus b times a minus b. So it's 5 times a plus b times a minus b. So let me write the b's down."}, {"video_title": "Example 2 Factoring a difference of squares with leading coefficient other than 1 Khan Academy.mp3", "Sentence": "So let me write this. So it's going to be 5 times a plus b times a minus b. So it's 5 times a plus b times a minus b. So let me write the b's down. Plus b and minus b. And we're done. 5 times 3x plus 5 times 3x minus 5 is 45x squared minus 125 factored out."}, {"video_title": "Horizontal & vertical lines Mathematics I High School Math Khan Academy.mp3", "Sentence": "Once you get the hang of it, you might not have to draw a graph, but for explanatory purposes, it might be useful. So negative four comma six, that's gonna be in the second quadrant. So if this is my x-axis, that is my y-axis. Let's see, I'm gonna go negative four in the x direction. So one, two, three, four, negative four, and then one, two, three, four, five, six in the y direction. So the point that we care about is going to be right over there, negative four comma six. And they're saying, what is the equation of the horizontal line?"}, {"video_title": "Horizontal & vertical lines Mathematics I High School Math Khan Academy.mp3", "Sentence": "Let's see, I'm gonna go negative four in the x direction. So one, two, three, four, negative four, and then one, two, three, four, five, six in the y direction. So the point that we care about is going to be right over there, negative four comma six. And they're saying, what is the equation of the horizontal line? It is a horizontal line. So it's just gonna go straight left, right like this. That is what the line would actually look like."}, {"video_title": "Horizontal & vertical lines Mathematics I High School Math Khan Academy.mp3", "Sentence": "And they're saying, what is the equation of the horizontal line? It is a horizontal line. So it's just gonna go straight left, right like this. That is what the line would actually look like. So what is that equation? Well, for any x, y is going to be equal to six. This is the equation y is equal to six."}, {"video_title": "Horizontal & vertical lines Mathematics I High School Math Khan Academy.mp3", "Sentence": "That is what the line would actually look like. So what is that equation? Well, for any x, y is going to be equal to six. This is the equation y is equal to six. Doesn't matter what x you input here, you're gonna get y equals six. It just stays constant right over there. So the equation is y is equal to six."}, {"video_title": "Horizontal & vertical lines Mathematics I High School Math Khan Academy.mp3", "Sentence": "This is the equation y is equal to six. Doesn't matter what x you input here, you're gonna get y equals six. It just stays constant right over there. So the equation is y is equal to six. Let's do another one of these. So here we are asked, what is the slope of the line y is equal to negative four? So let's visualize it, and then in the future you might not have to draw it like this."}, {"video_title": "Horizontal & vertical lines Mathematics I High School Math Khan Academy.mp3", "Sentence": "So the equation is y is equal to six. Let's do another one of these. So here we are asked, what is the slope of the line y is equal to negative four? So let's visualize it, and then in the future you might not have to draw it like this. But let's just draw our axes again. X axis, y axis, and the slope of the line y equals negative four, so for whatever x you have, y is gonna be negative four. Let's say that's negative four right over there."}, {"video_title": "Horizontal & vertical lines Mathematics I High School Math Khan Academy.mp3", "Sentence": "So let's visualize it, and then in the future you might not have to draw it like this. But let's just draw our axes again. X axis, y axis, and the slope of the line y equals negative four, so for whatever x you have, y is gonna be negative four. Let's say that's negative four right over there. And so the line is y, the line is y equals negative four. So I can draw it like this. So what's the slope of that?"}, {"video_title": "Horizontal & vertical lines Mathematics I High School Math Khan Academy.mp3", "Sentence": "Let's say that's negative four right over there. And so the line is y, the line is y equals negative four. So I can draw it like this. So what's the slope of that? Well, slope is change in y for a given change in x. And here no matter what I change my x, y doesn't change. It stays at negative four."}, {"video_title": "Horizontal & vertical lines Mathematics I High School Math Khan Academy.mp3", "Sentence": "So what's the slope of that? Well, slope is change in y for a given change in x. And here no matter what I change my x, y doesn't change. It stays at negative four. My change in y over change in x, doesn't matter what my change in x is. My change in y is always going to be zero, it's constant. So the slope here is going to be equal to zero."}, {"video_title": "Horizontal & vertical lines Mathematics I High School Math Khan Academy.mp3", "Sentence": "It stays at negative four. My change in y over change in x, doesn't matter what my change in x is. My change in y is always going to be zero, it's constant. So the slope here is going to be equal to zero. Y doesn't change no matter how much you change x. Let's do another one of these, this is fun. Alright, so now they are asking us, what is the slope of the line x equals negative three?"}, {"video_title": "Horizontal & vertical lines Mathematics I High School Math Khan Academy.mp3", "Sentence": "So the slope here is going to be equal to zero. Y doesn't change no matter how much you change x. Let's do another one of these, this is fun. Alright, so now they are asking us, what is the slope of the line x equals negative three? Well let me graph that one. So, let's again draw my axes real fast. X axis, y axis, x is equal to negative three."}, {"video_title": "Horizontal & vertical lines Mathematics I High School Math Khan Academy.mp3", "Sentence": "Alright, so now they are asking us, what is the slope of the line x equals negative three? Well let me graph that one. So, let's again draw my axes real fast. X axis, y axis, x is equal to negative three. So negative one, negative two, negative three. And so this line is gonna look, let me, is gonna look like this no matter what y, or you could say no matter what y is, x is gonna be equal to negative three. So it would look like this, x is equal to negative three."}, {"video_title": "Horizontal & vertical lines Mathematics I High School Math Khan Academy.mp3", "Sentence": "X axis, y axis, x is equal to negative three. So negative one, negative two, negative three. And so this line is gonna look, let me, is gonna look like this no matter what y, or you could say no matter what y is, x is gonna be equal to negative three. So it would look like this, x is equal to negative three. So what's the slope here? Well, it's undefined. A vertical line has an undefined slope."}, {"video_title": "Horizontal & vertical lines Mathematics I High School Math Khan Academy.mp3", "Sentence": "So it would look like this, x is equal to negative three. So what's the slope here? Well, it's undefined. A vertical line has an undefined slope. Remember, you wanna do, what's your change in y for a change in x? Change in y for a change in x. Well, you could think about what's the slope as you approach this, but once again, that could be, some people would say maybe it's infinite, maybe it's negative infinity, but that's why it's undefined."}, {"video_title": "Horizontal & vertical lines Mathematics I High School Math Khan Academy.mp3", "Sentence": "A vertical line has an undefined slope. Remember, you wanna do, what's your change in y for a change in x? Change in y for a change in x. Well, you could think about what's the slope as you approach this, but once again, that could be, some people would say maybe it's infinite, maybe it's negative infinity, but that's why it's undefined. A vertical line is going to have an undefined slope. So go with undefined. Let's do one more."}, {"video_title": "Horizontal & vertical lines Mathematics I High School Math Khan Academy.mp3", "Sentence": "Well, you could think about what's the slope as you approach this, but once again, that could be, some people would say maybe it's infinite, maybe it's negative infinity, but that's why it's undefined. A vertical line is going to have an undefined slope. So go with undefined. Let's do one more. What is the equation of the vertical line through negative five comma negative two? So let me do this one without even drawing it, and then I'll draw it right after that. So if we're talking about a vertical, if we're talking about a vertical line, that means that x doesn't change."}, {"video_title": "Horizontal & vertical lines Mathematics I High School Math Khan Academy.mp3", "Sentence": "Let's do one more. What is the equation of the vertical line through negative five comma negative two? So let me do this one without even drawing it, and then I'll draw it right after that. So if we're talking about a vertical, if we're talking about a vertical line, that means that x doesn't change. X doesn't change. If we were talking about a horizontal line, then we'd say y doesn't change. So if x doesn't change, that means that x is just going to be equal to some constant value."}, {"video_title": "Horizontal & vertical lines Mathematics I High School Math Khan Academy.mp3", "Sentence": "So if we're talking about a vertical, if we're talking about a vertical line, that means that x doesn't change. X doesn't change. If we were talking about a horizontal line, then we'd say y doesn't change. So if x doesn't change, that means that x is just going to be equal to some constant value. Well, if it contains the point negative five comma negative two, so if it has the point where x is equal to negative five, and if x never changes, it's a vertical line, well, that means this equation has to be x is equal to negative five. And we can draw that out if it helps. So let me draw that out."}, {"video_title": "Horizontal & vertical lines Mathematics I High School Math Khan Academy.mp3", "Sentence": "So if x doesn't change, that means that x is just going to be equal to some constant value. Well, if it contains the point negative five comma negative two, so if it has the point where x is equal to negative five, and if x never changes, it's a vertical line, well, that means this equation has to be x is equal to negative five. And we can draw that out if it helps. So let me draw that out. So, whoops, let me make sure that's a straight line. Okay, so we have x and we have y. And so we have the point negative five comma negative two."}, {"video_title": "Horizontal & vertical lines Mathematics I High School Math Khan Academy.mp3", "Sentence": "So let me draw that out. So, whoops, let me make sure that's a straight line. Okay, so we have x and we have y. And so we have the point negative five comma negative two. So negative one, two, three, four, five. Negative one, two. So we wanna go have a vertical line that goes through that point."}, {"video_title": "Horizontal & vertical lines Mathematics I High School Math Khan Academy.mp3", "Sentence": "And so we have the point negative five comma negative two. So negative one, two, three, four, five. Negative one, two. So we wanna go have a vertical line that goes through that point. So a vertical line, well, that just goes straight up and down. So it's just going to look like this. And so notice, x never changes."}, {"video_title": "Horizontal & vertical lines Mathematics I High School Math Khan Academy.mp3", "Sentence": "So we wanna go have a vertical line that goes through that point. So a vertical line, well, that just goes straight up and down. So it's just going to look like this. And so notice, x never changes. No matter what y is, x is equal to negative five. This has an undefined slope. It is a vertical line."}, {"video_title": "Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3", "Sentence": "So just as a reminder, slope is defined as rise over run, or you could view that rise as just change in y and run as just change in x. So you can see the triangles here, that's the delta symbol, it literally means change in. Or another way, and you might see this formula and it tends to be really complicated, but just remember it's just these two things over here. Sometimes slope will be specified with the variable m, and they'll say that m is the same thing, and this is really the same thing as change in y. They'll write y2 minus y1 over x2 minus x1. So the notation tends to be kind of complicated, but all this means is you take the y value of your end point and subtract from it the y value of your starting point. That'll essentially give you your change in y."}, {"video_title": "Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3", "Sentence": "Sometimes slope will be specified with the variable m, and they'll say that m is the same thing, and this is really the same thing as change in y. They'll write y2 minus y1 over x2 minus x1. So the notation tends to be kind of complicated, but all this means is you take the y value of your end point and subtract from it the y value of your starting point. That'll essentially give you your change in y. And it says take the x value of your end point and subtract from that the x value of your starting point, and that'll give you change in x. So whatever of these work for you, let's actually figure out the slope of the line that goes through these two points. So we're starting at, and actually we could do it both ways."}, {"video_title": "Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3", "Sentence": "That'll essentially give you your change in y. And it says take the x value of your end point and subtract from that the x value of your starting point, and that'll give you change in x. So whatever of these work for you, let's actually figure out the slope of the line that goes through these two points. So we're starting at, and actually we could do it both ways. We could start at this point and go to that point and calculate the slope, or we could start at this point and go to that point and calculate the slope. So let's do it both ways. So let's say that our starting point is the point 4, 2, and let's say that our end point is negative 3, 16."}, {"video_title": "Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3", "Sentence": "So we're starting at, and actually we could do it both ways. We could start at this point and go to that point and calculate the slope, or we could start at this point and go to that point and calculate the slope. So let's do it both ways. So let's say that our starting point is the point 4, 2, and let's say that our end point is negative 3, 16. So what is the change in x over here? What is the change in x in this scenario? So we're going from 4 to negative 3."}, {"video_title": "Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3", "Sentence": "So let's say that our starting point is the point 4, 2, and let's say that our end point is negative 3, 16. So what is the change in x over here? What is the change in x in this scenario? So we're going from 4 to negative 3. If something goes from 4 to negative 3, what was its change? Well, you have to go down 4 to get to 0, and then you have to go down another 3 to get to negative 3. So our change in x here is negative 7."}, {"video_title": "Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3", "Sentence": "So we're going from 4 to negative 3. If something goes from 4 to negative 3, what was its change? Well, you have to go down 4 to get to 0, and then you have to go down another 3 to get to negative 3. So our change in x here is negative 7. Actually, let me write it this way. Our change in x is equal to negative 3 minus 4, which is equal to negative 7. If I'm going from 4 to negative 3, I went down by 7."}, {"video_title": "Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3", "Sentence": "So our change in x here is negative 7. Actually, let me write it this way. Our change in x is equal to negative 3 minus 4, which is equal to negative 7. If I'm going from 4 to negative 3, I went down by 7. Our change in x is negative 7. Let's do the same thing for the change in y. Notice, I implicitly used this formula over here."}, {"video_title": "Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3", "Sentence": "If I'm going from 4 to negative 3, I went down by 7. Our change in x is negative 7. Let's do the same thing for the change in y. Notice, I implicitly used this formula over here. Our change in x was this value, our end point, our end x value, minus our starting x value. Let's do the same thing for our change in y. Our change in y, if we're starting at 2 and we go to 16, that means we moved up 14."}, {"video_title": "Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3", "Sentence": "Notice, I implicitly used this formula over here. Our change in x was this value, our end point, our end x value, minus our starting x value. Let's do the same thing for our change in y. Our change in y, if we're starting at 2 and we go to 16, that means we moved up 14. Or another way you could say it, you could take your ending y value and subtract from that your starting y value, and you get 14. So what is the slope over here? The slope is just change in y over change in x."}, {"video_title": "Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3", "Sentence": "Our change in y, if we're starting at 2 and we go to 16, that means we moved up 14. Or another way you could say it, you could take your ending y value and subtract from that your starting y value, and you get 14. So what is the slope over here? The slope is just change in y over change in x. The slope over here is change in y over change in x, which is our change in y is 14, and our change in x is negative 7. If we want to simplify this, 14 divided by negative 7 is negative 2. What I want to show you is that we could have done it the other way around."}, {"video_title": "Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3", "Sentence": "The slope is just change in y over change in x. The slope over here is change in y over change in x, which is our change in y is 14, and our change in x is negative 7. If we want to simplify this, 14 divided by negative 7 is negative 2. What I want to show you is that we could have done it the other way around. We could have made this the starting point and this the end point. What we would have gotten is the negative values of each of these, but then they would have cancelled out and we would still get negative 2. Let's try it out."}, {"video_title": "Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3", "Sentence": "What I want to show you is that we could have done it the other way around. We could have made this the starting point and this the end point. What we would have gotten is the negative values of each of these, but then they would have cancelled out and we would still get negative 2. Let's try it out. Let's say that our start point was negative 3 comma 16, and let's say that our end point is the 4 comma 2. In this situation, what is our change in x? If I start at negative 3 and I go to 4, that means I went up 7."}, {"video_title": "Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3", "Sentence": "Let's try it out. Let's say that our start point was negative 3 comma 16, and let's say that our end point is the 4 comma 2. In this situation, what is our change in x? If I start at negative 3 and I go to 4, that means I went up 7. Or if you want to just calculate that, you would do 4 minus negative 3. Needless to say, we just went up 7. What is our change in y?"}, {"video_title": "Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3", "Sentence": "If I start at negative 3 and I go to 4, that means I went up 7. Or if you want to just calculate that, you would do 4 minus negative 3. Needless to say, we just went up 7. What is our change in y? Our change in y over here, or we could say our rise. If we start at 16 and we end at 2, that means we went down 14. Or you could just say 2 minus 16 is negative 14."}, {"video_title": "Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3", "Sentence": "What is our change in y? Our change in y over here, or we could say our rise. If we start at 16 and we end at 2, that means we went down 14. Or you could just say 2 minus 16 is negative 14. We went down by 14. This was our run. If you say rise over run, which is the same thing as change in y over change in x, our rise is negative 14 and our run here is 7."}, {"video_title": "Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3", "Sentence": "Or you could just say 2 minus 16 is negative 14. We went down by 14. This was our run. If you say rise over run, which is the same thing as change in y over change in x, our rise is negative 14 and our run here is 7. Notice, these are just the negatives of these values from when we swapped them. Once again, this is equal to negative 2. Let's just visualize this."}, {"video_title": "Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3", "Sentence": "If you say rise over run, which is the same thing as change in y over change in x, our rise is negative 14 and our run here is 7. Notice, these are just the negatives of these values from when we swapped them. Once again, this is equal to negative 2. Let's just visualize this. Let me do a quick graph here just to show you what a downward slope would look like. Let me draw our two points. This is my x-axis."}, {"video_title": "Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3", "Sentence": "Let's just visualize this. Let me do a quick graph here just to show you what a downward slope would look like. Let me draw our two points. This is my x-axis. That is my y-axis. This point over here, 4, 2. Let me graph it."}, {"video_title": "Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3", "Sentence": "This is my x-axis. That is my y-axis. This point over here, 4, 2. Let me graph it. We are going to go all the way up to 16. Let me save some space here. We have 1, 2, 3, 4."}, {"video_title": "Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3", "Sentence": "Let me graph it. We are going to go all the way up to 16. Let me save some space here. We have 1, 2, 3, 4. It is 4, 1, 2. 4, 2 is right over here. Then we have the point negative 3, 16."}, {"video_title": "Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3", "Sentence": "We have 1, 2, 3, 4. It is 4, 1, 2. 4, 2 is right over here. Then we have the point negative 3, 16. Let me draw that over here. We have negative 1, 2, 3. We have to go up to 16."}, {"video_title": "Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3", "Sentence": "Then we have the point negative 3, 16. Let me draw that over here. We have negative 1, 2, 3. We have to go up to 16. This is 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16. It goes right over here. This is negative 3, 16."}, {"video_title": "Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3", "Sentence": "We have to go up to 16. This is 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16. It goes right over here. This is negative 3, 16. The line that goes between them is going to look something like this. I will try my best to draw a relatively straight line. That line will keep going."}, {"video_title": "Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3", "Sentence": "This is negative 3, 16. The line that goes between them is going to look something like this. I will try my best to draw a relatively straight line. That line will keep going. The line will keep going. That is my best attempt. Notice it is downward sloping."}, {"video_title": "Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3", "Sentence": "That line will keep going. The line will keep going. That is my best attempt. Notice it is downward sloping. As you increase an x value, the line goes down. It is going from the top left to the bottom right. As x gets bigger, y gets smaller."}, {"video_title": "Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3", "Sentence": "Notice it is downward sloping. As you increase an x value, the line goes down. It is going from the top left to the bottom right. As x gets bigger, y gets smaller. That is what a downward sloping line looks like. Just to visualize our change in x's and our change in y's that we dealt with here. When we started at 4, 2 and ended at negative 3, 16, that was analogous to starting here and ending over there."}, {"video_title": "Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3", "Sentence": "As x gets bigger, y gets smaller. That is what a downward sloping line looks like. Just to visualize our change in x's and our change in y's that we dealt with here. When we started at 4, 2 and ended at negative 3, 16, that was analogous to starting here and ending over there. We said our change in x was negative 7. We had to move back. Our run, we had to move in the left direction by 7."}, {"video_title": "Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3", "Sentence": "When we started at 4, 2 and ended at negative 3, 16, that was analogous to starting here and ending over there. We said our change in x was negative 7. We had to move back. Our run, we had to move in the left direction by 7. That is why it was negative 7. Then we had to move in the y direction positive 14. That is why our rise was positive."}, {"video_title": "Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3", "Sentence": "Our run, we had to move in the left direction by 7. That is why it was negative 7. Then we had to move in the y direction positive 14. That is why our rise was positive. It was 14 over negative 7 or negative 2. When we did it the other way, we started at this point and ended at this point. We started at negative 3, 16 and ended at that point."}, {"video_title": "Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3", "Sentence": "That is why our rise was positive. It was 14 over negative 7 or negative 2. When we did it the other way, we started at this point and ended at this point. We started at negative 3, 16 and ended at that point. In that situation, our run was positive 7. Now we had to go down in the y direction since we switched the starting and the end point. Now we had to go down negative 14."}, {"video_title": "Evaluating fractional exponents fractional base Algebra I Khan Academy.mp3", "Sentence": "Fractional exponents can be a little daunting at first, so it never hurts to do as many examples as possible. So let's do a few. What if we had 25 over 9, and we wanted to raise it to the 1 half power? So we're essentially just saying, well, what is the principal square root of 25 over 9? So what number times itself is going to be 25 over 9? Well, we know 5 times 5 is 25, and 3 times 3 is 9. So why don't we just go with 5 over 3?"}, {"video_title": "Evaluating fractional exponents fractional base Algebra I Khan Academy.mp3", "Sentence": "So we're essentially just saying, well, what is the principal square root of 25 over 9? So what number times itself is going to be 25 over 9? Well, we know 5 times 5 is 25, and 3 times 3 is 9. So why don't we just go with 5 over 3? Because notice, if you have 5 over 3 times 5 over 3, that is going to be 25 over 9. Or another way of saying this, that 5 over 3 squared is equal to 25 over 9. So 25 over 9 to the 1 half is going to be equal to 5 thirds."}, {"video_title": "Evaluating fractional exponents fractional base Algebra I Khan Academy.mp3", "Sentence": "So why don't we just go with 5 over 3? Because notice, if you have 5 over 3 times 5 over 3, that is going to be 25 over 9. Or another way of saying this, that 5 over 3 squared is equal to 25 over 9. So 25 over 9 to the 1 half is going to be equal to 5 thirds. Now let's escalate things a little bit. Let's take a really hairy one. Let's raise 81 over 256 to the negative 1 fourth power."}, {"video_title": "Evaluating fractional exponents fractional base Algebra I Khan Academy.mp3", "Sentence": "So 25 over 9 to the 1 half is going to be equal to 5 thirds. Now let's escalate things a little bit. Let's take a really hairy one. Let's raise 81 over 256 to the negative 1 fourth power. I encourage you to pause this and try this on your own. So what's going on here? This negative, the first thing I always like to do is I want to get rid of this negative in the exponent."}, {"video_title": "Evaluating fractional exponents fractional base Algebra I Khan Academy.mp3", "Sentence": "Let's raise 81 over 256 to the negative 1 fourth power. I encourage you to pause this and try this on your own. So what's going on here? This negative, the first thing I always like to do is I want to get rid of this negative in the exponent. So let me just take the reciprocal of this and raise it to the positive. So I could just say that this is equal to 256 over 81 to the 1 fourth power. And so now I can say, well, what number times itself is going to be equal to 256?"}, {"video_title": "Evaluating fractional exponents fractional base Algebra I Khan Academy.mp3", "Sentence": "This negative, the first thing I always like to do is I want to get rid of this negative in the exponent. So let me just take the reciprocal of this and raise it to the positive. So I could just say that this is equal to 256 over 81 to the 1 fourth power. And so now I can say, well, what number times itself is going to be equal to 256? And what number times itself times itself times itself? Did I say that four times? Well, what number, if I take four of them and multiply, do I get 81?"}, {"video_title": "Evaluating fractional exponents fractional base Algebra I Khan Academy.mp3", "Sentence": "And so now I can say, well, what number times itself is going to be equal to 256? And what number times itself times itself times itself? Did I say that four times? Well, what number, if I take four of them and multiply, do I get 81? And one way to think about it, this is going to be the same thing. And we'll talk about this in more depth later on when we talk about exponent properties. But this is going to be the exact same thing as 256 to the 1 fourth over 81 to the 1 fourth."}, {"video_title": "Evaluating fractional exponents fractional base Algebra I Khan Academy.mp3", "Sentence": "Well, what number, if I take four of them and multiply, do I get 81? And one way to think about it, this is going to be the same thing. And we'll talk about this in more depth later on when we talk about exponent properties. But this is going to be the exact same thing as 256 to the 1 fourth over 81 to the 1 fourth. You, in fact, saw it over here. This over here was the same thing as the square root of 25 over the square root of 9, or 25 to the 1 half over 9 to the 1 half. So we're just doing that over here."}, {"video_title": "Evaluating fractional exponents fractional base Algebra I Khan Academy.mp3", "Sentence": "But this is going to be the exact same thing as 256 to the 1 fourth over 81 to the 1 fourth. You, in fact, saw it over here. This over here was the same thing as the square root of 25 over the square root of 9, or 25 to the 1 half over 9 to the 1 half. So we're just doing that over here. So we still have to think about what number this is. And this is a little bit of, there's no easy way to do this. You kind of have to just play around a little bit to come up with it."}, {"video_title": "Evaluating fractional exponents fractional base Algebra I Khan Academy.mp3", "Sentence": "So we're just doing that over here. So we still have to think about what number this is. And this is a little bit of, there's no easy way to do this. You kind of have to just play around a little bit to come up with it. But 4 might jump out at you if you recognize that 16 times 16 is 256. We know that 4 to the fourth power, or you're about to know this, is 4 times 4 times 4 times 4. And 4 times 4 is 16, times 4 is 64, times 4 is equal to 256."}, {"video_title": "Evaluating fractional exponents fractional base Algebra I Khan Academy.mp3", "Sentence": "You kind of have to just play around a little bit to come up with it. But 4 might jump out at you if you recognize that 16 times 16 is 256. We know that 4 to the fourth power, or you're about to know this, is 4 times 4 times 4 times 4. And 4 times 4 is 16, times 4 is 64, times 4 is equal to 256. So 4 to the fourth is 256. Or we could say 4 is equal to 256 to the 1 fourth power. Fair enough."}, {"video_title": "Evaluating fractional exponents fractional base Algebra I Khan Academy.mp3", "Sentence": "And 4 times 4 is 16, times 4 is 64, times 4 is equal to 256. So 4 to the fourth is 256. Or we could say 4 is equal to 256 to the 1 fourth power. Fair enough. Now what about 81? Well, 3 might jump out at you. We know that 3 to the fourth power is equal to 3 times 3 times 3 times 3, which is equal to 81."}, {"video_title": "Evaluating fractional exponents fractional base Algebra I Khan Academy.mp3", "Sentence": "Fair enough. Now what about 81? Well, 3 might jump out at you. We know that 3 to the fourth power is equal to 3 times 3 times 3 times 3, which is equal to 81. So 3 is equal to 81 to the 1 fourth. So this top number, 256 to the 1 fourth, that's just 4. 81 to the 1 fourth, that is just 3."}, {"video_title": "Radius and center for a circle equation in standard form Algebra II Khan Academy.mp3", "Sentence": "What are its center hk and its radius r? So let's just remind ourselves what a circle is. You have some point. Let's call that hk. The circle is a set of all points that are equidistant from that point. So let's take the set of all points that are, say, r away from hk. So let's say that this distance right over here, this distance right over here is r. And so we want all the set of points that are exactly r away."}, {"video_title": "Radius and center for a circle equation in standard form Algebra II Khan Academy.mp3", "Sentence": "Let's call that hk. The circle is a set of all points that are equidistant from that point. So let's take the set of all points that are, say, r away from hk. So let's say that this distance right over here, this distance right over here is r. And so we want all the set of points that are exactly r away. So all the points x comma y that are exactly r away. And so you could imagine you could rotate around and all of these points are going to be exactly r away. And I'm going to try my best to draw at least a somewhat perfect looking circle."}, {"video_title": "Radius and center for a circle equation in standard form Algebra II Khan Academy.mp3", "Sentence": "So let's say that this distance right over here, this distance right over here is r. And so we want all the set of points that are exactly r away. So all the points x comma y that are exactly r away. And so you could imagine you could rotate around and all of these points are going to be exactly r away. And I'm going to try my best to draw at least a somewhat perfect looking circle. I won't be able to do a perfect job of it. But you get a sense. All of these are exactly r away, at least if I were to draw it properly."}, {"video_title": "Radius and center for a circle equation in standard form Algebra II Khan Academy.mp3", "Sentence": "And I'm going to try my best to draw at least a somewhat perfect looking circle. I won't be able to do a perfect job of it. But you get a sense. All of these are exactly r away, at least if I were to draw it properly. They are r away. So how do we find an equation in terms of r and hk and x and y that describes all these points? Well, we know how to find the distance between two points on a coordinate plane."}, {"video_title": "Radius and center for a circle equation in standard form Algebra II Khan Academy.mp3", "Sentence": "All of these are exactly r away, at least if I were to draw it properly. They are r away. So how do we find an equation in terms of r and hk and x and y that describes all these points? Well, we know how to find the distance between two points on a coordinate plane. And in fact, it comes straight out of the Pythagorean theorem. If we were to draw a vertical line right over here that essentially is the change in the vertical axis between these two points, up here we're at y, here we're k. So this distance is going to be y minus k. We can do the exact same thing on the horizontal axis. This x-coordinate is x, while this x-coordinate is h. So this is going to be x minus h is this distance."}, {"video_title": "Radius and center for a circle equation in standard form Algebra II Khan Academy.mp3", "Sentence": "Well, we know how to find the distance between two points on a coordinate plane. And in fact, it comes straight out of the Pythagorean theorem. If we were to draw a vertical line right over here that essentially is the change in the vertical axis between these two points, up here we're at y, here we're k. So this distance is going to be y minus k. We can do the exact same thing on the horizontal axis. This x-coordinate is x, while this x-coordinate is h. So this is going to be x minus h is this distance. And this is a right triangle because by definition, we're saying, hey, we're measuring vertical distance here. We're measuring horizontal distance here. So these two things are perpendicular."}, {"video_title": "Radius and center for a circle equation in standard form Algebra II Khan Academy.mp3", "Sentence": "This x-coordinate is x, while this x-coordinate is h. So this is going to be x minus h is this distance. And this is a right triangle because by definition, we're saying, hey, we're measuring vertical distance here. We're measuring horizontal distance here. So these two things are perpendicular. And so from the Pythagorean theorem, we know that this squared plus this squared must be equal to our distance squared. And this is where the distance formula comes from. So we know that x minus h squared plus y minus k squared must be equal to r squared."}, {"video_title": "Radius and center for a circle equation in standard form Algebra II Khan Academy.mp3", "Sentence": "So these two things are perpendicular. And so from the Pythagorean theorem, we know that this squared plus this squared must be equal to our distance squared. And this is where the distance formula comes from. So we know that x minus h squared plus y minus k squared must be equal to r squared. This is the equation for the set. This describes any x and y that satisfies this equation will sit on this circle. Now with that out of the way, let's go answer their question."}, {"video_title": "Radius and center for a circle equation in standard form Algebra II Khan Academy.mp3", "Sentence": "So we know that x minus h squared plus y minus k squared must be equal to r squared. This is the equation for the set. This describes any x and y that satisfies this equation will sit on this circle. Now with that out of the way, let's go answer their question. The equation of the circle is this thing. And this looks awfully close to what we just wrote. We just have to make sure that we don't get confused with the negatives."}, {"video_title": "Radius and center for a circle equation in standard form Algebra II Khan Academy.mp3", "Sentence": "Now with that out of the way, let's go answer their question. The equation of the circle is this thing. And this looks awfully close to what we just wrote. We just have to make sure that we don't get confused with the negatives. Remember, it has to be in the form x minus h, y minus k. So let's write it a little bit differently. Instead of x plus 3 squared, we can write that as x minus negative 3 squared. And then plus, well, this is already in the form, plus y minus 4 squared is equal to, instead of 49, we can just call that 7 squared."}, {"video_title": "Radius and center for a circle equation in standard form Algebra II Khan Academy.mp3", "Sentence": "We just have to make sure that we don't get confused with the negatives. Remember, it has to be in the form x minus h, y minus k. So let's write it a little bit differently. Instead of x plus 3 squared, we can write that as x minus negative 3 squared. And then plus, well, this is already in the form, plus y minus 4 squared is equal to, instead of 49, we can just call that 7 squared. And so now it becomes pretty clear that our h is negative 3. I wanted to do that in the red color. That our h is negative 3, and that our k is positive 4, and that our r is 7."}, {"video_title": "Radius and center for a circle equation in standard form Algebra II Khan Academy.mp3", "Sentence": "And then plus, well, this is already in the form, plus y minus 4 squared is equal to, instead of 49, we can just call that 7 squared. And so now it becomes pretty clear that our h is negative 3. I wanted to do that in the red color. That our h is negative 3, and that our k is positive 4, and that our r is 7. So we could say h comma k is equal to negative 3 comma positive 4. You might say, hey, there's a negative 4 here. No, but look, it's minus k, minus 4."}, {"video_title": "Radius and center for a circle equation in standard form Algebra II Khan Academy.mp3", "Sentence": "That our h is negative 3, and that our k is positive 4, and that our r is 7. So we could say h comma k is equal to negative 3 comma positive 4. You might say, hey, there's a negative 4 here. No, but look, it's minus k, minus 4. So k is 4. Likewise, it's minus h. You might say, hey, maybe h is a positive 3. But no, you're subtracting the h. So you'd say minus negative 3."}, {"video_title": "Solving absolute value equations.mp3", "Sentence": "And before we work some examples, let's just think about it graphically, why that is the case. So that's my x-axis, that's my y-axis. And let's say I had the graph of y is equal to something, I'll just call that a times x minus something else, let's just call that b, plus something else. This is the general, or this you could view this as a general absolute value equation. Well, this will have a graph something like this, I'm just gonna draw it very roughly. It's going to have this type of shape. It could also be downward pointing if a right over here were negative."}, {"video_title": "Solving absolute value equations.mp3", "Sentence": "This is the general, or this you could view this as a general absolute value equation. Well, this will have a graph something like this, I'm just gonna draw it very roughly. It's going to have this type of shape. It could also be downward pointing if a right over here were negative. But a graph like this, it obviously would keep going. If you said, when does this expression on the right equal zero or another way of thinking about it, for what x values does y equal zero? So y equals zero is right over here."}, {"video_title": "Solving absolute value equations.mp3", "Sentence": "It could also be downward pointing if a right over here were negative. But a graph like this, it obviously would keep going. If you said, when does this expression on the right equal zero or another way of thinking about it, for what x values does y equal zero? So y equals zero is right over here. And you could say, you could see from this graph that y is never equal to zero. So there's no x values that would make this left hand equal zero. So in that situation, you would have zero solutions if you had zero is equal to all of this stuff right over here."}, {"video_title": "Solving absolute value equations.mp3", "Sentence": "So y equals zero is right over here. And you could say, you could see from this graph that y is never equal to zero. So there's no x values that would make this left hand equal zero. So in that situation, you would have zero solutions if you had zero is equal to all of this stuff right over here. Now, let's say that this value right over here is when y is equal to two. So if you had two is equal to all of this stuff right over here, well, you would have exactly one x value that makes that true. Now, let's say this over here is five."}, {"video_title": "Solving absolute value equations.mp3", "Sentence": "So in that situation, you would have zero solutions if you had zero is equal to all of this stuff right over here. Now, let's say that this value right over here is when y is equal to two. So if you had two is equal to all of this stuff right over here, well, you would have exactly one x value that makes that true. Now, let's say this over here is five. So if you had the equation five is equal to all of this business, well, there would be two x values that make that true. So depending on the equation, you're going to have zero, one, or two solutions. Now, let's do some examples algebraically to see that happening."}, {"video_title": "Solving absolute value equations.mp3", "Sentence": "Now, let's say this over here is five. So if you had the equation five is equal to all of this business, well, there would be two x values that make that true. So depending on the equation, you're going to have zero, one, or two solutions. Now, let's do some examples algebraically to see that happening. So let's say I had the equation negative four is equal to negative five times the absolute value of x minus three, and then all of that plus two. Pause this video and try to solve this absolute value equation. All right, now let's do this together."}, {"video_title": "Solving absolute value equations.mp3", "Sentence": "Now, let's do some examples algebraically to see that happening. So let's say I had the equation negative four is equal to negative five times the absolute value of x minus three, and then all of that plus two. Pause this video and try to solve this absolute value equation. All right, now let's do this together. So what I try to do is isolate the absolute value part the way that you would manipulate any algebraic equation. So we can start off by subtracting two from both sides, so that we get rid of that plus two over there. And we are going to get negative four minus two is negative six is equal to negative five times the absolute value of x minus three, and then that's gone right over there."}, {"video_title": "Solving absolute value equations.mp3", "Sentence": "All right, now let's do this together. So what I try to do is isolate the absolute value part the way that you would manipulate any algebraic equation. So we can start off by subtracting two from both sides, so that we get rid of that plus two over there. And we are going to get negative four minus two is negative six is equal to negative five times the absolute value of x minus three, and then that's gone right over there. Now, let's divide both sides of this equation by negative five so divide by negative five, divide by negative five. And of course, we're going to get 6 5ths is equal to the absolute value of x minus three. Now, this is where it gets a little bit interesting."}, {"video_title": "Solving absolute value equations.mp3", "Sentence": "And we are going to get negative four minus two is negative six is equal to negative five times the absolute value of x minus three, and then that's gone right over there. Now, let's divide both sides of this equation by negative five so divide by negative five, divide by negative five. And of course, we're going to get 6 5ths is equal to the absolute value of x minus three. Now, this is where it gets a little bit interesting. If the absolute value of something is equal to 6 5ths, that means that that something, that x minus three could be equal to 6 5ths, but it also means that x minus three could be equal to negative 6 5ths. How do I know that? Well, if x minus three is equal to negative 6 5ths, you take the absolute value of that, you get 6 5ths."}, {"video_title": "Solving absolute value equations.mp3", "Sentence": "Now, this is where it gets a little bit interesting. If the absolute value of something is equal to 6 5ths, that means that that something, that x minus three could be equal to 6 5ths, but it also means that x minus three could be equal to negative 6 5ths. How do I know that? Well, if x minus three is equal to negative 6 5ths, you take the absolute value of that, you get 6 5ths. If it's equal to 6 5ths, and you take the absolute value of that, well, then you're going to get 6 5ths again. So these are two possible solutions. So let's solve it."}, {"video_title": "Solving absolute value equations.mp3", "Sentence": "Well, if x minus three is equal to negative 6 5ths, you take the absolute value of that, you get 6 5ths. If it's equal to 6 5ths, and you take the absolute value of that, well, then you're going to get 6 5ths again. So these are two possible solutions. So let's solve it. So if we add three to both sides of this equation, we're going to get x is equal to three plus 6 5ths. If you add three to both sides of this equation, you get x is equal to three minus 6 5ths. So three is the same thing as 15 over five."}, {"video_title": "Solving absolute value equations.mp3", "Sentence": "So let's solve it. So if we add three to both sides of this equation, we're going to get x is equal to three plus 6 5ths. If you add three to both sides of this equation, you get x is equal to three minus 6 5ths. So three is the same thing as 15 over five. 15 over five plus 6 over five is equal to 21 5ths in this situation. And over here, this is equal to 15 over five minus six over five, which is equal to nine over five. So here, we clearly had two solutions."}, {"video_title": "Solving absolute value equations.mp3", "Sentence": "So three is the same thing as 15 over five. 15 over five plus 6 over five is equal to 21 5ths in this situation. And over here, this is equal to 15 over five minus six over five, which is equal to nine over five. So here, we clearly had two solutions. It was kind of like this scenario right over here. Two x values gave us that y value. Let's look at two more examples."}, {"video_title": "Solving absolute value equations.mp3", "Sentence": "So here, we clearly had two solutions. It was kind of like this scenario right over here. Two x values gave us that y value. Let's look at two more examples. So let's say that I had the equation six minus the absolute value of two x plus three is equal to six. Pause this video and see if you can solve this. All right, once again, let's try to isolate the absolute value."}, {"video_title": "Solving absolute value equations.mp3", "Sentence": "Let's look at two more examples. So let's say that I had the equation six minus the absolute value of two x plus three is equal to six. Pause this video and see if you can solve this. All right, once again, let's try to isolate the absolute value. Let's subtract six from both sides. And we are going to get, on the left-hand side, we're just going to have negative absolute value of two x plus three is going to be equal to zero. Well, we can multiply both sides by negative one or divide both sides by negative one, and you're going to get the absolute value of two x plus three is still going to be equal to zero."}, {"video_title": "Solving absolute value equations.mp3", "Sentence": "All right, once again, let's try to isolate the absolute value. Let's subtract six from both sides. And we are going to get, on the left-hand side, we're just going to have negative absolute value of two x plus three is going to be equal to zero. Well, we can multiply both sides by negative one or divide both sides by negative one, and you're going to get the absolute value of two x plus three is still going to be equal to zero. And so that means that two x plus three could be equal to zero, and that's it. Because it's not like there's a positive version of zero and a negative version of zero. There's just zero."}, {"video_title": "Solving absolute value equations.mp3", "Sentence": "Well, we can multiply both sides by negative one or divide both sides by negative one, and you're going to get the absolute value of two x plus three is still going to be equal to zero. And so that means that two x plus three could be equal to zero, and that's it. Because it's not like there's a positive version of zero and a negative version of zero. There's just zero. If the absolute value of something is zero, well, that thing inside the absolute value sign must be equal to zero. And then you just solve from here. If you subtract three from both sides, we get two x is equal to negative three."}, {"video_title": "Solving absolute value equations.mp3", "Sentence": "There's just zero. If the absolute value of something is zero, well, that thing inside the absolute value sign must be equal to zero. And then you just solve from here. If you subtract three from both sides, we get two x is equal to negative three. Divide both sides by two, and we get x is equal to negative 3 1\u20442. And here we had one solution. And a good hint is is that when we isolated the absolute value sign here, on the right-hand side, you got a zero."}, {"video_title": "Solving absolute value equations.mp3", "Sentence": "If you subtract three from both sides, we get two x is equal to negative three. Divide both sides by two, and we get x is equal to negative 3 1\u20442. And here we had one solution. And a good hint is is that when we isolated the absolute value sign here, on the right-hand side, you got a zero. If on the right-hand side you got a positive number, you'd have two solutions. If on the right-hand side you get a zero, you're going to have one solution. And as we're about to see, if on the right-hand side, when you isolate the absolute value, you get a negative value, well, it's impossible to take the absolute value of something and get a negative value, and then you would have no solutions."}, {"video_title": "Solving absolute value equations.mp3", "Sentence": "And a good hint is is that when we isolated the absolute value sign here, on the right-hand side, you got a zero. If on the right-hand side you got a positive number, you'd have two solutions. If on the right-hand side you get a zero, you're going to have one solution. And as we're about to see, if on the right-hand side, when you isolate the absolute value, you get a negative value, well, it's impossible to take the absolute value of something and get a negative value, and then you would have no solutions. But let's look at an example of that. And I'm going to do this in purple just for the sake of variety. So let's add three times the absolute value of negative 4x plus seven is equal to one."}, {"video_title": "Solving absolute value equations.mp3", "Sentence": "And as we're about to see, if on the right-hand side, when you isolate the absolute value, you get a negative value, well, it's impossible to take the absolute value of something and get a negative value, and then you would have no solutions. But let's look at an example of that. And I'm going to do this in purple just for the sake of variety. So let's add three times the absolute value of negative 4x plus seven is equal to one. Well, let's subtract seven from both sides, and we are going to get three times the absolute value of negative 4x is equal to negative six. Let's divide both sides by three. And we are going to get the absolute value of negative 4x is equal to negative two."}, {"video_title": "Solving absolute value equations.mp3", "Sentence": "So let's add three times the absolute value of negative 4x plus seven is equal to one. Well, let's subtract seven from both sides, and we are going to get three times the absolute value of negative 4x is equal to negative six. Let's divide both sides by three. And we are going to get the absolute value of negative 4x is equal to negative two. Well, how can I take the absolute value of anything and get a negative two? This is impossible. This will have no solution."}, {"video_title": "Solve by completing the square Integer solutions Algebra I Khan Academy.mp3", "Sentence": "What I want you to do is pause this video and see if you can solve it. What x values satisfy the equation? All right, now let's work through this together. So one technique could be just, let's just try to complete the square here on the left-hand side. So to do that, let me write it this way, x squared minus eight x, and then I have, I'll write the plus one out here, is equal to 85. Now, if I wanna complete the square, I just have to think, what can I add to both sides of this equation that could make this part of the left-hand expression a perfect square? Well, if I look at this negative eight coefficient on the first degree term, I could say, okay, let me take half of negative eight, that would be negative four, and then negative four squared is going to be positive 16."}, {"video_title": "Solve by completing the square Integer solutions Algebra I Khan Academy.mp3", "Sentence": "So one technique could be just, let's just try to complete the square here on the left-hand side. So to do that, let me write it this way, x squared minus eight x, and then I have, I'll write the plus one out here, is equal to 85. Now, if I wanna complete the square, I just have to think, what can I add to both sides of this equation that could make this part of the left-hand expression a perfect square? Well, if I look at this negative eight coefficient on the first degree term, I could say, okay, let me take half of negative eight, that would be negative four, and then negative four squared is going to be positive 16. So I'm gonna add a positive 16 on the left-hand side. And if I want, I could then subtract a 16 from the left-hand side, or I could add a 16 on the right-hand side. Notice, I've just done the same thing to both sides of this equation."}, {"video_title": "Solve by completing the square Integer solutions Algebra I Khan Academy.mp3", "Sentence": "Well, if I look at this negative eight coefficient on the first degree term, I could say, okay, let me take half of negative eight, that would be negative four, and then negative four squared is going to be positive 16. So I'm gonna add a positive 16 on the left-hand side. And if I want, I could then subtract a 16 from the left-hand side, or I could add a 16 on the right-hand side. Notice, I've just done the same thing to both sides of this equation. And why was that useful? Well, now what I've just put in parentheses is a perfect square. This is the same thing as x minus four squared."}, {"video_title": "Solve by completing the square Integer solutions Algebra I Khan Academy.mp3", "Sentence": "Notice, I've just done the same thing to both sides of this equation. And why was that useful? Well, now what I've just put in parentheses is a perfect square. This is the same thing as x minus four squared. It was by design. We looked at that negative eight, half of that is negative four, you square it, you get 16, and you can verify x minus four times x minus four is indeed equal to this. And then we have plus one is going to be equal to, what's 85 plus 16?"}, {"video_title": "Solve by completing the square Integer solutions Algebra I Khan Academy.mp3", "Sentence": "This is the same thing as x minus four squared. It was by design. We looked at that negative eight, half of that is negative four, you square it, you get 16, and you can verify x minus four times x minus four is indeed equal to this. And then we have plus one is going to be equal to, what's 85 plus 16? That is 101. And now we wanna get rid of this one on the left-hand side, and the easiest way we can do that is subtract one from both sides. That way we'll just isolate that x minus four squared."}, {"video_title": "Solve by completing the square Integer solutions Algebra I Khan Academy.mp3", "Sentence": "And then we have plus one is going to be equal to, what's 85 plus 16? That is 101. And now we wanna get rid of this one on the left-hand side, and the easiest way we can do that is subtract one from both sides. That way we'll just isolate that x minus four squared. And we are left with x minus four squared, four squared, these cancel out, is going to be equal to 100. Now if something squared is equal to 100, that means that the something is equal to the positive or the negative square root of 100, or that that something, x minus four, is equal to positive or negative 10, positive or negative 10. All I did is took the plus or minus square root of 100."}, {"video_title": "Solve by completing the square Integer solutions Algebra I Khan Academy.mp3", "Sentence": "That way we'll just isolate that x minus four squared. And we are left with x minus four squared, four squared, these cancel out, is going to be equal to 100. Now if something squared is equal to 100, that means that the something is equal to the positive or the negative square root of 100, or that that something, x minus four, is equal to positive or negative 10, positive or negative 10. All I did is took the plus or minus square root of 100. And this makes sense. If I took positive 10 squared, I'll get 100. If I take negative 10 squared, I get 100."}, {"video_title": "Solve by completing the square Integer solutions Algebra I Khan Academy.mp3", "Sentence": "All I did is took the plus or minus square root of 100. And this makes sense. If I took positive 10 squared, I'll get 100. If I take negative 10 squared, I get 100. So x minus four could be either one of those. And now I just add four to both sides of this equation. And then what do I get?"}, {"video_title": "Solve by completing the square Integer solutions Algebra I Khan Academy.mp3", "Sentence": "If I take negative 10 squared, I get 100. So x minus four could be either one of those. And now I just add four to both sides of this equation. And then what do I get? I get x is equal to four plus or minus 10. Or another way of thinking about it, I could write it as x is equal to, four plus 10 is 14, and then four minus 10 is equal to negative six. So these are two ways to solve it."}, {"video_title": "Solve by completing the square Integer solutions Algebra I Khan Academy.mp3", "Sentence": "And then what do I get? I get x is equal to four plus or minus 10. Or another way of thinking about it, I could write it as x is equal to, four plus 10 is 14, and then four minus 10 is equal to negative six. So these are two ways to solve it. But there's other ways to solve this equation. We could, right from the get-go, try to subtract 85 from both sides. Some people feel more comfortable solving quadratics if they have the quadratic expression be equal to zero."}, {"video_title": "Solve by completing the square Integer solutions Algebra I Khan Academy.mp3", "Sentence": "So these are two ways to solve it. But there's other ways to solve this equation. We could, right from the get-go, try to subtract 85 from both sides. Some people feel more comfortable solving quadratics if they have the quadratic expression be equal to zero. And if you did that, you would get x squared minus eight x minus 84 is equal to zero. All I did is I subtracted 85 from both sides of this equation to get this right over here. Now this one, we can approach in two different ways."}, {"video_title": "Solve by completing the square Integer solutions Algebra I Khan Academy.mp3", "Sentence": "Some people feel more comfortable solving quadratics if they have the quadratic expression be equal to zero. And if you did that, you would get x squared minus eight x minus 84 is equal to zero. All I did is I subtracted 85 from both sides of this equation to get this right over here. Now this one, we can approach in two different ways. We can complete the square again, or we could just try to factor. If we complete the square, we're going to see something very similar to this. Actually, let me just do that really fast."}, {"video_title": "Solve by completing the square Integer solutions Algebra I Khan Academy.mp3", "Sentence": "Now this one, we can approach in two different ways. We can complete the square again, or we could just try to factor. If we complete the square, we're going to see something very similar to this. Actually, let me just do that really fast. If I look at this part right over there, I could say x squared minus eight x. And then once again, half of negative eight is negative four. That squared is plus 16."}, {"video_title": "Solve by completing the square Integer solutions Algebra I Khan Academy.mp3", "Sentence": "Actually, let me just do that really fast. If I look at this part right over there, I could say x squared minus eight x. And then once again, half of negative eight is negative four. That squared is plus 16. And then I'd have minus 84. So let me do that in that blue color so we can keep track. Minus 84."}, {"video_title": "Solve by completing the square Integer solutions Algebra I Khan Academy.mp3", "Sentence": "That squared is plus 16. And then I'd have minus 84. So let me do that in that blue color so we can keep track. Minus 84. And then if I added 16 on the left-hand side, I could either add that to the right-hand side so both sides have 16 added to it, or if I want to maintain the equality, I could just subtract 16 from the left-hand side. So I've added 16, subtracted 16. I haven't changed the left-hand side's value."}, {"video_title": "Solve by completing the square Integer solutions Algebra I Khan Academy.mp3", "Sentence": "Minus 84. And then if I added 16 on the left-hand side, I could either add that to the right-hand side so both sides have 16 added to it, or if I want to maintain the equality, I could just subtract 16 from the left-hand side. So I've added 16, subtracted 16. I haven't changed the left-hand side's value. And then that would be equal to zero. This part right over here, this is x minus four squared. This part right over here is minus 100 is equal to zero."}, {"video_title": "Solve by completing the square Integer solutions Algebra I Khan Academy.mp3", "Sentence": "I haven't changed the left-hand side's value. And then that would be equal to zero. This part right over here, this is x minus four squared. This part right over here is minus 100 is equal to zero. And then you add 100 to both sides of this and you get exactly this step right over here. Now, another way that we could have approached it without completing the square, we could have said x squared minus eight x minus 84 is equal to zero. And think about what two numbers, if I multiply them, I get negative 84."}, {"video_title": "Solve by completing the square Integer solutions Algebra I Khan Academy.mp3", "Sentence": "This part right over here is minus 100 is equal to zero. And then you add 100 to both sides of this and you get exactly this step right over here. Now, another way that we could have approached it without completing the square, we could have said x squared minus eight x minus 84 is equal to zero. And think about what two numbers, if I multiply them, I get negative 84. So they'd have to have different signs since when I take their product, I get a negative number. And when I add them together, I get negative eight. And there we could just look at the factorization of negative 84, of 84 generally."}, {"video_title": "Solve by completing the square Integer solutions Algebra I Khan Academy.mp3", "Sentence": "And think about what two numbers, if I multiply them, I get negative 84. So they'd have to have different signs since when I take their product, I get a negative number. And when I add them together, I get negative eight. And there we could just look at the factorization of negative 84, of 84 generally. It could be two times, let's just think about 84. 84 could be two times 42. And obviously one of them would have to be negative, one of them would have to be positive in order to get to negative 84."}, {"video_title": "Solve by completing the square Integer solutions Algebra I Khan Academy.mp3", "Sentence": "And there we could just look at the factorization of negative 84, of 84 generally. It could be two times, let's just think about 84. 84 could be two times 42. And obviously one of them would have to be negative, one of them would have to be positive in order to get to negative 84. But the difference between these two numbers, if one was positive and one is negative, is a lot more than eight. So that doesn't work. So let's try, let's see, I'll do a few in my head."}, {"video_title": "Solve by completing the square Integer solutions Algebra I Khan Academy.mp3", "Sentence": "And obviously one of them would have to be negative, one of them would have to be positive in order to get to negative 84. But the difference between these two numbers, if one was positive and one is negative, is a lot more than eight. So that doesn't work. So let's try, let's see, I'll do a few in my head. Three times 28, but still that difference is way more than eight. Four times, four times, let's see, four times 21, now that difference between four and 21 is still larger than eight. Let's see, five doesn't go to it."}, {"video_title": "Solve by completing the square Integer solutions Algebra I Khan Academy.mp3", "Sentence": "So let's try, let's see, I'll do a few in my head. Three times 28, but still that difference is way more than eight. Four times, four times, let's see, four times 21, now that difference between four and 21 is still larger than eight. Let's see, five doesn't go to it. Six times 14, that's interesting now. Okay, so let's think about this. So six times 14 is equal to 84."}, {"video_title": "Solve by completing the square Integer solutions Algebra I Khan Academy.mp3", "Sentence": "Let's see, five doesn't go to it. Six times 14, that's interesting now. Okay, so let's think about this. So six times 14 is equal to 84. One of them has to be negative. And since when we take the sum of the two numbers, we get a negative number, that means the larger one is negative. So let's see, six times negative 14 is negative 84."}, {"video_title": "Solve by completing the square Integer solutions Algebra I Khan Academy.mp3", "Sentence": "So six times 14 is equal to 84. One of them has to be negative. And since when we take the sum of the two numbers, we get a negative number, that means the larger one is negative. So let's see, six times negative 14 is negative 84. Six plus negative 14 is indeed equal to negative eight. So we can factor this as x plus six times x minus 14 is equal to zero. And so the product of two things is equal to zero."}, {"video_title": "Solve by completing the square Integer solutions Algebra I Khan Academy.mp3", "Sentence": "So let's see, six times negative 14 is negative 84. Six plus negative 14 is indeed equal to negative eight. So we can factor this as x plus six times x minus 14 is equal to zero. And so the product of two things is equal to zero. That means if either of them is equal to zero, that would make the entire expression equal to zero. So we could say x plus six is equal to zero or x minus 14 is equal to zero. Subtract six from both sides here, we get x is equal to negative six or add 14 to both sides here or x is equal to 14."}, {"video_title": "Intervals and interval notation Functions Algebra I Khan Academy.mp3", "Sentence": "So right over here I have a number line. Now let's say I wanted to talk about the interval on the number line that goes from, let's say, from negative three to two. So I care about this, let me do this in a different color. Let's say I care about this interval right over here. So I care about all of the numbers from negative three, negative three to two. So in order to be more precise, I have to be clear. Am I including negative three and two, or am I not including negative three and two?"}, {"video_title": "Intervals and interval notation Functions Algebra I Khan Academy.mp3", "Sentence": "Let's say I care about this interval right over here. So I care about all of the numbers from negative three, negative three to two. So in order to be more precise, I have to be clear. Am I including negative three and two, or am I not including negative three and two? Or maybe I'm just including one of them. So if I'm including negative three and two, then I would fill them in. So this right over here, I'm filling negative three and two in, which means that negative three and two are part of this interval."}, {"video_title": "Intervals and interval notation Functions Algebra I Khan Academy.mp3", "Sentence": "Am I including negative three and two, or am I not including negative three and two? Or maybe I'm just including one of them. So if I'm including negative three and two, then I would fill them in. So this right over here, I'm filling negative three and two in, which means that negative three and two are part of this interval. And when you include the end points, this is called a closed interval. Closed interval. Closed interval."}, {"video_title": "Intervals and interval notation Functions Algebra I Khan Academy.mp3", "Sentence": "So this right over here, I'm filling negative three and two in, which means that negative three and two are part of this interval. And when you include the end points, this is called a closed interval. Closed interval. Closed interval. And I just showed you how I can depict it on a number line by actually filling in the end points. And there's multiple ways to talk about this interval mathematically. I could say that this is all of the, let's say that this is, let's say this number line is showing different values for x. I could say these are all of the x's that are between negative three, negative three, and two, and two."}, {"video_title": "Intervals and interval notation Functions Algebra I Khan Academy.mp3", "Sentence": "Closed interval. And I just showed you how I can depict it on a number line by actually filling in the end points. And there's multiple ways to talk about this interval mathematically. I could say that this is all of the, let's say that this is, let's say this number line is showing different values for x. I could say these are all of the x's that are between negative three, negative three, and two, and two. And notice, I have negative three is less than or equal to x. So that's telling us that x could be equal to, that x could be equal to negative three. And then we have x is less than or equal to positive two."}, {"video_title": "Intervals and interval notation Functions Algebra I Khan Academy.mp3", "Sentence": "I could say that this is all of the, let's say that this is, let's say this number line is showing different values for x. I could say these are all of the x's that are between negative three, negative three, and two, and two. And notice, I have negative three is less than or equal to x. So that's telling us that x could be equal to, that x could be equal to negative three. And then we have x is less than or equal to positive two. So that means that x could be equal to positive two. So it is a closed interval. Now another way that we could depict this closed interval is we could say, okay, we're talking about the interval between, and we could use brackets because it's a closed interval, negative three and two."}, {"video_title": "Intervals and interval notation Functions Algebra I Khan Academy.mp3", "Sentence": "And then we have x is less than or equal to positive two. So that means that x could be equal to positive two. So it is a closed interval. Now another way that we could depict this closed interval is we could say, okay, we're talking about the interval between, and we could use brackets because it's a closed interval, negative three and two. And once again, I'm using brackets here. These brackets tell us that we include. So this bracket on the left say that we include negative three, and this bracket on the right says that we include positive two in our interval."}, {"video_title": "Intervals and interval notation Functions Algebra I Khan Academy.mp3", "Sentence": "Now another way that we could depict this closed interval is we could say, okay, we're talking about the interval between, and we could use brackets because it's a closed interval, negative three and two. And once again, I'm using brackets here. These brackets tell us that we include. So this bracket on the left say that we include negative three, and this bracket on the right says that we include positive two in our interval. Now sometimes you might see things written a little bit more math. You might see something like x is a member of the real numbers such that, so, and I could put these curly brackets around like this. These curly brackets say that we're talking about a set of values, and we're saying the set of all x's that are a member of the real numbers, so this is just fancy math notation to say member of the real numbers."}, {"video_title": "Intervals and interval notation Functions Algebra I Khan Academy.mp3", "Sentence": "So this bracket on the left say that we include negative three, and this bracket on the right says that we include positive two in our interval. Now sometimes you might see things written a little bit more math. You might see something like x is a member of the real numbers such that, so, and I could put these curly brackets around like this. These curly brackets say that we're talking about a set of values, and we're saying the set of all x's that are a member of the real numbers, so this is just fancy math notation to say member of the real numbers. I'm using the Greek letter epsilon right over here to say member of the real numbers such that, this little, this vertical line here means such that negative three is less than x is less than, negative three is less than or equal to x is less than or equal to two. I could also write it this way. I could write x is a member of the real numbers such that x is a member, such that x is a member of this closed set."}, {"video_title": "Intervals and interval notation Functions Algebra I Khan Academy.mp3", "Sentence": "These curly brackets say that we're talking about a set of values, and we're saying the set of all x's that are a member of the real numbers, so this is just fancy math notation to say member of the real numbers. I'm using the Greek letter epsilon right over here to say member of the real numbers such that, this little, this vertical line here means such that negative three is less than x is less than, negative three is less than or equal to x is less than or equal to two. I could also write it this way. I could write x is a member of the real numbers such that x is a member, such that x is a member of this closed set. I'm including the end points here. So these are all different ways of denoting or depicting the same interval. Let's do some more examples here."}, {"video_title": "Intervals and interval notation Functions Algebra I Khan Academy.mp3", "Sentence": "I could write x is a member of the real numbers such that x is a member, such that x is a member of this closed set. I'm including the end points here. So these are all different ways of denoting or depicting the same interval. Let's do some more examples here. Let me draw a number line again. So number line. Now let me do, let me just do an open interval, an open interval just so that we see, we clearly can see the difference."}, {"video_title": "Intervals and interval notation Functions Algebra I Khan Academy.mp3", "Sentence": "Let's do some more examples here. Let me draw a number line again. So number line. Now let me do, let me just do an open interval, an open interval just so that we see, we clearly can see the difference. Let's say that I want to talk about the values between negative one and four. Negative, let me do this in a different color. The values between negative one and four, but I don't want to include negative one and four."}, {"video_title": "Intervals and interval notation Functions Algebra I Khan Academy.mp3", "Sentence": "Now let me do, let me just do an open interval, an open interval just so that we see, we clearly can see the difference. Let's say that I want to talk about the values between negative one and four. Negative, let me do this in a different color. The values between negative one and four, but I don't want to include negative one and four. So this is going to be an open interval. So I'm not going to include four and I'm not going to include negative one. Notice I have open circles here."}, {"video_title": "Intervals and interval notation Functions Algebra I Khan Academy.mp3", "Sentence": "The values between negative one and four, but I don't want to include negative one and four. So this is going to be an open interval. So I'm not going to include four and I'm not going to include negative one. Notice I have open circles here. Over here I had closed circles. The closed circles told me that I included negative three and two. Now I have open circles here."}, {"video_title": "Intervals and interval notation Functions Algebra I Khan Academy.mp3", "Sentence": "Notice I have open circles here. Over here I had closed circles. The closed circles told me that I included negative three and two. Now I have open circles here. So that says that I'm not, it's all the values in between negative one and four. So negative one or negative.99999999 is going to be included, but negative one is not going to be included. And 3.99999999 is going to be included, but four is not going to be included."}, {"video_title": "Intervals and interval notation Functions Algebra I Khan Academy.mp3", "Sentence": "Now I have open circles here. So that says that I'm not, it's all the values in between negative one and four. So negative one or negative.99999999 is going to be included, but negative one is not going to be included. And 3.99999999 is going to be included, but four is not going to be included. So how would we, what would be the notation for this? Well here we could say, x is going to be a member of the real numbers such that negative one, I'm not going to say less than or equal to because x can't be equal to negative one. So negative one is strictly less than x, is strictly less than four."}, {"video_title": "Intervals and interval notation Functions Algebra I Khan Academy.mp3", "Sentence": "And 3.99999999 is going to be included, but four is not going to be included. So how would we, what would be the notation for this? Well here we could say, x is going to be a member of the real numbers such that negative one, I'm not going to say less than or equal to because x can't be equal to negative one. So negative one is strictly less than x, is strictly less than four. Notice, not less than or equal because I can't be equal to four. Four is not included. So that's one way to say it."}, {"video_title": "Intervals and interval notation Functions Algebra I Khan Academy.mp3", "Sentence": "So negative one is strictly less than x, is strictly less than four. Notice, not less than or equal because I can't be equal to four. Four is not included. So that's one way to say it. Or another way, I could write it like this. x is a member of the real numbers such that x is a member of, now the interval is from negative one to four, but I'm not going to use these brackets. These brackets say, hey, let me include the endpoints, but I'm not going to include them."}, {"video_title": "Intervals and interval notation Functions Algebra I Khan Academy.mp3", "Sentence": "So that's one way to say it. Or another way, I could write it like this. x is a member of the real numbers such that x is a member of, now the interval is from negative one to four, but I'm not going to use these brackets. These brackets say, hey, let me include the endpoints, but I'm not going to include them. So I'm going to put the parentheses right over here. Parentheses. So this tells us that we're dealing with an open interval."}, {"video_title": "Intervals and interval notation Functions Algebra I Khan Academy.mp3", "Sentence": "These brackets say, hey, let me include the endpoints, but I'm not going to include them. So I'm going to put the parentheses right over here. Parentheses. So this tells us that we're dealing with an open interval. So this right over here, let me make it clear, this is an open, an open interval. Now you're probably wondering, okay, in this case, both endpoints were included, it's a closed interval. In this case, both endpoints were excluded, it's an open interval."}, {"video_title": "Intervals and interval notation Functions Algebra I Khan Academy.mp3", "Sentence": "So this tells us that we're dealing with an open interval. So this right over here, let me make it clear, this is an open, an open interval. Now you're probably wondering, okay, in this case, both endpoints were included, it's a closed interval. In this case, both endpoints were excluded, it's an open interval. Can you have things that have one endpoint included and one endpoint excluded? And the answer is absolutely. Let's see an example of that."}, {"video_title": "Intervals and interval notation Functions Algebra I Khan Academy.mp3", "Sentence": "In this case, both endpoints were excluded, it's an open interval. Can you have things that have one endpoint included and one endpoint excluded? And the answer is absolutely. Let's see an example of that. So I'll get another number line here. Another number line. And let's say that we want to, actually let me do it the other way around."}, {"video_title": "Intervals and interval notation Functions Algebra I Khan Academy.mp3", "Sentence": "Let's see an example of that. So I'll get another number line here. Another number line. And let's say that we want to, actually let me do it the other way around. Let me write it first, and then I'll graph it. So let's say we're thinking about all of the x's that are a member of the real numbers, such that, let's say negative four, let's say negative four is not included, is less than x, is less than or equal to negative one. So now negative one is included."}, {"video_title": "Intervals and interval notation Functions Algebra I Khan Academy.mp3", "Sentence": "And let's say that we want to, actually let me do it the other way around. Let me write it first, and then I'll graph it. So let's say we're thinking about all of the x's that are a member of the real numbers, such that, let's say negative four, let's say negative four is not included, is less than x, is less than or equal to negative one. So now negative one is included. So we're not going to include negative four. Negative four is strictly less than, not less than or equal to. So x can't be equal to negative four, so open circle there."}, {"video_title": "Intervals and interval notation Functions Algebra I Khan Academy.mp3", "Sentence": "So now negative one is included. So we're not going to include negative four. Negative four is strictly less than, not less than or equal to. So x can't be equal to negative four, so open circle there. But x could be equal to negative one. It has to be less than or equal to negative one. So it could be equal to negative one, so I'm going to fill that in right over there."}, {"video_title": "Intervals and interval notation Functions Algebra I Khan Academy.mp3", "Sentence": "So x can't be equal to negative four, so open circle there. But x could be equal to negative one. It has to be less than or equal to negative one. So it could be equal to negative one, so I'm going to fill that in right over there. And then it's everything in between. If I want to write it with this notation, I could write x is a member of the real numbers, such that x is a member of the interval. So it's going to go between negative four and negative one."}, {"video_title": "Intervals and interval notation Functions Algebra I Khan Academy.mp3", "Sentence": "So it could be equal to negative one, so I'm going to fill that in right over there. And then it's everything in between. If I want to write it with this notation, I could write x is a member of the real numbers, such that x is a member of the interval. So it's going to go between negative four and negative one. But we're not including negative four. We have an open circle here, so I'm going to put a parentheses on that side. But we are including negative one."}, {"video_title": "Intervals and interval notation Functions Algebra I Khan Academy.mp3", "Sentence": "So it's going to go between negative four and negative one. But we're not including negative four. We have an open circle here, so I'm going to put a parentheses on that side. But we are including negative one. We are including negative one. So we put a bracket on that side. And so that would be, that right over there would be the notation."}, {"video_title": "Intervals and interval notation Functions Algebra I Khan Academy.mp3", "Sentence": "But we are including negative one. We are including negative one. So we put a bracket on that side. And so that would be, that right over there would be the notation. Now there's other things that you could do with interval notation. You could say, well hey, everything except for some value. So let me give another example."}, {"video_title": "Intervals and interval notation Functions Algebra I Khan Academy.mp3", "Sentence": "And so that would be, that right over there would be the notation. Now there's other things that you could do with interval notation. You could say, well hey, everything except for some value. So let me give another example. Let's get another example here. Let's say that we want to talk about all the real numbers except for one. So we want to include all of the real numbers except for one."}, {"video_title": "Intervals and interval notation Functions Algebra I Khan Academy.mp3", "Sentence": "So let me give another example. Let's get another example here. Let's say that we want to talk about all the real numbers except for one. So we want to include all of the real numbers except for one. So we're going to exclude one right over here, so open circle, but it can be any other real number. So how would we denote this? Well, we could write, x is a member of the real numbers such that x does not equal one."}, {"video_title": "Intervals and interval notation Functions Algebra I Khan Academy.mp3", "Sentence": "So we want to include all of the real numbers except for one. So we're going to exclude one right over here, so open circle, but it can be any other real number. So how would we denote this? Well, we could write, x is a member of the real numbers such that x does not equal one. So here I'm saying, look, x can be a member of the real numbers, but x cannot be equal to one. It can be anything else, but it cannot be equal to one. And there's other ways of denoting this exact same interval."}, {"video_title": "Intervals and interval notation Functions Algebra I Khan Academy.mp3", "Sentence": "Well, we could write, x is a member of the real numbers such that x does not equal one. So here I'm saying, look, x can be a member of the real numbers, but x cannot be equal to one. It can be anything else, but it cannot be equal to one. And there's other ways of denoting this exact same interval. You could say, x is a member of the real numbers such that x is less than one, or x is greater than one. So you could write it just like that, or you could do something interesting. This is the one that I would use."}, {"video_title": "Intervals and interval notation Functions Algebra I Khan Academy.mp3", "Sentence": "And there's other ways of denoting this exact same interval. You could say, x is a member of the real numbers such that x is less than one, or x is greater than one. So you could write it just like that, or you could do something interesting. This is the one that I would use. This is the shortest, and it makes it very clear. You're saying, hey, everything except for one. But you could even do something fancy."}, {"video_title": "Intervals and interval notation Functions Algebra I Khan Academy.mp3", "Sentence": "This is the one that I would use. This is the shortest, and it makes it very clear. You're saying, hey, everything except for one. But you could even do something fancy. Like you could say, x is a member of the real numbers such that x is a member of the set going from negative infinity to one, not including one, or x is a member of the set going from, or a member of the interval going from one, not including one, all the way to positive, all the way to positive infinity. And when you're talking about negative infinity or positive infinity, you always put a parentheses. And the view there is you can never include everything all the way up to infinity."}, {"video_title": "Intervals and interval notation Functions Algebra I Khan Academy.mp3", "Sentence": "But you could even do something fancy. Like you could say, x is a member of the real numbers such that x is a member of the set going from negative infinity to one, not including one, or x is a member of the set going from, or a member of the interval going from one, not including one, all the way to positive, all the way to positive infinity. And when you're talking about negative infinity or positive infinity, you always put a parentheses. And the view there is you can never include everything all the way up to infinity. It needs to be at least open at that end point, because infinity just keeps going on and on. So you always want to put a parentheses if you're talking about infinity or negative infinity. It's not really an endpoint."}, {"video_title": "Intervals and interval notation Functions Algebra I Khan Academy.mp3", "Sentence": "And the view there is you can never include everything all the way up to infinity. It needs to be at least open at that end point, because infinity just keeps going on and on. So you always want to put a parentheses if you're talking about infinity or negative infinity. It's not really an endpoint. It keeps going on and on forever. So you use the notation for open interval, at least at that end. And notice we're not including one either."}, {"video_title": "Finding dimensions of triangle from area Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "The height of a triangle is 4 inches less than the length of the base. The area of the triangle is 30 inches squared. Find the height and base. Use the formula area equals 1 half base times height for the area of a triangle. Okay, so let's think about it a little bit. We have the base. Let me draw a triangle here."}, {"video_title": "Finding dimensions of triangle from area Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Use the formula area equals 1 half base times height for the area of a triangle. Okay, so let's think about it a little bit. We have the base. Let me draw a triangle here. So this is our triangle. And let's say that the length of this bottom side, that's the base. Let's call that b."}, {"video_title": "Finding dimensions of triangle from area Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Let me draw a triangle here. So this is our triangle. And let's say that the length of this bottom side, that's the base. Let's call that b. And then this is the height. This is the height right over here. And then the area is equal to 1 half base times height."}, {"video_title": "Finding dimensions of triangle from area Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Let's call that b. And then this is the height. This is the height right over here. And then the area is equal to 1 half base times height. Now in this first sentence, they tell us that the height of a triangle is 4 inches less than the length of the base. So the height is equal to the base minus 4. That's what that first sentence tells us."}, {"video_title": "Finding dimensions of triangle from area Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "And then the area is equal to 1 half base times height. Now in this first sentence, they tell us that the height of a triangle is 4 inches less than the length of the base. So the height is equal to the base minus 4. That's what that first sentence tells us. The area of the triangle is 30 inches squared. So if we take 1 half the base times the height, we'll get 30 inches squared. Or we could say that 30 inches squared is equal to 1 half times the base times the height."}, {"video_title": "Finding dimensions of triangle from area Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "That's what that first sentence tells us. The area of the triangle is 30 inches squared. So if we take 1 half the base times the height, we'll get 30 inches squared. Or we could say that 30 inches squared is equal to 1 half times the base times the height. Now instead of putting an h in for height, we know that the height is the same thing as 4 less than the base. So let's put that in there. 4 less than the base."}, {"video_title": "Finding dimensions of triangle from area Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Or we could say that 30 inches squared is equal to 1 half times the base times the height. Now instead of putting an h in for height, we know that the height is the same thing as 4 less than the base. So let's put that in there. 4 less than the base. And then let's see what we get here. We get 30 is equal to 1 half times b over 2 times b minus 4. I just multiplied the 1 half times the b."}, {"video_title": "Finding dimensions of triangle from area Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "4 less than the base. And then let's see what we get here. We get 30 is equal to 1 half times b over 2 times b minus 4. I just multiplied the 1 half times the b. Now let's distribute the b over 2. 30 is equal to b squared over 2. Be careful."}, {"video_title": "Finding dimensions of triangle from area Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "I just multiplied the 1 half times the b. Now let's distribute the b over 2. 30 is equal to b squared over 2. Be careful. b over 2 times b is just b squared over 2. And then b over 2 times negative 4 is negative 2b. Now just to get rid of this fraction here, let's multiply both sides of this equation by 2."}, {"video_title": "Finding dimensions of triangle from area Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Be careful. b over 2 times b is just b squared over 2. And then b over 2 times negative 4 is negative 2b. Now just to get rid of this fraction here, let's multiply both sides of this equation by 2. So let's multiply that side by 2 and let's multiply that side by 2. On the left-hand side, you get 60. On the right-hand side, 2 times b squared over 2 is just b squared."}, {"video_title": "Finding dimensions of triangle from area Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Now just to get rid of this fraction here, let's multiply both sides of this equation by 2. So let's multiply that side by 2 and let's multiply that side by 2. On the left-hand side, you get 60. On the right-hand side, 2 times b squared over 2 is just b squared. Negative 2b times 2 is negative 4b. And now we have a quadratic here. And the best way to solve a quadratic, we have a second degree term right here, is to get all of the terms on one side of the equation having them equal 0."}, {"video_title": "Finding dimensions of triangle from area Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "On the right-hand side, 2 times b squared over 2 is just b squared. Negative 2b times 2 is negative 4b. And now we have a quadratic here. And the best way to solve a quadratic, we have a second degree term right here, is to get all of the terms on one side of the equation having them equal 0. So let's subtract 60 from both sides of this equation. Let's subtract 60 from both sides. And we get 0 is equal to b squared minus 4b minus 60."}, {"video_title": "Finding dimensions of triangle from area Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "And the best way to solve a quadratic, we have a second degree term right here, is to get all of the terms on one side of the equation having them equal 0. So let's subtract 60 from both sides of this equation. Let's subtract 60 from both sides. And we get 0 is equal to b squared minus 4b minus 60. And so what we need to do here is just factor this thing right now, or factor it, and then know that if I have the product of some things and that equals 0, that means that either one or both of those things need to be equal to 0. So we need to factor b squared minus 4b minus 60. So what we want to do, we want to find 2 numbers whose sum is negative 4 and whose product is negative 60."}, {"video_title": "Finding dimensions of triangle from area Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "And we get 0 is equal to b squared minus 4b minus 60. And so what we need to do here is just factor this thing right now, or factor it, and then know that if I have the product of some things and that equals 0, that means that either one or both of those things need to be equal to 0. So we need to factor b squared minus 4b minus 60. So what we want to do, we want to find 2 numbers whose sum is negative 4 and whose product is negative 60. So we want to find 2 numbers whose sum is equal to negative 4 and whose product is equal to negative 60. Now, given that their product is negative, we know there are different signs, and this tells us that their absolute values are going to be 4 apart, that one is going to be 4 less than the other. So you can look at the factors of 60."}, {"video_title": "Finding dimensions of triangle from area Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So what we want to do, we want to find 2 numbers whose sum is negative 4 and whose product is negative 60. So we want to find 2 numbers whose sum is equal to negative 4 and whose product is equal to negative 60. Now, given that their product is negative, we know there are different signs, and this tells us that their absolute values are going to be 4 apart, that one is going to be 4 less than the other. So you can look at the factors of 60. 1 and 60 are too far apart. If you made one of them negative, you would either get positive 59 as a sum or negative 59 as a sum. 2 and 30, still too far apart."}, {"video_title": "Finding dimensions of triangle from area Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So you can look at the factors of 60. 1 and 60 are too far apart. If you made one of them negative, you would either get positive 59 as a sum or negative 59 as a sum. 2 and 30, still too far apart. 3 and 20, sorry, still too far apart. If you had made one negative, you would either get negative 17 or positive 17. Then you can have 4 and 15, still too far apart."}, {"video_title": "Finding dimensions of triangle from area Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "2 and 30, still too far apart. 3 and 20, sorry, still too far apart. If you had made one negative, you would either get negative 17 or positive 17. Then you can have 4 and 15, still too far apart. If you made one of them negative, their sum would be either negative 11 or positive 11. Then you have 5 and 12, still seems too far apart from each other. One of them is negative, then you would either have their sum being positive 7 or negative 7."}, {"video_title": "Finding dimensions of triangle from area Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Then you can have 4 and 15, still too far apart. If you made one of them negative, their sum would be either negative 11 or positive 11. Then you have 5 and 12, still seems too far apart from each other. One of them is negative, then you would either have their sum being positive 7 or negative 7. Then you have 6 and 10. Now this looks interesting. They're 4 apart."}, {"video_title": "Finding dimensions of triangle from area Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "One of them is negative, then you would either have their sum being positive 7 or negative 7. Then you have 6 and 10. Now this looks interesting. They're 4 apart. So if we make, and we want the larger absolute magnitude number to be negative so that their sum is negative. So if we make it 6 and negative 10, their sum will be negative 4 and their product is negative 60. So that works."}, {"video_title": "Finding dimensions of triangle from area Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "They're 4 apart. So if we make, and we want the larger absolute magnitude number to be negative so that their sum is negative. So if we make it 6 and negative 10, their sum will be negative 4 and their product is negative 60. So that works. So you can literally say that this is equal to B plus 6 times B minus 10. B plus the A times B minus, or plus B minus the B. And let me be very careful here."}, {"video_title": "Finding dimensions of triangle from area Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So that works. So you can literally say that this is equal to B plus 6 times B minus 10. B plus the A times B minus, or plus B minus the B. And let me be very careful here. This B over here, I want to make it very clear, is different than the B that we're using in the equation. I just used this B here to say, look, we're looking for two numbers that add up to this second term right over here. It's a different B. I could have said X plus Y is equal to negative 4 and X times Y is equal to negative 60."}, {"video_title": "Finding dimensions of triangle from area Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "And let me be very careful here. This B over here, I want to make it very clear, is different than the B that we're using in the equation. I just used this B here to say, look, we're looking for two numbers that add up to this second term right over here. It's a different B. I could have said X plus Y is equal to negative 4 and X times Y is equal to negative 60. In fact, let me do it that way just so we don't get confused. So we could write X plus Y is equal to negative 4 and then we have X times Y is equal to negative 60. So we have B plus 6 times B plus Y. X is 6, Y is negative 10."}, {"video_title": "Finding dimensions of triangle from area Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "It's a different B. I could have said X plus Y is equal to negative 4 and X times Y is equal to negative 60. In fact, let me do it that way just so we don't get confused. So we could write X plus Y is equal to negative 4 and then we have X times Y is equal to negative 60. So we have B plus 6 times B plus Y. X is 6, Y is negative 10. And that is equal to 0. And you could, well, let's just solve this right here and then we'll go back and show you that you could also factor this by grouping. But just from this, we know that either one of these is equal to 0."}, {"video_title": "Finding dimensions of triangle from area Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So we have B plus 6 times B plus Y. X is 6, Y is negative 10. And that is equal to 0. And you could, well, let's just solve this right here and then we'll go back and show you that you could also factor this by grouping. But just from this, we know that either one of these is equal to 0. Either B plus 6 is equal to 0 or B minus 10 is equal to 0. If we subtract 6 from both sides of this equation, we get B is equal to negative 6. Or if you add 10 to both sides of this equation, you get B is equal to 10."}, {"video_title": "Finding dimensions of triangle from area Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "But just from this, we know that either one of these is equal to 0. Either B plus 6 is equal to 0 or B minus 10 is equal to 0. If we subtract 6 from both sides of this equation, we get B is equal to negative 6. Or if you add 10 to both sides of this equation, you get B is equal to 10. And those are our two solutions. You could put them back in and verify that they satisfy our constraints. Now, the other way that you could solve this, and we're going to get the exact same answer, is you could just break up this negative 4B into its constituents."}, {"video_title": "Finding dimensions of triangle from area Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Or if you add 10 to both sides of this equation, you get B is equal to 10. And those are our two solutions. You could put them back in and verify that they satisfy our constraints. Now, the other way that you could solve this, and we're going to get the exact same answer, is you could just break up this negative 4B into its constituents. So you could have broken this up into 0 is equal to B squared. And then you could have broken it up into plus 6B minus 10B minus 60. And then factor it by grouping."}, {"video_title": "Finding dimensions of triangle from area Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Now, the other way that you could solve this, and we're going to get the exact same answer, is you could just break up this negative 4B into its constituents. So you could have broken this up into 0 is equal to B squared. And then you could have broken it up into plus 6B minus 10B minus 60. And then factor it by grouping. Group these first two terms, group these second two terms. And you're going to add them together. The first one you can factor out a B."}, {"video_title": "Finding dimensions of triangle from area Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "And then factor it by grouping. Group these first two terms, group these second two terms. And you're going to add them together. The first one you can factor out a B. So you have B times B plus 6. The second one you can factor out a negative 10. So minus 10 times B plus 6."}, {"video_title": "Finding dimensions of triangle from area Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "The first one you can factor out a B. So you have B times B plus 6. The second one you can factor out a negative 10. So minus 10 times B plus 6. All that's equal to 0. And now you can factor out a B plus 6. So if you factor out a B plus 6 here, you get 0 is equal to B minus 10 times B plus 6."}, {"video_title": "Finding dimensions of triangle from area Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So minus 10 times B plus 6. All that's equal to 0. And now you can factor out a B plus 6. So if you factor out a B plus 6 here, you get 0 is equal to B minus 10 times B plus 6. We're literally just factoring out this out of the expression. You're just left with a B minus 10. It's the same thing that we did in one step over here."}, {"video_title": "Finding dimensions of triangle from area Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So if you factor out a B plus 6 here, you get 0 is equal to B minus 10 times B plus 6. We're literally just factoring out this out of the expression. You're just left with a B minus 10. It's the same thing that we did in one step over here. Whatever works for you. But either way, the solutions are either B is equal to negative 6 or B is equal to 10. And we have to be careful here."}, {"video_title": "Finding dimensions of triangle from area Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "It's the same thing that we did in one step over here. Whatever works for you. But either way, the solutions are either B is equal to negative 6 or B is equal to 10. And we have to be careful here. Remember, this is a word problem. We can't just state, oh, B could be negative 6 or B could be 10. We have to think about whether this makes sense in the context of the actual problem."}, {"video_title": "Finding dimensions of triangle from area Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "And we have to be careful here. Remember, this is a word problem. We can't just state, oh, B could be negative 6 or B could be 10. We have to think about whether this makes sense in the context of the actual problem. We're talking about lengths of triangles or lengths of the sides of triangles. We can't have a negative length. So because of that, the base of a triangle can't have length negative 6."}, {"video_title": "Finding dimensions of triangle from area Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "We have to think about whether this makes sense in the context of the actual problem. We're talking about lengths of triangles or lengths of the sides of triangles. We can't have a negative length. So because of that, the base of a triangle can't have length negative 6. So we can cross that out. So we actually only have one solution here. Almost made a careless mistake."}, {"video_title": "Finding dimensions of triangle from area Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So because of that, the base of a triangle can't have length negative 6. So we can cross that out. So we actually only have one solution here. Almost made a careless mistake. Forgot that we were dealing with a word problem. The only possible base is 10. And let's see."}, {"video_title": "Finding dimensions of triangle from area Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Almost made a careless mistake. Forgot that we were dealing with a word problem. The only possible base is 10. And let's see. They say find the height and the base. Once again, not done. So the base we're saying is 10."}, {"video_title": "Finding dimensions of triangle from area Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "And let's see. They say find the height and the base. Once again, not done. So the base we're saying is 10. The height is 4 inches less. It's B minus 4. So the height is 6."}, {"video_title": "Verifying inverse functions by composition not inverse High School Math Khan Academy.mp3", "Sentence": "Let's say that f of x is equal to two x minus three, and g of x, g of x, is equal to 1 1\u20442 x plus three. What I want to do in this video is evaluate what f of g of x is, f of g of x, and then I want to evaluate what g of f of x is. So first I want to evaluate f of g of x, and then I'm going to evaluate the other way around. I'm going to evaluate g of f of x. But let's evaluate f of g of x first. And like always, I encourage you to pause the video and see if you can work through it. So this is going to be equal to, f of g of x is going to be equal to, well wherever we see the x in our definition for f of x, the input now is g of x."}, {"video_title": "Verifying inverse functions by composition not inverse High School Math Khan Academy.mp3", "Sentence": "I'm going to evaluate g of f of x. But let's evaluate f of g of x first. And like always, I encourage you to pause the video and see if you can work through it. So this is going to be equal to, f of g of x is going to be equal to, well wherever we see the x in our definition for f of x, the input now is g of x. So we'd replace it with a g of x. It's going to be two times g of x, two times g of x minus three, minus three, and this is going to be equal to two times, well g of x is all of that business, two times 1 1\u20442 x plus three, and then we have the minus three, minus three, and now we can distribute this two, two times 1 1\u20442 x is just going to be equal to x, two times three is going to be six, so x plus six minus three. And so this is going to be, this is going to equal x plus three."}, {"video_title": "Verifying inverse functions by composition not inverse High School Math Khan Academy.mp3", "Sentence": "So this is going to be equal to, f of g of x is going to be equal to, well wherever we see the x in our definition for f of x, the input now is g of x. So we'd replace it with a g of x. It's going to be two times g of x, two times g of x minus three, minus three, and this is going to be equal to two times, well g of x is all of that business, two times 1 1\u20442 x plus three, and then we have the minus three, minus three, and now we can distribute this two, two times 1 1\u20442 x is just going to be equal to x, two times three is going to be six, so x plus six minus three. And so this is going to be, this is going to equal x plus three. X plus three, all right, interesting. That's f of g of x. Now let's think about what g of f of x is going to be."}, {"video_title": "Verifying inverse functions by composition not inverse High School Math Khan Academy.mp3", "Sentence": "And so this is going to be, this is going to equal x plus three. X plus three, all right, interesting. That's f of g of x. Now let's think about what g of f of x is going to be. So g of, our input instead of being, instead of calling our input x, we're gonna call our input f of x. So g of f of x is going to be equal to, is going to be equal to, 1 1\u20442 times our input, which in this case is f of x, 1 1\u20442 times f of x plus three, plus three. You can view the x up here as the placeholder for whatever our input happens to be, and now our input is going to be f of x."}, {"video_title": "Verifying inverse functions by composition not inverse High School Math Khan Academy.mp3", "Sentence": "Now let's think about what g of f of x is going to be. So g of, our input instead of being, instead of calling our input x, we're gonna call our input f of x. So g of f of x is going to be equal to, is going to be equal to, 1 1\u20442 times our input, which in this case is f of x, 1 1\u20442 times f of x plus three, plus three. You can view the x up here as the placeholder for whatever our input happens to be, and now our input is going to be f of x. And so this is going to be equal to 1 1\u20442 times, what is f of x? It is two x minus three. So two times x minus three, and then we have a plus three."}, {"video_title": "Verifying inverse functions by composition not inverse High School Math Khan Academy.mp3", "Sentence": "You can view the x up here as the placeholder for whatever our input happens to be, and now our input is going to be f of x. And so this is going to be equal to 1 1\u20442 times, what is f of x? It is two x minus three. So two times x minus three, and then we have a plus three. And now we can distribute the 1 1\u20442. 1 1\u20442 times two x is going to be x. 1 1\u20442 times negative three is negative 3\u20442, and then we have a plus three."}, {"video_title": "Verifying inverse functions by composition not inverse High School Math Khan Academy.mp3", "Sentence": "So two times x minus three, and then we have a plus three. And now we can distribute the 1 1\u20442. 1 1\u20442 times two x is going to be x. 1 1\u20442 times negative three is negative 3\u20442, and then we have a plus three. So let's see, three is the same thing as 6\u20442, so 6\u20442 minus 3\u20442 is going to be 3\u20442. So this is going to be equal to x plus 3\u20442. So notice, we definitely got different things for f of g of x and g of f of x, and we also didn't do a round trip."}, {"video_title": "Verifying inverse functions by composition not inverse High School Math Khan Academy.mp3", "Sentence": "1 1\u20442 times negative three is negative 3\u20442, and then we have a plus three. So let's see, three is the same thing as 6\u20442, so 6\u20442 minus 3\u20442 is going to be 3\u20442. So this is going to be equal to x plus 3\u20442. So notice, we definitely got different things for f of g of x and g of f of x, and we also didn't do a round trip. We didn't go back to x. So we know that these are not inverses of each other. In fact, we just had to do either this or that to know that they're not inverses of each other."}, {"video_title": "Verifying inverse functions by composition not inverse High School Math Khan Academy.mp3", "Sentence": "So notice, we definitely got different things for f of g of x and g of f of x, and we also didn't do a round trip. We didn't go back to x. So we know that these are not inverses of each other. In fact, we just had to do either this or that to know that they're not inverses of each other. So these are not inverses. So let me write it this way. f of x does not equal the inverse of g of x."}, {"video_title": "Verifying inverse functions by composition not inverse High School Math Khan Academy.mp3", "Sentence": "In fact, we just had to do either this or that to know that they're not inverses of each other. So these are not inverses. So let me write it this way. f of x does not equal the inverse of g of x. Does not equal the inverse of g of x. And g of x does not equal the inverse of f of x. Does not equal the inverse of f of x."}, {"video_title": "Verifying inverse functions by composition not inverse High School Math Khan Academy.mp3", "Sentence": "f of x does not equal the inverse of g of x. Does not equal the inverse of g of x. And g of x does not equal the inverse of f of x. Does not equal the inverse of f of x. In order for them to be inverses, if you have an x value right over here, and if you apply g to it, if you input it into g, and that takes you to g of x, so that takes you to g of x right over here, so that's the function g, and then you apply f to it, you would have to get back to the same place. So g inverse would get us back to the same place. And clearly we did not get back to the same place."}, {"video_title": "Verifying inverse functions by composition not inverse High School Math Khan Academy.mp3", "Sentence": "Does not equal the inverse of f of x. In order for them to be inverses, if you have an x value right over here, and if you apply g to it, if you input it into g, and that takes you to g of x, so that takes you to g of x right over here, so that's the function g, and then you apply f to it, you would have to get back to the same place. So g inverse would get us back to the same place. And clearly we did not get back to the same place. We didn't get back to x, we got back to x plus three. Same thing over here. We see that we did not get back to x, we got to x plus 3 1\u20442."}, {"video_title": "Adding complex numbers Imaginary and complex numbers Algebra II Khan Academy.mp3", "Sentence": "And as we'll see when we're adding complex numbers, you can only add the real parts to each other and you can only add the imaginary parts to each other. So let's add the real parts. So we have a 5 plus a 3. And then the imaginary parts, we have a 2i, so plus 2i. And then we have a negative 7i, or we're subtracting 7i. So minus 7i right over here. And 5 plus 3, that's pretty straightforward."}, {"video_title": "Adding complex numbers Imaginary and complex numbers Algebra II Khan Academy.mp3", "Sentence": "And then the imaginary parts, we have a 2i, so plus 2i. And then we have a negative 7i, or we're subtracting 7i. So minus 7i right over here. And 5 plus 3, that's pretty straightforward. That's just going to be 8. And then if I have 2 of something, and from that I subtract 7 of that something, and in this case the something is the imaginary unit, the number i. If I have 2i's and I take away 7i's, then I have negative 5i's."}, {"video_title": "Adding complex numbers Imaginary and complex numbers Algebra II Khan Academy.mp3", "Sentence": "And 5 plus 3, that's pretty straightforward. That's just going to be 8. And then if I have 2 of something, and from that I subtract 7 of that something, and in this case the something is the imaginary unit, the number i. If I have 2i's and I take away 7i's, then I have negative 5i's. 2 minus 7 is negative 5. So then I have negative 5i. So when you add these two complex numbers, you get 8 minus 5i."}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So we're told this is the graph of function f right over here, and then they tell us that function g is defined as g of x is equal to 1 3rd f of x. What is the graph of g? And if we were doing this on Khan Academy, this is a screenshot from our mobile app, it has multiple choices, but I thought we would just try to sketch it. So pause this video, maybe in your mind, imagine what you think the graph of g is going to look like, or at least how you would tackle it. All right, so g of x is equal to 1 3rd f of x. So for example, we can see here that f of three is equal to negative three. So g of three should be 1 3rd that, so it should be negative one."}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So pause this video, maybe in your mind, imagine what you think the graph of g is going to look like, or at least how you would tackle it. All right, so g of x is equal to 1 3rd f of x. So for example, we can see here that f of three is equal to negative three. So g of three should be 1 3rd that, so it should be negative one. Likewise, so g of three would be right over there. And likewise, g of negative three, what would that be? Well, f of negative three is three."}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So g of three should be 1 3rd that, so it should be negative one. Likewise, so g of three would be right over there. And likewise, g of negative three, what would that be? Well, f of negative three is three. So g of negative three is going to be 1 3rd that, or it's going to be equal to one. F of zero is zero, 1 3rd of that is still zero, so g of zero is still going to be right over there. And we know that's going to happen there and there as well."}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Well, f of negative three is three. So g of negative three is going to be 1 3rd that, or it's going to be equal to one. F of zero is zero, 1 3rd of that is still zero, so g of zero is still going to be right over there. And we know that's going to happen there and there as well. And so we already have a sense of what this graph is going to look like. The function g is going to look something like this. I'm just connecting the dots."}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And we know that's going to happen there and there as well. And so we already have a sense of what this graph is going to look like. The function g is going to look something like this. I'm just connecting the dots. And they did give us some dots that we can use as reference points. So the graph of g is going to look something like this. It gets a little bit flattened out or a little bit squooshed or smooshed a little bit to look something like that, and you would pick the choice that looks like that."}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "I'm just connecting the dots. And they did give us some dots that we can use as reference points. So the graph of g is going to look something like this. It gets a little bit flattened out or a little bit squooshed or smooshed a little bit to look something like that, and you would pick the choice that looks like that. Let's do another example. So here, we're told this is the graph of f of x, and it's defined by this expression. What is the graph of g of x, and g of x is this?"}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "It gets a little bit flattened out or a little bit squooshed or smooshed a little bit to look something like that, and you would pick the choice that looks like that. Let's do another example. So here, we're told this is the graph of f of x, and it's defined by this expression. What is the graph of g of x, and g of x is this? So pause this video and think about it again. All right, now the key realization is is it looks like g of x is, if you were to take all the terms of f of x and multiply it by two, or at least if you were to multiply the absolute value by two, and then if you were to multiply this negative two by two. So it looks like g of x is equal to two times f of x."}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "What is the graph of g of x, and g of x is this? So pause this video and think about it again. All right, now the key realization is is it looks like g of x is, if you were to take all the terms of f of x and multiply it by two, or at least if you were to multiply the absolute value by two, and then if you were to multiply this negative two by two. So it looks like g of x is equal to two times f of x. And we could even set up a little table here. This is another way that we can think about it. We can think about x."}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So it looks like g of x is equal to two times f of x. And we could even set up a little table here. This is another way that we can think about it. We can think about x. We can think about f of x. And now we can think about g of x, which should be two times that. So we can see that when x is equal to zero, f of x is equal to one."}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "We can think about x. We can think about f of x. And now we can think about g of x, which should be two times that. So we can see that when x is equal to zero, f of x is equal to one. So g of x should be equal to two, because it's two times f of x. So g of x is going to be equal to, or g of zero, I should say, is going to be equal to two. What about when at x equals, well, say when x equals three?"}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So we can see that when x is equal to zero, f of x is equal to one. So g of x should be equal to two, because it's two times f of x. So g of x is going to be equal to, or g of zero, I should say, is going to be equal to two. What about when at x equals, well, say when x equals three? When x equals three, f of x is negative two. G of x is going to be two times that, because it's two times f of x. So it's going to be negative four."}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "What about when at x equals, well, say when x equals three? When x equals three, f of x is negative two. G of x is going to be two times that, because it's two times f of x. So it's going to be negative four. So g of x, or I should say g of three, is going to be negative four. It's going to be right over there. And then maybe let's think about one more point."}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So it's going to be negative four. So g of x, or I should say g of three, is going to be negative four. It's going to be right over there. And then maybe let's think about one more point. So f of five is equal to zero. G of five is going to be two times that, which is still going to be equal to zero. So it's going to be right over there."}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And then maybe let's think about one more point. So f of five is equal to zero. G of five is going to be two times that, which is still going to be equal to zero. So it's going to be right over there. And so the graph is going to look something like this. I'm just really just connecting, I'm just connecting the dots, trying to draw some straight lines. It's going to look something like this."}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So it's going to be right over there. And so the graph is going to look something like this. I'm just really just connecting, I'm just connecting the dots, trying to draw some straight lines. It's going to look something like this. You can see it's kind of stretched in the vertical direction. So if you were doing this on Khan Academy, it'd be multiple choice. You'd look for the graph that looks like that."}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "It's going to look something like this. You can see it's kind of stretched in the vertical direction. So if you were doing this on Khan Academy, it'd be multiple choice. You'd look for the graph that looks like that. Let's do a few more examples. So here we're given a function g is a vertically scaled version of f. So we can see that g is a vertically scaled version of f. The functions are graphed where f is a solid and g is dash. Yeah, we see that."}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "You'd look for the graph that looks like that. Let's do a few more examples. So here we're given a function g is a vertically scaled version of f. So we can see that g is a vertically scaled version of f. The functions are graphed where f is a solid and g is dash. Yeah, we see that. What is the equation of g in terms of f? So pause this video and try to think about it. Well, the way that I would tackle this is once again, let's do it with a table."}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Yeah, we see that. What is the equation of g in terms of f? So pause this video and try to think about it. Well, the way that I would tackle this is once again, let's do it with a table. Let's see the relationship between f and g. So this column is x. This column is f of x. And then this column is g of x."}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Well, the way that I would tackle this is once again, let's do it with a table. Let's see the relationship between f and g. So this column is x. This column is f of x. And then this column is g of x. Make another column right over here. And so let's see some interesting points. So when, actually, I could pick zero, but zero is maybe less interesting than this point over here."}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And then this column is g of x. Make another column right over here. And so let's see some interesting points. So when, actually, I could pick zero, but zero is maybe less interesting than this point over here. So this is when x is equal to negative three. F of negative three is negative three. What is g of negative three?"}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So when, actually, I could pick zero, but zero is maybe less interesting than this point over here. So this is when x is equal to negative three. F of negative three is negative three. What is g of negative three? It looks like it is negative nine. When f is, when x is zero, f of zero is negative two. What is g of zero?"}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "What is g of negative three? It looks like it is negative nine. When f is, when x is zero, f of zero is negative two. What is g of zero? It is equal to negative six. And so we already see a pattern forming. Whatever f is, g is three times that."}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "What is g of zero? It is equal to negative six. And so we already see a pattern forming. Whatever f is, g is three times that. Whatever f is, g is three times that. And so we don't even actually need these big columns, but we can see that g of x is equal to three times f of x. So that is the equation of g in terms of f."}, {"video_title": "What is a function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So I could have something like f of x, and x tends to be the variable most used for an input into the function, and the name of a function, f tends to be the most used variable, but we'll see that you can use others. But we could have f of x is equal to, let's say, x squared, if x is even, and let's say it is equal to x plus 5, if x is odd. So what would happen if we input 2 into this function? And the way that we would denote inputting 2 is that we would want to evaluate f of 2. This is saying let's input 2 into our function f. And everywhere we see this x here, this variable, you can kind of use this as a placeholder, let's replace it with our input. So let's see, if 2 is even, do 2 squared. If 2 is odd, do 2 plus 5."}, {"video_title": "What is a function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And the way that we would denote inputting 2 is that we would want to evaluate f of 2. This is saying let's input 2 into our function f. And everywhere we see this x here, this variable, you can kind of use this as a placeholder, let's replace it with our input. So let's see, if 2 is even, do 2 squared. If 2 is odd, do 2 plus 5. Well, 2 is even, so we're going to do 2 squared. So in this case, f of 2 is going to be 2 squared, or 4. Now, what would f of 3 be?"}, {"video_title": "What is a function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "If 2 is odd, do 2 plus 5. Well, 2 is even, so we're going to do 2 squared. So in this case, f of 2 is going to be 2 squared, or 4. Now, what would f of 3 be? Well, once again, everywhere we see this variable, we'll replace it with our input. So f of 3. So 3 squared if 3 is even, 3 plus 5 if 3 is odd."}, {"video_title": "What is a function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Now, what would f of 3 be? Well, once again, everywhere we see this variable, we'll replace it with our input. So f of 3. So 3 squared if 3 is even, 3 plus 5 if 3 is odd. Well, 3 is odd, so it's going to be 3 plus 5. It is going to be equal to 8. Now, you might say, okay, that's neat, Sal."}, {"video_title": "What is a function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So 3 squared if 3 is even, 3 plus 5 if 3 is odd. Well, 3 is odd, so it's going to be 3 plus 5. It is going to be equal to 8. Now, you might say, okay, that's neat, Sal. This was kind of an interesting way to define a function, a way to kind of munch on these numbers. But I could have done this with traditional equations in some way, especially if you allowed me to use the squirrely bracket thing. What can a function do that maybe my traditional toolkits might have not been as expressive about?"}, {"video_title": "What is a function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Now, you might say, okay, that's neat, Sal. This was kind of an interesting way to define a function, a way to kind of munch on these numbers. But I could have done this with traditional equations in some way, especially if you allowed me to use the squirrely bracket thing. What can a function do that maybe my traditional toolkits might have not been as expressive about? Well, you could even do a function like this. You could have, and let me not use f and x anymore, just to show you that the notation is more general than that. So I could say h of a is equal to the next largest number that starts with the same letter as variable a, and we're going to assume that we are dealing in English."}, {"video_title": "What is a function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "What can a function do that maybe my traditional toolkits might have not been as expressive about? Well, you could even do a function like this. You could have, and let me not use f and x anymore, just to show you that the notation is more general than that. So I could say h of a is equal to the next largest number that starts with the same letter as variable a, and we're going to assume that we are dealing in English. So given that, what is h of 2 going to be? Well, 2 starts with a t. What's the next largest number that starts with a t? Well, it's going to be equal to 3."}, {"video_title": "What is a function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So I could say h of a is equal to the next largest number that starts with the same letter as variable a, and we're going to assume that we are dealing in English. So given that, what is h of 2 going to be? Well, 2 starts with a t. What's the next largest number that starts with a t? Well, it's going to be equal to 3. Now, what would h of, I don't know, let's think about this. What would h of 8 be equal to? Well, 8 starts with an e. The next largest number that starts with an e, so it's not 9, 10, it would be 11."}, {"video_title": "What is a function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Well, it's going to be equal to 3. Now, what would h of, I don't know, let's think about this. What would h of 8 be equal to? Well, 8 starts with an e. The next largest number that starts with an e, so it's not 9, 10, it would be 11. And so now you see it's a very, very, very general tool. This function is, this h function that we just defined, we'll look at it, we'll look at the letter that the number starts with in English, so it's doing this really, really, really, really wacky thing. Now, not all functions have to be this wacky."}, {"video_title": "What is a function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Well, 8 starts with an e. The next largest number that starts with an e, so it's not 9, 10, it would be 11. And so now you see it's a very, very, very general tool. This function is, this h function that we just defined, we'll look at it, we'll look at the letter that the number starts with in English, so it's doing this really, really, really, really wacky thing. Now, not all functions have to be this wacky. In fact, you have already been dealing with functions. You have seen things like y is equal to x plus 1. This can be viewed as a function."}, {"video_title": "What is a function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Now, not all functions have to be this wacky. In fact, you have already been dealing with functions. You have seen things like y is equal to x plus 1. This can be viewed as a function. We could write this as y is a function of x, which is equal to x plus 1. If you give as an input, as you give as an input, so let me write it this way. For example, when x is 0, we could say f of 0 is equal to, well, you take 0, you add 1, it's equal to 1. f of 2 is equal to 2."}, {"video_title": "What is a function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "This can be viewed as a function. We could write this as y is a function of x, which is equal to x plus 1. If you give as an input, as you give as an input, so let me write it this way. For example, when x is 0, we could say f of 0 is equal to, well, you take 0, you add 1, it's equal to 1. f of 2 is equal to 2. You've already done this before. You've done things where you said, look, let me make a table of x, let me take and put our y's there. When x is 0, y is 1."}, {"video_title": "What is a function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "For example, when x is 0, we could say f of 0 is equal to, well, you take 0, you add 1, it's equal to 1. f of 2 is equal to 2. You've already done this before. You've done things where you said, look, let me make a table of x, let me take and put our y's there. When x is 0, y is 1. When x is 2, I'm sorry, this was when x is, I made a little mistake, where f of 2 is equal to 3. You've done this before with tables where you say, look, x and y. When x is 0, y is 1."}, {"video_title": "What is a function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "When x is 0, y is 1. When x is 2, I'm sorry, this was when x is, I made a little mistake, where f of 2 is equal to 3. You've done this before with tables where you say, look, x and y. When x is 0, y is 1. When x is 2, y is 3. You might say, well, what was the whole point of using the function notation here to say f of x is equal to x plus 1? The whole point is to think in these more general terms."}, {"video_title": "What is a function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "When x is 0, y is 1. When x is 2, y is 3. You might say, well, what was the whole point of using the function notation here to say f of x is equal to x plus 1? The whole point is to think in these more general terms. For something like this, you didn't really have to introduce function notations, but it doesn't hurt to introduce function notations because it makes it very clear that a function takes an input, takes my x, and this definition, it munches on it, it says, okay, x plus 1, and then it produces 1 more than it. Here, whatever the input is, the output is 1 more than that original function. I know what you're asking."}, {"video_title": "What is a function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "The whole point is to think in these more general terms. For something like this, you didn't really have to introduce function notations, but it doesn't hurt to introduce function notations because it makes it very clear that a function takes an input, takes my x, and this definition, it munches on it, it says, okay, x plus 1, and then it produces 1 more than it. Here, whatever the input is, the output is 1 more than that original function. I know what you're asking. All right, what is not a function then? Remember, we said a function is something that takes an input and produces only one possible output for that given input. For example, something like, and let me look at a visual way of thinking about a function this time, or a relationship, I should say."}, {"video_title": "What is a function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "I know what you're asking. All right, what is not a function then? Remember, we said a function is something that takes an input and produces only one possible output for that given input. For example, something like, and let me look at a visual way of thinking about a function this time, or a relationship, I should say. Let's say that's our y-axis, and this right over here is our x-axis. Let me draw a circle here that has radius 2. It's a circle of radius 2."}, {"video_title": "What is a function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "For example, something like, and let me look at a visual way of thinking about a function this time, or a relationship, I should say. Let's say that's our y-axis, and this right over here is our x-axis. Let me draw a circle here that has radius 2. It's a circle of radius 2. This is negative 2. This is positive 2. This is negative 2."}, {"video_title": "What is a function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "It's a circle of radius 2. This is negative 2. This is positive 2. This is negative 2. My circle, it's centered at the origin. It has radius 2, so that's my best attempt at drawing the circle. Let me fill it in."}, {"video_title": "What is a function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "This is negative 2. My circle, it's centered at the origin. It has radius 2, so that's my best attempt at drawing the circle. Let me fill it in. This is a circle. The equation of a circle, of this circle, is going to be x squared plus y squared is equal to the radius squared, is equal to 2 squared, or it's equal to 4. The question is, is this relationship between x and y, here I've expressed it as equation, here I've visually drawn all of the x's and y's that satisfy this equation, is this relationship between x and y a function?"}, {"video_title": "What is a function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Let me fill it in. This is a circle. The equation of a circle, of this circle, is going to be x squared plus y squared is equal to the radius squared, is equal to 2 squared, or it's equal to 4. The question is, is this relationship between x and y, here I've expressed it as equation, here I've visually drawn all of the x's and y's that satisfy this equation, is this relationship between x and y a function? We can see visually that it's not going to be a function. You pick a given x, let's say x is equal to 1. There's two possible y's that are associated with it, this y up here and this y down here."}, {"video_title": "What is a function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "The question is, is this relationship between x and y, here I've expressed it as equation, here I've visually drawn all of the x's and y's that satisfy this equation, is this relationship between x and y a function? We can see visually that it's not going to be a function. You pick a given x, let's say x is equal to 1. There's two possible y's that are associated with it, this y up here and this y down here. We can even solve for that by looking at the equation. When x is equal to 1, we get 1 squared plus y squared is equal to 4. 1 plus y squared is equal to 4."}, {"video_title": "What is a function Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "There's two possible y's that are associated with it, this y up here and this y down here. We can even solve for that by looking at the equation. When x is equal to 1, we get 1 squared plus y squared is equal to 4. 1 plus y squared is equal to 4. Or, subtracting 1 from both sides, y squared is equal to 3, or y is equal to the positive or the negative square root of 3. This right over here is the positive square root of 3, and this right over here is the negative square root of 3. This situation, this relationship, where I inputted a 1 into my little box here, and associated with the 1, I associate both a positive square root of 3 and a negative square root of 3, this is not a function."}, {"video_title": "Converting linear equations to slope-intercept form 8th grade Khan Academy.mp3", "Sentence": "We're asked to convert these linear equations into slope-intercept form, and then graph them on a single coordinate plane. We have our coordinate plane over here. And just as a bit of review, slope-intercept form is the form y is equal to mx plus b, where m is the slope and b is the intercept. That's why it's called slope-intercept form. So we just have to algebraically manipulate these equations into this form. So let's start with line A. Line A, it's in standard form right now."}, {"video_title": "Converting linear equations to slope-intercept form 8th grade Khan Academy.mp3", "Sentence": "That's why it's called slope-intercept form. So we just have to algebraically manipulate these equations into this form. So let's start with line A. Line A, it's in standard form right now. It's 4x plus 2y is equal to negative 8. The first thing I'd like to do is get rid of this 4x from the left-hand side, and the best way to do that is to subtract 4x from both sides of this equation. So let me subtract 4x from both sides."}, {"video_title": "Converting linear equations to slope-intercept form 8th grade Khan Academy.mp3", "Sentence": "Line A, it's in standard form right now. It's 4x plus 2y is equal to negative 8. The first thing I'd like to do is get rid of this 4x from the left-hand side, and the best way to do that is to subtract 4x from both sides of this equation. So let me subtract 4x from both sides. The left-hand side of the equation, these two 4x's cancel out, and I'm just left with 2y is equal to, and on the right-hand side, I have negative 4x minus 8, or negative 8 minus 4x, however you want to view it. Now, we're almost at slope-intercept form. We just have to get rid of this 2, and the best way to do that that I can think of is divide both sides of this equation by 2."}, {"video_title": "Converting linear equations to slope-intercept form 8th grade Khan Academy.mp3", "Sentence": "So let me subtract 4x from both sides. The left-hand side of the equation, these two 4x's cancel out, and I'm just left with 2y is equal to, and on the right-hand side, I have negative 4x minus 8, or negative 8 minus 4x, however you want to view it. Now, we're almost at slope-intercept form. We just have to get rid of this 2, and the best way to do that that I can think of is divide both sides of this equation by 2. So let's divide both sides by 2. So we divide the left-hand side by 2, and then divide the right-hand side by 2. You have to divide every term by 2."}, {"video_title": "Converting linear equations to slope-intercept form 8th grade Khan Academy.mp3", "Sentence": "We just have to get rid of this 2, and the best way to do that that I can think of is divide both sides of this equation by 2. So let's divide both sides by 2. So we divide the left-hand side by 2, and then divide the right-hand side by 2. You have to divide every term by 2. And then we are left with y is equal to negative 4 divided by 2 is negative 2x. Negative 8 divided by 2 is negative 4, negative 2x minus 4. So this is line A, and let me graph it right now."}, {"video_title": "Converting linear equations to slope-intercept form 8th grade Khan Academy.mp3", "Sentence": "You have to divide every term by 2. And then we are left with y is equal to negative 4 divided by 2 is negative 2x. Negative 8 divided by 2 is negative 4, negative 2x minus 4. So this is line A, and let me graph it right now. So line A, its y-intercept is negative 4, so the point 0, negative 4 is on this graph. If x is equal to 0, y is going to be equal to negative 4. You can just substitute that in the graph."}, {"video_title": "Converting linear equations to slope-intercept form 8th grade Khan Academy.mp3", "Sentence": "So this is line A, and let me graph it right now. So line A, its y-intercept is negative 4, so the point 0, negative 4 is on this graph. If x is equal to 0, y is going to be equal to negative 4. You can just substitute that in the graph. So 0, 1, 2, 3, 4, that's the point 0, negative 4. That's the y-intercept for line A. And then the slope is negative 2x."}, {"video_title": "Converting linear equations to slope-intercept form 8th grade Khan Academy.mp3", "Sentence": "You can just substitute that in the graph. So 0, 1, 2, 3, 4, that's the point 0, negative 4. That's the y-intercept for line A. And then the slope is negative 2x. So that means that if I change x by positive 1, that y goes down by negative 2. So let's do that. So if I go over 1 in the positive direction, I have to go down 2."}, {"video_title": "Converting linear equations to slope-intercept form 8th grade Khan Academy.mp3", "Sentence": "And then the slope is negative 2x. So that means that if I change x by positive 1, that y goes down by negative 2. So let's do that. So if I go over 1 in the positive direction, I have to go down 2. That's what a negative slope is going to do, negative 2 slope. If I go over 2, I'm going to have to go down 4. If I go back negative 1, so if I go in the x direction negative 1, that means in the y direction I go positive 2, because 2 divided by negative 1 is still negative 2."}, {"video_title": "Converting linear equations to slope-intercept form 8th grade Khan Academy.mp3", "Sentence": "So if I go over 1 in the positive direction, I have to go down 2. That's what a negative slope is going to do, negative 2 slope. If I go over 2, I'm going to have to go down 4. If I go back negative 1, so if I go in the x direction negative 1, that means in the y direction I go positive 2, because 2 divided by negative 1 is still negative 2. So I go over here. If I go back 2, I'm going to go up 4. Let me just do that a couple of times."}, {"video_title": "Converting linear equations to slope-intercept form 8th grade Khan Academy.mp3", "Sentence": "If I go back negative 1, so if I go in the x direction negative 1, that means in the y direction I go positive 2, because 2 divided by negative 1 is still negative 2. So I go over here. If I go back 2, I'm going to go up 4. Let me just do that a couple of times. Back 2 and then up 4. So this line is going to look like this. I'll do my best to draw it."}, {"video_title": "Converting linear equations to slope-intercept form 8th grade Khan Academy.mp3", "Sentence": "Let me just do that a couple of times. Back 2 and then up 4. So this line is going to look like this. I'll do my best to draw it. That's a decent job. That is line A right there. Let's do line B."}, {"video_title": "Converting linear equations to slope-intercept form 8th grade Khan Academy.mp3", "Sentence": "I'll do my best to draw it. That's a decent job. That is line A right there. Let's do line B. They say 4x is equal to negative 8. You might be saying, how do I get that into slope-intercept form? I don't see a y."}, {"video_title": "Converting linear equations to slope-intercept form 8th grade Khan Academy.mp3", "Sentence": "Let's do line B. They say 4x is equal to negative 8. You might be saying, how do I get that into slope-intercept form? I don't see a y. The answer is you won't be able to, because this can't be put into slope-intercept form, but we can simplify it. Let's divide both sides of this equation by 4. You divide both sides of this equation by 4, and you get x is equal to negative 2."}, {"video_title": "Converting linear equations to slope-intercept form 8th grade Khan Academy.mp3", "Sentence": "I don't see a y. The answer is you won't be able to, because this can't be put into slope-intercept form, but we can simplify it. Let's divide both sides of this equation by 4. You divide both sides of this equation by 4, and you get x is equal to negative 2. This just means, I don't care what your y is, x is always going to be equal to negative 2. So x is equal to negative 2 is right there. Negative 1, negative 2."}, {"video_title": "Converting linear equations to slope-intercept form 8th grade Khan Academy.mp3", "Sentence": "You divide both sides of this equation by 4, and you get x is equal to negative 2. This just means, I don't care what your y is, x is always going to be equal to negative 2. So x is equal to negative 2 is right there. Negative 1, negative 2. And x is just always going to be equal to negative 2. x is always going to be equal to negative 2 in both directions. This is a bit of a review. This is the x axis."}, {"video_title": "Converting linear equations to slope-intercept form 8th grade Khan Academy.mp3", "Sentence": "Negative 1, negative 2. And x is just always going to be equal to negative 2. x is always going to be equal to negative 2 in both directions. This is a bit of a review. This is the x axis. That's the y axis. I forgot to label them. Let's do this last character."}, {"video_title": "Converting linear equations to slope-intercept form 8th grade Khan Academy.mp3", "Sentence": "This is the x axis. That's the y axis. I forgot to label them. Let's do this last character. 2y is equal to negative 8. Line C. We have 2y is equal to negative 8. We can divide both sides of this equation by 2."}, {"video_title": "Converting linear equations to slope-intercept form 8th grade Khan Academy.mp3", "Sentence": "Let's do this last character. 2y is equal to negative 8. Line C. We have 2y is equal to negative 8. We can divide both sides of this equation by 2. We get y is equal to negative 4. You might say, hey Sal, that doesn't look like this form, slope-intercept form. But it is."}, {"video_title": "Converting linear equations to slope-intercept form 8th grade Khan Academy.mp3", "Sentence": "We can divide both sides of this equation by 2. We get y is equal to negative 4. You might say, hey Sal, that doesn't look like this form, slope-intercept form. But it is. It's just that the slope is 0. We could rewrite this as y is equal to 0x minus 4. Where the y-intercept is negative 4 and the slope is 0. y-intercept is negative 4. y-intercept is negative 4 right there."}, {"video_title": "Converting linear equations to slope-intercept form 8th grade Khan Academy.mp3", "Sentence": "But it is. It's just that the slope is 0. We could rewrite this as y is equal to 0x minus 4. Where the y-intercept is negative 4 and the slope is 0. y-intercept is negative 4. y-intercept is negative 4 right there. And the slope is 0. So if you move an arbitrary amount in the x direction, the y is not going to change. It's just going to stay at negative 4."}, {"video_title": "Converting linear equations to slope-intercept form 8th grade Khan Academy.mp3", "Sentence": "Where the y-intercept is negative 4 and the slope is 0. y-intercept is negative 4. y-intercept is negative 4 right there. And the slope is 0. So if you move an arbitrary amount in the x direction, the y is not going to change. It's just going to stay at negative 4. Let me do that a little bit neater. y is just going to stay at negative 4. Or you could just interpret it as y is equal to negative 4 no matter what x is."}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy.mp3", "Sentence": "So a function we can view as something, so I'll put a function in this box here, and it takes inputs, and for a given input, it's going to produce an output, which we call f of x. So for example, let's say that we have the function, let's say we have the function f of x is equal to two over x. So in this case, if, so let me see, so if that's my function f, if I were to input the number three, well, f of three that we're going to output, we know how to figure that out. We've defined it right over here. It's going to be equal to two over three. It's going to be equal to two over three. So we were able, for that input, we were able to find an output."}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy.mp3", "Sentence": "We've defined it right over here. It's going to be equal to two over three. It's going to be equal to two over three. So we were able, for that input, we were able to find an output. If our input was pi, then we input into our function, and then f of pi, when x is pi, we're going to output f of pi, which is equal to two over pi. So we could write this as two over pi. So we were able to find the output pretty easily."}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy.mp3", "Sentence": "So we were able, for that input, we were able to find an output. If our input was pi, then we input into our function, and then f of pi, when x is pi, we're going to output f of pi, which is equal to two over pi. So we could write this as two over pi. So we were able to find the output pretty easily. But now let's do something interesting. Let's attempt to input zero into the function. If we input zero, does the function tell us what we need to output?"}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy.mp3", "Sentence": "So we were able to find the output pretty easily. But now let's do something interesting. Let's attempt to input zero into the function. If we input zero, does the function tell us what we need to output? Does this definition tell us what we need to output? So if I attempt to put x equals zero, then this definition would say, f of zero would be two over zero, but two over zero is undefined. Let me write this, two over zero."}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy.mp3", "Sentence": "If we input zero, does the function tell us what we need to output? Does this definition tell us what we need to output? So if I attempt to put x equals zero, then this definition would say, f of zero would be two over zero, but two over zero is undefined. Let me write this, two over zero. This is undefined. This function definition does not tell us what to actually do with zero. It gives us an undefined answer."}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy.mp3", "Sentence": "Let me write this, two over zero. This is undefined. This function definition does not tell us what to actually do with zero. It gives us an undefined answer. So this function is not defined here. It gives a question mark. So this gets to the essence of what domain is."}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy.mp3", "Sentence": "It gives us an undefined answer. So this function is not defined here. It gives a question mark. So this gets to the essence of what domain is. Domain is the set of all inputs over which the function is defined. So the domain of this function f would be all real numbers except for x equals zero. So let me write down these big ideas."}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy.mp3", "Sentence": "So this gets to the essence of what domain is. Domain is the set of all inputs over which the function is defined. So the domain of this function f would be all real numbers except for x equals zero. So let me write down these big ideas. This is the domain, a domain of a function. Actually, let me write that out. Domain of a function, a domain of a function is the set of all inputs, inputs over which the function is defined, over which the function is defined, or the function has defined outputs, over which the function has defined outputs."}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy.mp3", "Sentence": "So let me write down these big ideas. This is the domain, a domain of a function. Actually, let me write that out. Domain of a function, a domain of a function is the set of all inputs, inputs over which the function is defined, over which the function is defined, or the function has defined outputs, over which the function has defined outputs. So the domain for this f in particular, so the domain for this one, if I wanted to say its domain, I could say, look, it's going to be the set, and this curly brackets, these are kind of typical mathy set notation. I could say, okay, it's going to be the set of, and I'm going to put curly brackets like that. Well, x can be a member, this little symbol means a member of the real numbers, but it can't just be any real number."}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy.mp3", "Sentence": "Domain of a function, a domain of a function is the set of all inputs, inputs over which the function is defined, over which the function is defined, or the function has defined outputs, over which the function has defined outputs. So the domain for this f in particular, so the domain for this one, if I wanted to say its domain, I could say, look, it's going to be the set, and this curly brackets, these are kind of typical mathy set notation. I could say, okay, it's going to be the set of, and I'm going to put curly brackets like that. Well, x can be a member, this little symbol means a member of the real numbers, but it can't just be any real number. It can be most of the real numbers except it cannot be zero because we don't know this definition. It's undefined when you put an input of zero. So x is a member of the real numbers, and when we write real numbers, we write it with this kind of double stroke right over here."}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy.mp3", "Sentence": "Well, x can be a member, this little symbol means a member of the real numbers, but it can't just be any real number. It can be most of the real numbers except it cannot be zero because we don't know this definition. It's undefined when you put an input of zero. So x is a member of the real numbers, and when we write real numbers, we write it with this kind of double stroke right over here. That's the set of all real numbers such that, but we have to put the exception, zero is not, x equals zero is not a member of that domain, such that x does not equal zero. Now let's make this a little bit more concrete by doing some more examples. The more examples we do, hopefully the clearer this will become."}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy.mp3", "Sentence": "So x is a member of the real numbers, and when we write real numbers, we write it with this kind of double stroke right over here. That's the set of all real numbers such that, but we have to put the exception, zero is not, x equals zero is not a member of that domain, such that x does not equal zero. Now let's make this a little bit more concrete by doing some more examples. The more examples we do, hopefully the clearer this will become. So let's say we have another function. Just to be clear, we don't always have to use f's and x's. We could say, let's say we have g of y is equal to the square root of y minus six."}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy.mp3", "Sentence": "The more examples we do, hopefully the clearer this will become. So let's say we have another function. Just to be clear, we don't always have to use f's and x's. We could say, let's say we have g of y is equal to the square root of y minus six. So what's the domain here? What is the set of all inputs over which this function g is defined? So here we are inputting a y into function g, and we're going to output g of y."}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy.mp3", "Sentence": "We could say, let's say we have g of y is equal to the square root of y minus six. So what's the domain here? What is the set of all inputs over which this function g is defined? So here we are inputting a y into function g, and we're going to output g of y. Well, it's going to be defined as long as whatever we have under the radical right over here is non-negative. If this becomes negative, our traditional principal root operator here is not defined. We need something that, if this ended up being a negative number, hey, how do you take the principal root of a negative number?"}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy.mp3", "Sentence": "So here we are inputting a y into function g, and we're going to output g of y. Well, it's going to be defined as long as whatever we have under the radical right over here is non-negative. If this becomes negative, our traditional principal root operator here is not defined. We need something that, if this ended up being a negative number, hey, how do you take the principal root of a negative number? And we're just saying this is kind of the traditional principal root operator. So y minus six needs to be greater than or equal to zero in order for g to be defined for that input y, or you could say, add six to both sides, y needs to be greater than or equal to six. Or you could say g is defined for any inputs y that are greater than or equal to six."}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy.mp3", "Sentence": "We need something that, if this ended up being a negative number, hey, how do you take the principal root of a negative number? And we're just saying this is kind of the traditional principal root operator. So y minus six needs to be greater than or equal to zero in order for g to be defined for that input y, or you could say, add six to both sides, y needs to be greater than or equal to six. Or you could say g is defined for any inputs y that are greater than or equal to six. So we could say the domain here, we could say that the domain here is the set of all y's that are a member of the real numbers such that y, such that they're also greater than or equal, such that they're also greater than or equal to six. So hopefully this is starting to make some sense. And we're always used to functions defined this way, but you could even see functions that are defined in fairly exotic ways."}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy.mp3", "Sentence": "Or you could say g is defined for any inputs y that are greater than or equal to six. So we could say the domain here, we could say that the domain here is the set of all y's that are a member of the real numbers such that y, such that they're also greater than or equal, such that they're also greater than or equal to six. So hopefully this is starting to make some sense. And we're always used to functions defined this way, but you could even see functions that are defined in fairly exotic ways. You could see a function, let me say h of x. h of x could be defined as, it literally could be defined as, well, h of x is gonna be one if x is equal to pi, and it's equal to zero if x is equal to three. Now what's the domain here? And I encourage you to pause the video and think about it."}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy.mp3", "Sentence": "And we're always used to functions defined this way, but you could even see functions that are defined in fairly exotic ways. You could see a function, let me say h of x. h of x could be defined as, it literally could be defined as, well, h of x is gonna be one if x is equal to pi, and it's equal to zero if x is equal to three. Now what's the domain here? And I encourage you to pause the video and think about it. Well, this function is actually only defined for two inputs. If you, we know h of pi, if you input pi into it, we know you're gonna output one. And we know that if you input three into it, h of three, when x equals three, you're going to, let me put some commas here, you're gonna get a zero."}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy.mp3", "Sentence": "And I encourage you to pause the video and think about it. Well, this function is actually only defined for two inputs. If you, we know h of pi, if you input pi into it, we know you're gonna output one. And we know that if you input three into it, h of three, when x equals three, you're going to, let me put some commas here, you're gonna get a zero. But if you input anything else, what's h of four going to be? Well, it hasn't defined it, it's undefined. What's h of negative one going to be?"}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy.mp3", "Sentence": "And we know that if you input three into it, h of three, when x equals three, you're going to, let me put some commas here, you're gonna get a zero. But if you input anything else, what's h of four going to be? Well, it hasn't defined it, it's undefined. What's h of negative one going to be? It hasn't defined it. So the domain here, the domain of h is literally, it's just literally going to be the two valid inputs that x can be are three and pi. Three and pi."}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy.mp3", "Sentence": "What's h of negative one going to be? It hasn't defined it. So the domain here, the domain of h is literally, it's just literally going to be the two valid inputs that x can be are three and pi. Three and pi. These are the only valid inputs. These are the only two numbers over which this function is actually defined. So this hopefully starts to give you a flavor of why we care about domains."}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy.mp3", "Sentence": "Three and pi. These are the only valid inputs. These are the only two numbers over which this function is actually defined. So this hopefully starts to give you a flavor of why we care about domains. Not all functions are defined over all real numbers. Some are defined for only a small subset of real numbers, or for some other thing, or only whole numbers, or natural numbers, or positive numbers, or negative numbers, or they have exceptions. So we'll see that as we do more and more examples."}, {"video_title": "Applying rational equations 2 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "The year before, Bill painted it by himself, but took twice as long as Anya did. How long did Anya and Bill take when each was painting the deck alone? So let's define some variables here. Let's define A as number of hours for Anya to paint a deck. So we could say that Anya paints or has 8 hours for one deck. Or we could invert this and say that she can do 1 over A decks per hour. Now let's do another variable for Bill, just like that."}, {"video_title": "Applying rational equations 2 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Let's define A as number of hours for Anya to paint a deck. So we could say that Anya paints or has 8 hours for one deck. Or we could invert this and say that she can do 1 over A decks per hour. Now let's do another variable for Bill, just like that. Let's define B as the number of hours for Bill to paint a deck. So for Bill, it takes him B hours per deck, per one deck. And this is the same thing as saying that he can do 1 deck per every B hours."}, {"video_title": "Applying rational equations 2 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Now let's do another variable for Bill, just like that. Let's define B as the number of hours for Bill to paint a deck. So for Bill, it takes him B hours per deck, per one deck. And this is the same thing as saying that he can do 1 deck per every B hours. Or another way to think of it is he can do 1 over B deck per hour. Now, when they work together, Anya and Bill stained a large porch deck in 8 hours. So let's write this down."}, {"video_title": "Applying rational equations 2 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And this is the same thing as saying that he can do 1 deck per every B hours. Or another way to think of it is he can do 1 over B deck per hour. Now, when they work together, Anya and Bill stained a large porch deck in 8 hours. So let's write this down. Anya plus Bill. This information right here. Anya and Bill stained a large porch deck in 8 hours."}, {"video_title": "Applying rational equations 2 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So let's write this down. Anya plus Bill. This information right here. Anya and Bill stained a large porch deck in 8 hours. Or 8 hours per one deck, which is the same thing as saying 1 deck per 8 hours. And this is going to be the combination of each of their rates. So this 1 deck per 8 hours is going to be equal to Anya's rate, 1 over A decks per hour, plus Bill's rate, plus 1 over B decks per hour."}, {"video_title": "Applying rational equations 2 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Anya and Bill stained a large porch deck in 8 hours. Or 8 hours per one deck, which is the same thing as saying 1 deck per 8 hours. And this is going to be the combination of each of their rates. So this 1 deck per 8 hours is going to be equal to Anya's rate, 1 over A decks per hour, plus Bill's rate, plus 1 over B decks per hour. So we have one equation set up. Let me scroll down a little bit. And I won't write the units anymore."}, {"video_title": "Applying rational equations 2 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So this 1 deck per 8 hours is going to be equal to Anya's rate, 1 over A decks per hour, plus Bill's rate, plus 1 over B decks per hour. So we have one equation set up. Let me scroll down a little bit. And I won't write the units anymore. So 1 over 1 over 8 is equal to 1 over A plus 1 over B. Now, we have two unknowns, so we need another equation if we're going to solve these. It tells us right here, the year before Bill painted it by himself, but took twice as long as Anya did."}, {"video_title": "Applying rational equations 2 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And I won't write the units anymore. So 1 over 1 over 8 is equal to 1 over A plus 1 over B. Now, we have two unknowns, so we need another equation if we're going to solve these. It tells us right here, the year before Bill painted it by himself, but took twice as long as Anya did. So the number of hours it takes Bill to paint the deck is twice as long as the number of hours it takes Anya to paint a deck. So B is equal to 2A. The number of hours Bill takes is twice as the number of hours Anya takes per deck."}, {"video_title": "Applying rational equations 2 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "It tells us right here, the year before Bill painted it by himself, but took twice as long as Anya did. So the number of hours it takes Bill to paint the deck is twice as long as the number of hours it takes Anya to paint a deck. So B is equal to 2A. The number of hours Bill takes is twice as the number of hours Anya takes per deck. So B is equal to 2A. So we can rewrite this equation as, I'll stick to the colors for now, we could say 1 eighth is equal to 1 over Anya. Instead of writing 1 over Bill, we would write, so plus 1 over, Bill is 2 times Anya."}, {"video_title": "Applying rational equations 2 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "The number of hours Bill takes is twice as the number of hours Anya takes per deck. So B is equal to 2A. So we can rewrite this equation as, I'll stick to the colors for now, we could say 1 eighth is equal to 1 over Anya. Instead of writing 1 over Bill, we would write, so plus 1 over, Bill is 2 times Anya. The number of hours Bill takes is 2 times the number of hours Anya takes. So 2 times Anya. And now we have one equation in one unknown, and we can solve for A."}, {"video_title": "Applying rational equations 2 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Instead of writing 1 over Bill, we would write, so plus 1 over, Bill is 2 times Anya. The number of hours Bill takes is 2 times the number of hours Anya takes. So 2 times Anya. And now we have one equation in one unknown, and we can solve for A. The easiest way to solve for A right here is if we just multiply both sides of the equation by 2A. So let's multiply both sides of this by 2A. Multiply the left side by 2A."}, {"video_title": "Applying rational equations 2 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And now we have one equation in one unknown, and we can solve for A. The easiest way to solve for A right here is if we just multiply both sides of the equation by 2A. So let's multiply both sides of this by 2A. Multiply the left side by 2A. Actually, let's multiply both sides by 8A, so we get rid of this 1 eighth as well. 8A, and then multiply the right hand side by 8A as well. The left hand side, 8A divided by 8 is just A."}, {"video_title": "Applying rational equations 2 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Multiply the left side by 2A. Actually, let's multiply both sides by 8A, so we get rid of this 1 eighth as well. 8A, and then multiply the right hand side by 8A as well. The left hand side, 8A divided by 8 is just A. A is equal to 1 over A times 8A is just going to be 8. 1 over 2A times 8A, 8A divided by 2A is 4. 8A over 2A is 4."}, {"video_title": "Applying rational equations 2 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "The left hand side, 8A divided by 8 is just A. A is equal to 1 over A times 8A is just going to be 8. 1 over 2A times 8A, 8A divided by 2A is 4. 8A over 2A is 4. So A, the number of hours it takes Anya to paint a deck, and I made it lowercase, which I shouldn't have. Well, these should all be capital As. So this is A is equal to 8A over A is 8."}, {"video_title": "Applying rational equations 2 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "8A over 2A is 4. So A, the number of hours it takes Anya to paint a deck, and I made it lowercase, which I shouldn't have. Well, these should all be capital As. So this is A is equal to 8A over A is 8. 8A over 2A is 4. So the number of hours it takes Anya to paint a deck, or A, is 12 hours. Now what are they asking over here?"}, {"video_title": "Applying rational equations 2 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So this is A is equal to 8A over A is 8. 8A over 2A is 4. So the number of hours it takes Anya to paint a deck, or A, is 12 hours. Now what are they asking over here? They're asking, how long did Anya and Bill take when each was painting alone? So we figured out Anya, it takes her 8 hours, and then we know that it takes Bill twice as long as Anya. Bill is 2 times A, so Bill is going to be 2 times 12."}, {"video_title": "Applying rational equations 2 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Now what are they asking over here? They're asking, how long did Anya and Bill take when each was painting alone? So we figured out Anya, it takes her 8 hours, and then we know that it takes Bill twice as long as Anya. Bill is 2 times A, so Bill is going to be 2 times 12. So Bill is equal to 2 times Anya, which is equal to 2 times 12 hours, which is equal to 24 hours. So when they're alone, it would take Anya 12 hours, it would take Bill 24 hours. When they do it combined, it takes 8 hours, which makes sense."}, {"video_title": "Applying rational equations 2 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Bill is 2 times A, so Bill is going to be 2 times 12. So Bill is equal to 2 times Anya, which is equal to 2 times 12 hours, which is equal to 24 hours. So when they're alone, it would take Anya 12 hours, it would take Bill 24 hours. When they do it combined, it takes 8 hours, which makes sense. Because if Bill took 12 hours by himself, combined they would take 6 hours. They would take half as long. But Bill isn't that efficient."}, {"video_title": "Applying rational equations 2 Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "When they do it combined, it takes 8 hours, which makes sense. Because if Bill took 12 hours by himself, combined they would take 6 hours. They would take half as long. But Bill isn't that efficient. He's not as efficient as Anya, so it takes them a little bit longer. It takes them 8 hours. But it makes sense that combined they're going to take less time than individually."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy (2).mp3", "Sentence": "So in a relation, you have a set of numbers that you can kind of view as the input into the relation. We call that the domain. You could view them as the set of numbers over which that relation is defined. And then you have a set of numbers that can be, you can view it as the output of the relation or what the numbers that can be associated with anything in domain. And we call that the range. And it's a fairly straightforward idea. So for example, let's say that the number one is in the domain and that we associate the number one with the number two in the range."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy (2).mp3", "Sentence": "And then you have a set of numbers that can be, you can view it as the output of the relation or what the numbers that can be associated with anything in domain. And we call that the range. And it's a fairly straightforward idea. So for example, let's say that the number one is in the domain and that we associate the number one with the number two in the range. So in this type of notation, you would say that the relation has one comma two, one comma two in its set of ordered pairs. These are two ways of saying the same thing. Now the relation can also say, hey, maybe if I have two, maybe that is associated with two as well."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy (2).mp3", "Sentence": "So for example, let's say that the number one is in the domain and that we associate the number one with the number two in the range. So in this type of notation, you would say that the relation has one comma two, one comma two in its set of ordered pairs. These are two ways of saying the same thing. Now the relation can also say, hey, maybe if I have two, maybe that is associated with two as well. So two is also associated with the number two. And so notice, I'm just building a bunch of associations. I've visually drawn them over here."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy (2).mp3", "Sentence": "Now the relation can also say, hey, maybe if I have two, maybe that is associated with two as well. So two is also associated with the number two. And so notice, I'm just building a bunch of associations. I've visually drawn them over here. Here I'm just doing them as ordered pairs. We could say that we have the number three. Three is in our domain."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy (2).mp3", "Sentence": "I've visually drawn them over here. Here I'm just doing them as ordered pairs. We could say that we have the number three. Three is in our domain. Our relation is defined for number three. And three is associated with, let's say, negative seven. So this is three and negative seven."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy (2).mp3", "Sentence": "Three is in our domain. Our relation is defined for number three. And three is associated with, let's say, negative seven. So this is three and negative seven. Now this type of relation right over here, where if you give me any member of the domain and I'm able to tell you exactly which member of the range is associated with it, this is also referred to as a function. And in a few seconds, I'll show you a relation that is not a function. Because over here, you pick any member of the domain, and the function really is just a relation."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy (2).mp3", "Sentence": "So this is three and negative seven. Now this type of relation right over here, where if you give me any member of the domain and I'm able to tell you exactly which member of the range is associated with it, this is also referred to as a function. And in a few seconds, I'll show you a relation that is not a function. Because over here, you pick any member of the domain, and the function really is just a relation. It's really just an association, sometimes called a mapping, between members of the domain and particular members of the range. So you give me any member of the domain, I'll tell you exactly which member of the range it maps to. You give me one, I say, hey, it definitely maps to two."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy (2).mp3", "Sentence": "Because over here, you pick any member of the domain, and the function really is just a relation. It's really just an association, sometimes called a mapping, between members of the domain and particular members of the range. So you give me any member of the domain, I'll tell you exactly which member of the range it maps to. You give me one, I say, hey, it definitely maps to two. You give me two, it definitely maps to two as well. You give me three, it's definitely associated with negative seven as well. So this relation is both a, it's obviously a relation, but it is also a function."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy (2).mp3", "Sentence": "You give me one, I say, hey, it definitely maps to two. You give me two, it definitely maps to two as well. You give me three, it's definitely associated with negative seven as well. So this relation is both a, it's obviously a relation, but it is also a function. Now to show you a relation that is not a function, imagine something like this. So once again, I'll draw a domain over here. And I do this big fuzzy cloud looking thing to show you that I'm not showing you all of the things in the domain."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy (2).mp3", "Sentence": "So this relation is both a, it's obviously a relation, but it is also a function. Now to show you a relation that is not a function, imagine something like this. So once again, I'll draw a domain over here. And I do this big fuzzy cloud looking thing to show you that I'm not showing you all of the things in the domain. I'm just picking specific examples. And let's say that this big fuzzy cloud looking thing is the range. And let's say in this relation, and I'll build it the same way that we built it over here."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy (2).mp3", "Sentence": "And I do this big fuzzy cloud looking thing to show you that I'm not showing you all of the things in the domain. I'm just picking specific examples. And let's say that this big fuzzy cloud looking thing is the range. And let's say in this relation, and I'll build it the same way that we built it over here. Let's say in this relation, one is associated with two. So let's build the set of ordered pairs. So one is associated with two."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy (2).mp3", "Sentence": "And let's say in this relation, and I'll build it the same way that we built it over here. Let's say in this relation, one is associated with two. So let's build the set of ordered pairs. So one is associated with two. Let's say that two is associated with negative three. So you'd have two, negative three over there. And let's say on top of that, we also associate one with the number four."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy (2).mp3", "Sentence": "So one is associated with two. Let's say that two is associated with negative three. So you'd have two, negative three over there. And let's say on top of that, we also associate one with the number four. So we also create an association with one with the number four. So we have the ordered pair one, comma, four. Now this is a relationship."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy (2).mp3", "Sentence": "And let's say on top of that, we also associate one with the number four. So we also create an association with one with the number four. So we have the ordered pair one, comma, four. Now this is a relationship. We have, it's defined for certain, if this was the whole relationship, then the entire domain is just the numbers one, two, actually just the numbers one and two. It's definitely a relation, but this is no longer a function. And the reason why it's no longer a function is if you tell me, OK, I'm giving you one in the domain."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy (2).mp3", "Sentence": "Now this is a relationship. We have, it's defined for certain, if this was the whole relationship, then the entire domain is just the numbers one, two, actually just the numbers one and two. It's definitely a relation, but this is no longer a function. And the reason why it's no longer a function is if you tell me, OK, I'm giving you one in the domain. What member of the range is one associated with it? Over here you say, well, I don't know. Is one associated with two?"}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy (2).mp3", "Sentence": "And the reason why it's no longer a function is if you tell me, OK, I'm giving you one in the domain. What member of the range is one associated with it? Over here you say, well, I don't know. Is one associated with two? Or is it associated with four? It could be either one. So you don't have a clear association."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy (2).mp3", "Sentence": "Is one associated with two? Or is it associated with four? It could be either one. So you don't have a clear association. If I give you one here, you're like, I don't know. Do I hand you a two or a four? That's not what a function does."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy (2).mp3", "Sentence": "So you don't have a clear association. If I give you one here, you're like, I don't know. Do I hand you a two or a four? That's not what a function does. A function says, oh, if you give me one, I know I'm giving you a two. You give me a two, I know I'm giving you a two. Now with that out of the way, let's actually try to tackle the problem right over here."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy (2).mp3", "Sentence": "That's not what a function does. A function says, oh, if you give me one, I know I'm giving you a two. You give me a two, I know I'm giving you a two. Now with that out of the way, let's actually try to tackle the problem right over here. So let's think about its domain and let's think about its range. So the domain here, the possible, you could view them as x values or inputs into this thing that could be a function. That's definitely a relation."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy (2).mp3", "Sentence": "Now with that out of the way, let's actually try to tackle the problem right over here. So let's think about its domain and let's think about its range. So the domain here, the possible, you could view them as x values or inputs into this thing that could be a function. That's definitely a relation. You could have a negative 3. You could have a negative 2. You could have a 0."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy (2).mp3", "Sentence": "That's definitely a relation. You could have a negative 3. You could have a negative 2. You could have a 0. You could have a, well, we already listed a negative 2, so that's right over there. Or you could have a positive 3. Those are the possible values that this relation is defined for, that you could input into this relation and figure out what it outputs."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy (2).mp3", "Sentence": "You could have a 0. You could have a, well, we already listed a negative 2, so that's right over there. Or you could have a positive 3. Those are the possible values that this relation is defined for, that you could input into this relation and figure out what it outputs. Now the range here, these are the possible outputs or the numbers that are associated with the numbers in the domain. The range includes 2, 4, 5, 6, and 8. I could have drawn this with a cloud like this and I could have done this with a cloud like this."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy (2).mp3", "Sentence": "Those are the possible values that this relation is defined for, that you could input into this relation and figure out what it outputs. Now the range here, these are the possible outputs or the numbers that are associated with the numbers in the domain. The range includes 2, 4, 5, 6, and 8. I could have drawn this with a cloud like this and I could have done this with a cloud like this. But here we're showing the exact numbers in the domain and the range. And now let's draw the actual associations. So negative 3 is associated with 2 or it's mapped to 2."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy (2).mp3", "Sentence": "I could have drawn this with a cloud like this and I could have done this with a cloud like this. But here we're showing the exact numbers in the domain and the range. And now let's draw the actual associations. So negative 3 is associated with 2 or it's mapped to 2. So negative 3 maps to 2 based on this ordered pair right over there. Then we have negative 2 is associated with 4. So negative 2 is associated with 4 based on this ordered pair right over there."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy (2).mp3", "Sentence": "So negative 3 is associated with 2 or it's mapped to 2. So negative 3 maps to 2 based on this ordered pair right over there. Then we have negative 2 is associated with 4. So negative 2 is associated with 4 based on this ordered pair right over there. Actually, that first ordered pair, I don't want to get you confused. It should just be this ordered pair right over here. Negative 3 is associated with 2."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy (2).mp3", "Sentence": "So negative 2 is associated with 4 based on this ordered pair right over there. Actually, that first ordered pair, I don't want to get you confused. It should just be this ordered pair right over here. Negative 3 is associated with 2. Then we have negative 2. We do that in a different color. We have negative 2 is associated with 4."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy (2).mp3", "Sentence": "Negative 3 is associated with 2. Then we have negative 2. We do that in a different color. We have negative 2 is associated with 4. Negative 2 is associated with 4. We have 0 is associated with 5. Sometimes people say it's mapped to 5."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy (2).mp3", "Sentence": "We have negative 2 is associated with 4. Negative 2 is associated with 4. We have 0 is associated with 5. Sometimes people say it's mapped to 5. We have negative 2 is mapped to 6. Now this is interesting. Negative 2 is already mapped to something."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy (2).mp3", "Sentence": "Sometimes people say it's mapped to 5. We have negative 2 is mapped to 6. Now this is interesting. Negative 2 is already mapped to something. Now this ordered pair is saying it's also mapped to 6. And then finally, I'll do this in a color that I haven't used yet, although I've used almost all of them. We have 3 is mapped to 8."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy (2).mp3", "Sentence": "Negative 2 is already mapped to something. Now this ordered pair is saying it's also mapped to 6. And then finally, I'll do this in a color that I haven't used yet, although I've used almost all of them. We have 3 is mapped to 8. So the question here, is this a function? And for it to be a function, for any member of the domain, you have to know what it's going to map to. It can only map to one member of the range."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy (2).mp3", "Sentence": "We have 3 is mapped to 8. So the question here, is this a function? And for it to be a function, for any member of the domain, you have to know what it's going to map to. It can only map to one member of the range. So negative 3, if you put negative 3 as the input into the function, you know it's going to output 2. If you put negative 2 into the input of the function, all of a sudden you get confused. Do I output 4 or do I output 6?"}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy (2).mp3", "Sentence": "It can only map to one member of the range. So negative 3, if you put negative 3 as the input into the function, you know it's going to output 2. If you put negative 2 into the input of the function, all of a sudden you get confused. Do I output 4 or do I output 6? So you don't know if you output 4 or you output 6. And because there's this confusion, this is not a function. You have a member of the domain that maps to multiple members of the range."}, {"video_title": "Graphing exponential growth & decay Mathematics I High School Math Khan Academy.mp3", "Sentence": "And they ask us graph the following exponential function. And they give us the function, h of x is equal to 27 times 1 3rd to the x. So our initial value is 27, and 1 3rd is our common ratio. It's written in kind of standard exponential form. And they give us this little graphing tool where we can define these two points, and we can also define, we can define a horizontal asymptote to construct our function. And these three things are enough to define, to graph an exponential if we know that it is an exponential function. So let's think about it a little bit."}, {"video_title": "Graphing exponential growth & decay Mathematics I High School Math Khan Academy.mp3", "Sentence": "It's written in kind of standard exponential form. And they give us this little graphing tool where we can define these two points, and we can also define, we can define a horizontal asymptote to construct our function. And these three things are enough to define, to graph an exponential if we know that it is an exponential function. So let's think about it a little bit. So the easiest thing that I could think of is, well, let's think about its initial value. Its initial value is going to be when x equals zero. X equals zero, it's 1 3rd to the zero power, which is just one."}, {"video_title": "Graphing exponential growth & decay Mathematics I High School Math Khan Academy.mp3", "Sentence": "So let's think about it a little bit. So the easiest thing that I could think of is, well, let's think about its initial value. Its initial value is going to be when x equals zero. X equals zero, it's 1 3rd to the zero power, which is just one. And so you're just left with 27 times one, or just 27. That's why we call this number here when you're written it in this form. You call this the initial value."}, {"video_title": "Graphing exponential growth & decay Mathematics I High School Math Khan Academy.mp3", "Sentence": "X equals zero, it's 1 3rd to the zero power, which is just one. And so you're just left with 27 times one, or just 27. That's why we call this number here when you're written it in this form. You call this the initial value. So when x is equal to zero, h of x is equal to 27. And we're graphing y equals h of x. So now let's graph another point."}, {"video_title": "Graphing exponential growth & decay Mathematics I High School Math Khan Academy.mp3", "Sentence": "You call this the initial value. So when x is equal to zero, h of x is equal to 27. And we're graphing y equals h of x. So now let's graph another point. So let's think about it a little bit. When x is equal to one, when x is equal to one, what is h of x? Well, it's going to be 1 3rd to the first power, which is just 1 3rd."}, {"video_title": "Graphing exponential growth & decay Mathematics I High School Math Khan Academy.mp3", "Sentence": "So now let's graph another point. So let's think about it a little bit. When x is equal to one, when x is equal to one, what is h of x? Well, it's going to be 1 3rd to the first power, which is just 1 3rd. And so 1 3rd times 27 is going to be nine. So when x is one, h of one is nine. And we can verify, well, and now let's just think about, let's think about the asymptote."}, {"video_title": "Graphing exponential growth & decay Mathematics I High School Math Khan Academy.mp3", "Sentence": "Well, it's going to be 1 3rd to the first power, which is just 1 3rd. And so 1 3rd times 27 is going to be nine. So when x is one, h of one is nine. And we can verify, well, and now let's just think about, let's think about the asymptote. So what's going to happen here when x becomes really, really, really, really, really big? Well, if I take 1 3rd to like a really large exponent, to say to the 10th power, to the 100th power, or to the 1,000th power, this thing right over here is going to start approaching zero as x becomes much, much, much, much larger. And so something that is approaching zero times 27, well, that's going to approach zero as well."}, {"video_title": "Graphing exponential growth & decay Mathematics I High School Math Khan Academy.mp3", "Sentence": "And we can verify, well, and now let's just think about, let's think about the asymptote. So what's going to happen here when x becomes really, really, really, really, really big? Well, if I take 1 3rd to like a really large exponent, to say to the 10th power, to the 100th power, or to the 1,000th power, this thing right over here is going to start approaching zero as x becomes much, much, much, much larger. And so something that is approaching zero times 27, well, that's going to approach zero as well. So we're going to have a horizontal asymptote at zero. And you can verify that this works for more than just the two points we thought about. When x is equal to two, this is telling us that the graph y equals h of x goes through the point two comma three."}, {"video_title": "Graphing exponential growth & decay Mathematics I High School Math Khan Academy.mp3", "Sentence": "And so something that is approaching zero times 27, well, that's going to approach zero as well. So we're going to have a horizontal asymptote at zero. And you can verify that this works for more than just the two points we thought about. When x is equal to two, this is telling us that the graph y equals h of x goes through the point two comma three. So h of two should be equal to three. And you can verify that that is indeed the case. If x is two, 1 3rd squared is nine, oh, sorry, 1 3rd squared is 1 9th, times 27 is three."}, {"video_title": "Graphing exponential growth & decay Mathematics I High School Math Khan Academy.mp3", "Sentence": "When x is equal to two, this is telling us that the graph y equals h of x goes through the point two comma three. So h of two should be equal to three. And you can verify that that is indeed the case. If x is two, 1 3rd squared is nine, oh, sorry, 1 3rd squared is 1 9th, times 27 is three. And we see that right over here. When x is two, h of two is three. So I feel pretty good about that."}, {"video_title": "Graphing exponential growth & decay Mathematics I High School Math Khan Academy.mp3", "Sentence": "If x is two, 1 3rd squared is nine, oh, sorry, 1 3rd squared is 1 9th, times 27 is three. And we see that right over here. When x is two, h of two is three. So I feel pretty good about that. Let's do another one of these. So graph the following exponential function. So same logic."}, {"video_title": "Graphing exponential growth & decay Mathematics I High School Math Khan Academy.mp3", "Sentence": "So I feel pretty good about that. Let's do another one of these. So graph the following exponential function. So same logic. When x is zero, the g of zero is just going to boil down to that initial value. And so let me scroll down. The initial value is negative 30."}, {"video_title": "Graphing exponential growth & decay Mathematics I High School Math Khan Academy.mp3", "Sentence": "So same logic. When x is zero, the g of zero is just going to boil down to that initial value. And so let me scroll down. The initial value is negative 30. And so let's think about when x is equal to one. When x is equal to one, two to the 1st power is just two. And so two times negative 30 is negative 60."}, {"video_title": "Graphing exponential growth & decay Mathematics I High School Math Khan Academy.mp3", "Sentence": "The initial value is negative 30. And so let's think about when x is equal to one. When x is equal to one, two to the 1st power is just two. And so two times negative 30 is negative 60. So when x is equal to one, the value of the graph is negative 60. Now let's think about this asymptote where that should sit. So let's think about what happens when x becomes really, really, really, really, really, really negative."}, {"video_title": "Graphing exponential growth & decay Mathematics I High School Math Khan Academy.mp3", "Sentence": "And so two times negative 30 is negative 60. So when x is equal to one, the value of the graph is negative 60. Now let's think about this asymptote where that should sit. So let's think about what happens when x becomes really, really, really, really, really, really negative. So when x is really negative, so two to the negative one power is 1 1st, two to the negative two is 1 4th, two to the negative three is 1 8th. As you get larger and larger negative, or higher magnitude negative values, or in other words, as x becomes more and more and more negative, two to that power is going to approach zero. And so negative 30 times something approaching zero is going to approach zero."}, {"video_title": "Graphing exponential growth & decay Mathematics I High School Math Khan Academy.mp3", "Sentence": "So let's think about what happens when x becomes really, really, really, really, really, really negative. So when x is really negative, so two to the negative one power is 1 1st, two to the negative two is 1 4th, two to the negative three is 1 8th. As you get larger and larger negative, or higher magnitude negative values, or in other words, as x becomes more and more and more negative, two to that power is going to approach zero. And so negative 30 times something approaching zero is going to approach zero. So this asymptote's in the right place. Our horizontal asymptote, as x approaches negative infinity, as we move further and further to the left, the value of the function is going to approach zero. And we can see it kind of approaches zero from below."}, {"video_title": "Subtracting rational expressions factored denominators High School Math Khan Academy.mp3", "Sentence": "Pause this video and see if you can subtract this magenta rational expression from this yellow one. All right, now let's do this together. And the first thing that jumps out at you is that you realize that these don't have the same denominator and you would like them to have the same denominator. And so you might say, well, let me rewrite them so that they have a common denominator. And a common denominator that will work will be one that is divisible by each of these denominators. So it has all of the factors of each of these denominators. And lucky for us, each of these denominators are already factored."}, {"video_title": "Subtracting rational expressions factored denominators High School Math Khan Academy.mp3", "Sentence": "And so you might say, well, let me rewrite them so that they have a common denominator. And a common denominator that will work will be one that is divisible by each of these denominators. So it has all of the factors of each of these denominators. And lucky for us, each of these denominators are already factored. So let me just write the common denominator. I'll start rewriting the yellow expression. So you have the yellow expression."}, {"video_title": "Subtracting rational expressions factored denominators High School Math Khan Academy.mp3", "Sentence": "And lucky for us, each of these denominators are already factored. So let me just write the common denominator. I'll start rewriting the yellow expression. So you have the yellow expression. Actually, let me just make it clear. I'm gonna write both. So the yellow one, and then you're gonna subtract the magenta one."}, {"video_title": "Subtracting rational expressions factored denominators High School Math Khan Academy.mp3", "Sentence": "So you have the yellow expression. Actually, let me just make it clear. I'm gonna write both. So the yellow one, and then you're gonna subtract the magenta one. Whoops, I'm saying yellow but drawing in magenta. So you have the yellow expression, which I'm about to rewrite. Actually, I'm gonna make a longer line."}, {"video_title": "Subtracting rational expressions factored denominators High School Math Khan Academy.mp3", "Sentence": "So the yellow one, and then you're gonna subtract the magenta one. Whoops, I'm saying yellow but drawing in magenta. So you have the yellow expression, which I'm about to rewrite. Actually, I'm gonna make a longer line. So the yellow expression minus the magenta one. Minus the magenta one right over there. Now, as I mentioned, we wanna have a denominator that has all, the common denominator has to have, be divisible by both this yellow denominator and this magenta one."}, {"video_title": "Subtracting rational expressions factored denominators High School Math Khan Academy.mp3", "Sentence": "Actually, I'm gonna make a longer line. So the yellow expression minus the magenta one. Minus the magenta one right over there. Now, as I mentioned, we wanna have a denominator that has all, the common denominator has to have, be divisible by both this yellow denominator and this magenta one. So it's gotta have the z plus eight in it. It's gotta have the nine z minus five in it. And it's also gotta have both of these."}, {"video_title": "Subtracting rational expressions factored denominators High School Math Khan Academy.mp3", "Sentence": "Now, as I mentioned, we wanna have a denominator that has all, the common denominator has to have, be divisible by both this yellow denominator and this magenta one. So it's gotta have the z plus eight in it. It's gotta have the nine z minus five in it. And it's also gotta have both of these. Well, we already accounted for the nine z minus five, so it has to have, be divisible by z plus six. Z plus six. Notice, just by multiplying the denominator by z plus six, we're now divisible by both of these factors and both of these factors because nine z minus five was a factor common to both of them."}, {"video_title": "Subtracting rational expressions factored denominators High School Math Khan Academy.mp3", "Sentence": "And it's also gotta have both of these. Well, we already accounted for the nine z minus five, so it has to have, be divisible by z plus six. Z plus six. Notice, just by multiplying the denominator by z plus six, we're now divisible by both of these factors and both of these factors because nine z minus five was a factor common to both of them. And if you were just dealing with numbers when you were just adding or subtracting fractions, it works the exact same way. All right, so what will the numerator become? Well, we multiplied the denominator times z plus six, so we have to do the same thing to the numerator."}, {"video_title": "Subtracting rational expressions factored denominators High School Math Khan Academy.mp3", "Sentence": "Notice, just by multiplying the denominator by z plus six, we're now divisible by both of these factors and both of these factors because nine z minus five was a factor common to both of them. And if you were just dealing with numbers when you were just adding or subtracting fractions, it works the exact same way. All right, so what will the numerator become? Well, we multiplied the denominator times z plus six, so we have to do the same thing to the numerator. It's going to be negative z to the third times z plus six. Now let's focus over here. We had, well, we want the same denominator, so we could write this as z plus eight, z plus eight times z plus six times z plus six times nine z minus five."}, {"video_title": "Subtracting rational expressions factored denominators High School Math Khan Academy.mp3", "Sentence": "Well, we multiplied the denominator times z plus six, so we have to do the same thing to the numerator. It's going to be negative z to the third times z plus six. Now let's focus over here. We had, well, we want the same denominator, so we could write this as z plus eight, z plus eight times z plus six times z plus six times nine z minus five. And these are equivalent. I've just changed the order that we multiply in and that doesn't change their value. And if we multiplied the, so we had a three on top before, and if we multiplied the denominator times z plus eight, we also have to multiply the numerator times z plus eight."}, {"video_title": "Subtracting rational expressions factored denominators High School Math Khan Academy.mp3", "Sentence": "We had, well, we want the same denominator, so we could write this as z plus eight, z plus eight times z plus six times z plus six times nine z minus five. And these are equivalent. I've just changed the order that we multiply in and that doesn't change their value. And if we multiplied the, so we had a three on top before, and if we multiplied the denominator times z plus eight, we also have to multiply the numerator times z plus eight. Z plus eight. So there you go. And so this is going to be equal to, this is going to be equal to, actually, I'll just make a big line right over here."}, {"video_title": "Subtracting rational expressions factored denominators High School Math Khan Academy.mp3", "Sentence": "And if we multiplied the, so we had a three on top before, and if we multiplied the denominator times z plus eight, we also have to multiply the numerator times z plus eight. Z plus eight. So there you go. And so this is going to be equal to, this is going to be equal to, actually, I'll just make a big line right over here. This is all going to be equal to, we have our, probably don't need that much space. Let me see. Maybe that, maybe about that much."}, {"video_title": "Subtracting rational expressions factored denominators High School Math Khan Academy.mp3", "Sentence": "And so this is going to be equal to, this is going to be equal to, actually, I'll just make a big line right over here. This is all going to be equal to, we have our, probably don't need that much space. Let me see. Maybe that, maybe about that much. So I'm gonna have the same denominator. I'll just write it in neutral color now. Z plus eight times nine z minus five times z plus six."}, {"video_title": "Subtracting rational expressions factored denominators High School Math Khan Academy.mp3", "Sentence": "Maybe that, maybe about that much. So I'm gonna have the same denominator. I'll just write it in neutral color now. Z plus eight times nine z minus five times z plus six. So over here, just in this blue color, we want to distribute this negative z to the third. Negative z to the third times z is negative z to the fourth. Negative z to the third times six is minus six z to the third and now this negative sign right over here, actually, instead of saying negative z, negative of this entire thing, we could just say plus the negative of this or another way of thinking about it, you could view this as negative three times z plus eight."}, {"video_title": "Subtracting rational expressions factored denominators High School Math Khan Academy.mp3", "Sentence": "Z plus eight times nine z minus five times z plus six. So over here, just in this blue color, we want to distribute this negative z to the third. Negative z to the third times z is negative z to the fourth. Negative z to the third times six is minus six z to the third and now this negative sign right over here, actually, instead of saying negative z, negative of this entire thing, we could just say plus the negative of this or another way of thinking about it, you could view this as negative three times z plus eight. So we could just distribute that. So let's do that. So negative three times z is negative three z and negative three times eight is negative 24."}, {"video_title": "Subtracting rational expressions factored denominators High School Math Khan Academy.mp3", "Sentence": "Negative z to the third times six is minus six z to the third and now this negative sign right over here, actually, instead of saying negative z, negative of this entire thing, we could just say plus the negative of this or another way of thinking about it, you could view this as negative three times z plus eight. So we could just distribute that. So let's do that. So negative three times z is negative three z and negative three times eight is negative 24. And there you go. We are done. We found a common denominator and once you have a common denominator, you can just subtract or add the numerators and instead of viewing this as minus this entire thing, I viewed it as adding and then having a negative three in the numerator, distributing that and then these, I can't simplify it any further."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So before we even attempt to do this problem right here, let's just remind ourselves what a relation is and what type of relations can be functions. So in a relation, you have a set of numbers that you can kind of view as the input into the relation. We call that the domain. You could view them as the set of numbers over which that relation is defined. And then you have a set of numbers that can be, you can view it as the output of the relation or what the numbers that can be associated with anything in domain. And we call that the range. And it's a fairly straightforward idea."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "You could view them as the set of numbers over which that relation is defined. And then you have a set of numbers that can be, you can view it as the output of the relation or what the numbers that can be associated with anything in domain. And we call that the range. And it's a fairly straightforward idea. So for example, let's say that the number one is in the domain and that we associate the number one with the number two in the range. So in this type of notation, you would say that the relation has one comma two, one comma two in its set of ordered pairs. These are two ways of saying the same thing."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And it's a fairly straightforward idea. So for example, let's say that the number one is in the domain and that we associate the number one with the number two in the range. So in this type of notation, you would say that the relation has one comma two, one comma two in its set of ordered pairs. These are two ways of saying the same thing. Now the relation can also say, hey, maybe if I have two, maybe that is associated with two as well. So two is also associated with the number two. And so notice, I'm just building a bunch of associations."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "These are two ways of saying the same thing. Now the relation can also say, hey, maybe if I have two, maybe that is associated with two as well. So two is also associated with the number two. And so notice, I'm just building a bunch of associations. I've visually drawn them over here. Here I'm just doing them as ordered pairs. We could say that we have the number three."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And so notice, I'm just building a bunch of associations. I've visually drawn them over here. Here I'm just doing them as ordered pairs. We could say that we have the number three. Three is in our domain. Our relation is defined for number three. And three is associated with, let's say, negative seven."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "We could say that we have the number three. Three is in our domain. Our relation is defined for number three. And three is associated with, let's say, negative seven. So this is three and negative seven. Now this type of relation right over here, where if you give me any member of the domain and I'm able to tell you exactly which member of the range is associated with it, this is also referred to as a function. And in a few seconds, I'll show you a relation that is not a function."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And three is associated with, let's say, negative seven. So this is three and negative seven. Now this type of relation right over here, where if you give me any member of the domain and I'm able to tell you exactly which member of the range is associated with it, this is also referred to as a function. And in a few seconds, I'll show you a relation that is not a function. Because over here, you pick any member of the domain, and the function really is just a relation. It's really just an association, sometimes called a mapping, between members of the domain and particular members of the range. So you give me any member of the domain, I'll tell you exactly which member of the range it maps to."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And in a few seconds, I'll show you a relation that is not a function. Because over here, you pick any member of the domain, and the function really is just a relation. It's really just an association, sometimes called a mapping, between members of the domain and particular members of the range. So you give me any member of the domain, I'll tell you exactly which member of the range it maps to. You give me one, I say, hey, it definitely maps to two. You give me two, it definitely maps to two as well. You give me three, it's definitely associated with negative seven as well."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So you give me any member of the domain, I'll tell you exactly which member of the range it maps to. You give me one, I say, hey, it definitely maps to two. You give me two, it definitely maps to two as well. You give me three, it's definitely associated with negative seven as well. So this relation is both a, it's obviously a relation, but it is also a function. Now to show you a relation that is not a function, imagine something like this. So once again, I'll draw a domain over here."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "You give me three, it's definitely associated with negative seven as well. So this relation is both a, it's obviously a relation, but it is also a function. Now to show you a relation that is not a function, imagine something like this. So once again, I'll draw a domain over here. And I do this big fuzzy cloud looking thing to show you that I'm not showing you all of the things in the domain. I'm just picking specific examples. And let's say that this big fuzzy cloud looking thing is the range."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So once again, I'll draw a domain over here. And I do this big fuzzy cloud looking thing to show you that I'm not showing you all of the things in the domain. I'm just picking specific examples. And let's say that this big fuzzy cloud looking thing is the range. And let's say in this relation, and I'll build it the same way that we built it over here. Let's say in this relation, one is associated with two. So let's build the set of ordered pairs."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And let's say that this big fuzzy cloud looking thing is the range. And let's say in this relation, and I'll build it the same way that we built it over here. Let's say in this relation, one is associated with two. So let's build the set of ordered pairs. So one is associated with two. Let's say that two is associated with negative three. So you'd have two, negative three over there."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So let's build the set of ordered pairs. So one is associated with two. Let's say that two is associated with negative three. So you'd have two, negative three over there. And let's say on top of that, we also associate one with the number four. So we also create an association with one with the number four. So we have the ordered pair one, comma, four."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So you'd have two, negative three over there. And let's say on top of that, we also associate one with the number four. So we also create an association with one with the number four. So we have the ordered pair one, comma, four. Now this is a relationship. We have, it's defined for certain, if this was the whole relationship, then the entire domain is just the numbers one, two, actually just the numbers one and two. It's definitely a relation, but this is no longer a function."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So we have the ordered pair one, comma, four. Now this is a relationship. We have, it's defined for certain, if this was the whole relationship, then the entire domain is just the numbers one, two, actually just the numbers one and two. It's definitely a relation, but this is no longer a function. And the reason why it's no longer a function is if you tell me, OK, I'm giving you one in the domain. What member of the range is one associated with it? Over here you say, well, I don't know."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "It's definitely a relation, but this is no longer a function. And the reason why it's no longer a function is if you tell me, OK, I'm giving you one in the domain. What member of the range is one associated with it? Over here you say, well, I don't know. Is one associated with two? Or is it associated with four? It could be either one."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Over here you say, well, I don't know. Is one associated with two? Or is it associated with four? It could be either one. So you don't have a clear association. If I give you one here, you're like, I don't know. Do I hand you a two or a four?"}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "It could be either one. So you don't have a clear association. If I give you one here, you're like, I don't know. Do I hand you a two or a four? That's not what a function does. A function says, oh, if you give me one, I know I'm giving you a two. You give me a two, I know I'm giving you a two."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Do I hand you a two or a four? That's not what a function does. A function says, oh, if you give me one, I know I'm giving you a two. You give me a two, I know I'm giving you a two. Now with that out of the way, let's actually try to tackle the problem right over here. So let's think about its domain and let's think about its range. So the domain here, the possible, you could view them as x values or inputs into this thing that could be a function."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "You give me a two, I know I'm giving you a two. Now with that out of the way, let's actually try to tackle the problem right over here. So let's think about its domain and let's think about its range. So the domain here, the possible, you could view them as x values or inputs into this thing that could be a function. That's definitely a relation. You could have a negative 3. You could have a negative 2."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So the domain here, the possible, you could view them as x values or inputs into this thing that could be a function. That's definitely a relation. You could have a negative 3. You could have a negative 2. You could have a 0. You could have a, well, we already listed a negative 2, so that's right over there. Or you could have a positive 3."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "You could have a negative 2. You could have a 0. You could have a, well, we already listed a negative 2, so that's right over there. Or you could have a positive 3. Those are the possible values that this relation is defined for, that you could input into this relation and figure out what it outputs. Now the range here, these are the possible outputs or the numbers that are associated with the numbers in the domain. The range includes 2, 4, 5, 6, and 8."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Or you could have a positive 3. Those are the possible values that this relation is defined for, that you could input into this relation and figure out what it outputs. Now the range here, these are the possible outputs or the numbers that are associated with the numbers in the domain. The range includes 2, 4, 5, 6, and 8. I could have drawn this with a cloud like this and I could have done this with a cloud like this. But here we're showing the exact numbers in the domain and the range. And now let's draw the actual associations."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "The range includes 2, 4, 5, 6, and 8. I could have drawn this with a cloud like this and I could have done this with a cloud like this. But here we're showing the exact numbers in the domain and the range. And now let's draw the actual associations. So negative 3 is associated with 2 or it's mapped to 2. So negative 3 maps to 2 based on this ordered pair right over there. Then we have negative 2 is associated with 4."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And now let's draw the actual associations. So negative 3 is associated with 2 or it's mapped to 2. So negative 3 maps to 2 based on this ordered pair right over there. Then we have negative 2 is associated with 4. So negative 2 is associated with 4 based on this ordered pair right over there. Actually, that first ordered pair, I don't want to get you confused. It should just be this ordered pair right over here."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Then we have negative 2 is associated with 4. So negative 2 is associated with 4 based on this ordered pair right over there. Actually, that first ordered pair, I don't want to get you confused. It should just be this ordered pair right over here. Negative 3 is associated with 2. Then we have negative 2. We do that in a different color."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "It should just be this ordered pair right over here. Negative 3 is associated with 2. Then we have negative 2. We do that in a different color. We have negative 2 is associated with 4. Negative 2 is associated with 4. We have 0 is associated with 5."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "We do that in a different color. We have negative 2 is associated with 4. Negative 2 is associated with 4. We have 0 is associated with 5. Sometimes people say it's mapped to 5. We have negative 2 is mapped to 6. Now this is interesting."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "We have 0 is associated with 5. Sometimes people say it's mapped to 5. We have negative 2 is mapped to 6. Now this is interesting. Negative 2 is already mapped to something. Now this ordered pair is saying it's also mapped to 6. And then finally, I'll do this in a color that I haven't used yet, although I've used almost all of them."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Now this is interesting. Negative 2 is already mapped to something. Now this ordered pair is saying it's also mapped to 6. And then finally, I'll do this in a color that I haven't used yet, although I've used almost all of them. We have 3 is mapped to 8. So the question here, is this a function? And for it to be a function, for any member of the domain, you have to know what it's going to map to."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And then finally, I'll do this in a color that I haven't used yet, although I've used almost all of them. We have 3 is mapped to 8. So the question here, is this a function? And for it to be a function, for any member of the domain, you have to know what it's going to map to. It can only map to one member of the range. So negative 3, if you put negative 3 as the input into the function, you know it's going to output 2. If you put negative 2 into the input of the function, all of a sudden you get confused."}, {"video_title": "Relations and functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And for it to be a function, for any member of the domain, you have to know what it's going to map to. It can only map to one member of the range. So negative 3, if you put negative 3 as the input into the function, you know it's going to output 2. If you put negative 2 into the input of the function, all of a sudden you get confused. Do I output 4 or do I output 6? So you don't know if you output 4 or you output 6. And because there's this confusion, this is not a function."}, {"video_title": "Exponential model word problem medication dissolve High School Math Khan Academy.mp3", "Sentence": "Carlos has taken an initial dose of a prescription medication. The relationship between the elapsed time, t, in hours since he took the first dose, and the amount of medication, m of t, in milligrams in his bloodstream is modeled by the following function. All right. In how many hours will Carlos have one milligram of medication remaining in his bloodstream? So m of what t is equal to, so we need to essentially solve for m of t is equal to one milligram, because m of t outputs, whatever value it outputs is going to be in milligram. So let's just solve that. So m of t is, they give us a definition."}, {"video_title": "Exponential model word problem medication dissolve High School Math Khan Academy.mp3", "Sentence": "In how many hours will Carlos have one milligram of medication remaining in his bloodstream? So m of what t is equal to, so we need to essentially solve for m of t is equal to one milligram, because m of t outputs, whatever value it outputs is going to be in milligram. So let's just solve that. So m of t is, they give us a definition. Its model is an exponential function. 20 times e to the negative 0.8 t is equal to one. So let's see, we can divide both sides by 20, and so we will get e to the negative 0.8 t is equal to one over 20, one over 20, which we could write as 0.05, 0.05, I have a feeling we're gonna have to deal with decimals here regardless."}, {"video_title": "Exponential model word problem medication dissolve High School Math Khan Academy.mp3", "Sentence": "So m of t is, they give us a definition. Its model is an exponential function. 20 times e to the negative 0.8 t is equal to one. So let's see, we can divide both sides by 20, and so we will get e to the negative 0.8 t is equal to one over 20, one over 20, which we could write as 0.05, 0.05, I have a feeling we're gonna have to deal with decimals here regardless. And so how do we solve this? Well, one way to think about it, one way to think about it, what happens if we took the natural log of both sides? And just remember, a reminder, the natural log is the logarithm base e. So actually let me write this a little bit differently."}, {"video_title": "Exponential model word problem medication dissolve High School Math Khan Academy.mp3", "Sentence": "So let's see, we can divide both sides by 20, and so we will get e to the negative 0.8 t is equal to one over 20, one over 20, which we could write as 0.05, 0.05, I have a feeling we're gonna have to deal with decimals here regardless. And so how do we solve this? Well, one way to think about it, one way to think about it, what happens if we took the natural log of both sides? And just remember, a reminder, the natural log is the logarithm base e. So actually let me write this a little bit differently. This is zero, that is 0.05. So I'm gonna take the natural log of both sides. So ln, ln."}, {"video_title": "Exponential model word problem medication dissolve High School Math Khan Academy.mp3", "Sentence": "And just remember, a reminder, the natural log is the logarithm base e. So actually let me write this a little bit differently. This is zero, that is 0.05. So I'm gonna take the natural log of both sides. So ln, ln. So the natural log, this says, what power do I have to raise e to to get to e to the negative 0.8 t? Well, I've got to raise e to the, this simplifies to negative 0.8 t. Once again, natural log, this thing, let me clarify, ln of e to the negative 0.8 t, this is equivalent to if I were to write log base e of e to the negative 0.8 t. What power do I have to raise e to to get to e to the negative 0.8 t? We'll have to raise it to the negative 0.8 t power."}, {"video_title": "Exponential model word problem medication dissolve High School Math Khan Academy.mp3", "Sentence": "So ln, ln. So the natural log, this says, what power do I have to raise e to to get to e to the negative 0.8 t? Well, I've got to raise e to the, this simplifies to negative 0.8 t. Once again, natural log, this thing, let me clarify, ln of e to the negative 0.8 t, this is equivalent to if I were to write log base e of e to the negative 0.8 t. What power do I have to raise e to to get to e to the negative 0.8 t? We'll have to raise it to the negative 0.8 t power. So that's why the left-hand side's simplified to this, and that's going to be equal to the natural log, actually I'll just leave it in those terms, the natural log of 0.05, natural log of 0.05, all of that, and now we can divide both sides by negative 0.8 to solve for t. So let's do that. So we divide by negative 0.8, divide by negative 0.8, and so t is going to be equal to all of this business. On the left-hand side now we just have a t, and on the right-hand side we have all of this business, which I think a calculator will be valuable for."}, {"video_title": "Exponential model word problem medication dissolve High School Math Khan Academy.mp3", "Sentence": "We'll have to raise it to the negative 0.8 t power. So that's why the left-hand side's simplified to this, and that's going to be equal to the natural log, actually I'll just leave it in those terms, the natural log of 0.05, natural log of 0.05, all of that, and now we can divide both sides by negative 0.8 to solve for t. So let's do that. So we divide by negative 0.8, divide by negative 0.8, and so t is going to be equal to all of this business. On the left-hand side now we just have a t, and on the right-hand side we have all of this business, which I think a calculator will be valuable for. So let me get a calculator out, clear it out, and let's start with 0.05. Let's take the natural log, that's that button right over there, the natural log, we get that value, and we want to divide it by negative 0.8. So divide it by, divide it by 0.8 negative, so we're going to divide by 0.8 negative, is equal to, let's see, they want us to round to the nearest hundredth, so 3.74."}, {"video_title": "Exponential model word problem medication dissolve High School Math Khan Academy.mp3", "Sentence": "On the left-hand side now we just have a t, and on the right-hand side we have all of this business, which I think a calculator will be valuable for. So let me get a calculator out, clear it out, and let's start with 0.05. Let's take the natural log, that's that button right over there, the natural log, we get that value, and we want to divide it by negative 0.8. So divide it by, divide it by 0.8 negative, so we're going to divide by 0.8 negative, is equal to, let's see, they want us to round to the nearest hundredth, so 3.74. So it'll take 3.74, 7.4 hours for his dosage to go down to one milligram, where it actually started at 20 milligrams. When t equals zero, it's 20. After 3.74 hours, he's down in his bloodstream to one milligram, I guess his body has metabolized the rest of it in some way."}, {"video_title": "Rational equations intro Algebra 2 Khan Academy.mp3", "Sentence": "Pause this video and see if you can try this before we work through it together. All right, now let's work through this together. Now, the first thing that we might want to do, there's several ways that you could approach this, but the one thing I like to do is get rid of this x here in the denominator, and the easiest way I can think of doing that is by multiplying both sides of this equation by nine minus x. Now, when you do that, it's important that you then put the qualifier that the x cannot be equal to the value that would have made this denominator zero, because clearly if somehow you do all this algebraic manipulation and you got x is equal to nine, that still wouldn't be a valid solution, because if you were to substitute nine back into the original equation, you would be dividing by zero in the denominator. So let's just put that right over here. X cannot be equal to nine. And so then we can safely move ahead with our algebraic manipulations."}, {"video_title": "Rational equations intro Algebra 2 Khan Academy.mp3", "Sentence": "Now, when you do that, it's important that you then put the qualifier that the x cannot be equal to the value that would have made this denominator zero, because clearly if somehow you do all this algebraic manipulation and you got x is equal to nine, that still wouldn't be a valid solution, because if you were to substitute nine back into the original equation, you would be dividing by zero in the denominator. So let's just put that right over here. X cannot be equal to nine. And so then we can safely move ahead with our algebraic manipulations. So on the left-hand side, as long as x does not equal nine, if we multiply and divide by nine minus x, they cancel out, and we'll just be left with an x plus one. And on the right-hand side, if you multiply 2 3rds times nine minus x, we get 2 3rds times nine is six, and then 2 3rds times negative x is negative 2 3rds x. And once again, let's remind ourselves that x cannot be equal to nine."}, {"video_title": "Rational equations intro Algebra 2 Khan Academy.mp3", "Sentence": "And so then we can safely move ahead with our algebraic manipulations. So on the left-hand side, as long as x does not equal nine, if we multiply and divide by nine minus x, they cancel out, and we'll just be left with an x plus one. And on the right-hand side, if you multiply 2 3rds times nine minus x, we get 2 3rds times nine is six, and then 2 3rds times negative x is negative 2 3rds x. And once again, let's remind ourselves that x cannot be equal to nine. And then we can get all of our x's on the same side. So let's put that on the left. So let's add 2 3rds x to both sides."}, {"video_title": "Rational equations intro Algebra 2 Khan Academy.mp3", "Sentence": "And once again, let's remind ourselves that x cannot be equal to nine. And then we can get all of our x's on the same side. So let's put that on the left. So let's add 2 3rds x to both sides. So plus 2 3rds, 2 3rds x, plus 2 3rds x. And then what do we have? Well, on the left-hand side, we have one x, which is the same thing as 3 3rds x plus 2 3rds x is going to give us 5 3rds x plus one is equal to six, and then these characters cancel out."}, {"video_title": "Rational equations intro Algebra 2 Khan Academy.mp3", "Sentence": "So let's add 2 3rds x to both sides. So plus 2 3rds, 2 3rds x, plus 2 3rds x. And then what do we have? Well, on the left-hand side, we have one x, which is the same thing as 3 3rds x plus 2 3rds x is going to give us 5 3rds x plus one is equal to six, and then these characters cancel out. And then we can just subtract one from both sides, and we get 5 3rds x, 5 3rds x is equal to five. And then last but not least, we can multiply both sides of this equation times the reciprocal of 5 3rds, which is, of course, 3 5ths, and I'm doing that, so I just have an x isolated on the left-hand side. So times 3 5ths, and we are left with 3 5ths times 5 3rds is, of course, equal to one, so we're left with x is equal to five times 3 5ths is three."}, {"video_title": "Rational equations intro Algebra 2 Khan Academy.mp3", "Sentence": "Well, on the left-hand side, we have one x, which is the same thing as 3 3rds x plus 2 3rds x is going to give us 5 3rds x plus one is equal to six, and then these characters cancel out. And then we can just subtract one from both sides, and we get 5 3rds x, 5 3rds x is equal to five. And then last but not least, we can multiply both sides of this equation times the reciprocal of 5 3rds, which is, of course, 3 5ths, and I'm doing that, so I just have an x isolated on the left-hand side. So times 3 5ths, and we are left with 3 5ths times 5 3rds is, of course, equal to one, so we're left with x is equal to five times 3 5ths is three. And so we're feeling pretty good about x equals three, but we have to make sure that that's consistent with our original expression. Well, if we look up here, if you substitute back x equals three, you don't get a zero in the denominator. X is not equal to nine, x equals three is consistent with that, so we should feel good about our solution."}, {"video_title": "Dividing rational expressions Precalculus Khan Academy.mp3", "Sentence": "So if you are so inspired, I encourage you to pause the video and work on this on your own before we do this together. All right, now let's do this together. So this is completely analogous to dividing fractions. So if we were to divide the fraction six over 25 by the fraction 15 over nine, we know that we can rewrite this as six over 25 divided by 15 over nine, which we know is the same thing as six over 25 times nine over 15. And then we can factor the various numerators and the denominators. This is two times three. This is three times three."}, {"video_title": "Dividing rational expressions Precalculus Khan Academy.mp3", "Sentence": "So if we were to divide the fraction six over 25 by the fraction 15 over nine, we know that we can rewrite this as six over 25 divided by 15 over nine, which we know is the same thing as six over 25 times nine over 15. And then we can factor the various numerators and the denominators. This is two times three. This is three times three. This is five times five. This is five times three. Let's see, three in the numerator, three in the denominator."}, {"video_title": "Dividing rational expressions Precalculus Khan Academy.mp3", "Sentence": "This is three times three. This is five times five. This is five times three. Let's see, three in the numerator, three in the denominator. And actually that's about as far as we can get. So then we'll have two times three times three, which is going to be 18 in the numerator. And then in the denominator, we have five times five times five, which is 125."}, {"video_title": "Dividing rational expressions Precalculus Khan Academy.mp3", "Sentence": "Let's see, three in the numerator, three in the denominator. And actually that's about as far as we can get. So then we'll have two times three times three, which is going to be 18 in the numerator. And then in the denominator, we have five times five times five, which is 125. So we're going to do the exact same thing up here, but there's one extra complication. We have to keep track of the x values that would make this expression undefined in any way, because as we reduce to lowest terms, we might lose that information. But if we lose that information, then we have changed the expression."}, {"video_title": "Dividing rational expressions Precalculus Khan Academy.mp3", "Sentence": "And then in the denominator, we have five times five times five, which is 125. So we're going to do the exact same thing up here, but there's one extra complication. We have to keep track of the x values that would make this expression undefined in any way, because as we reduce to lowest terms, we might lose that information. But if we lose that information, then we have changed the expression. So we have to keep track of how we are constraining this domain. So first I can just rewrite this as x squared minus three x minus four, all of that over negative three x minus 15. And that's getting divided by, divided by this business, x squared minus 16 over we have x squared minus x minus 30."}, {"video_title": "Dividing rational expressions Precalculus Khan Academy.mp3", "Sentence": "But if we lose that information, then we have changed the expression. So we have to keep track of how we are constraining this domain. So first I can just rewrite this as x squared minus three x minus four, all of that over negative three x minus 15. And that's getting divided by, divided by this business, x squared minus 16 over we have x squared minus x minus 30. Now, the next thing we can do is we can factor the various numerators and denominators and think about which x values could get us into trouble. So x squared minus three x minus four, let's see, negative one, let's see, negative four times plus one would be negative four, and then these would add up to negative three. So I can rewrite this as x minus four times x plus one."}, {"video_title": "Dividing rational expressions Precalculus Khan Academy.mp3", "Sentence": "And that's getting divided by, divided by this business, x squared minus 16 over we have x squared minus x minus 30. Now, the next thing we can do is we can factor the various numerators and denominators and think about which x values could get us into trouble. So x squared minus three x minus four, let's see, negative one, let's see, negative four times plus one would be negative four, and then these would add up to negative three. So I can rewrite this as x minus four times x plus one. Rewrite it that way. And then I can rewrite what I have down here. I could factor out a negative three."}, {"video_title": "Dividing rational expressions Precalculus Khan Academy.mp3", "Sentence": "So I can rewrite this as x minus four times x plus one. Rewrite it that way. And then I can rewrite what I have down here. I could factor out a negative three. So I could write that as negative three times x plus five. And then I could write this one here. This is a difference of squares."}, {"video_title": "Dividing rational expressions Precalculus Khan Academy.mp3", "Sentence": "I could factor out a negative three. So I could write that as negative three times x plus five. And then I could write this one here. This is a difference of squares. It's going to be x plus four times x minus four. And then last but not least, this one over here, let's see, if I have a five and a six, negative six plus five is negative one, negative six times five is negative 30. So this is going to be x minus six times x plus five."}, {"video_title": "Dividing rational expressions Precalculus Khan Academy.mp3", "Sentence": "This is a difference of squares. It's going to be x plus four times x minus four. And then last but not least, this one over here, let's see, if I have a five and a six, negative six plus five is negative one, negative six times five is negative 30. So this is going to be x minus six times x plus five. Now, before I go any further, and the reason why I factored at that point is now we can think about what x values will get us in trouble. We know that any x values that make any of the denominators equal to zero, that would be undefined. So we wanna constrain our domain in that way."}, {"video_title": "Dividing rational expressions Precalculus Khan Academy.mp3", "Sentence": "So this is going to be x minus six times x plus five. Now, before I go any further, and the reason why I factored at that point is now we can think about what x values will get us in trouble. We know that any x values that make any of the denominators equal to zero, that would be undefined. So we wanna constrain our domain in that way. So we know, for example, that x cannot be equal to negative five. That would make this denominator equal to zero. Let me write that here."}, {"video_title": "Dividing rational expressions Precalculus Khan Academy.mp3", "Sentence": "So we wanna constrain our domain in that way. So we know, for example, that x cannot be equal to negative five. That would make this denominator equal to zero. Let me write that here. So x cannot be equal to negative five. We also know that x cannot be equal to six. X cannot be equal to six."}, {"video_title": "Dividing rational expressions Precalculus Khan Academy.mp3", "Sentence": "Let me write that here. So x cannot be equal to negative five. We also know that x cannot be equal to six. X cannot be equal to six. And this would also tell us that x cannot be equal to negative five, so I don't have to rewrite that again. But we're not done. So we've figured out the x values that make these denominators equal to zero."}, {"video_title": "Dividing rational expressions Precalculus Khan Academy.mp3", "Sentence": "X cannot be equal to six. And this would also tell us that x cannot be equal to negative five, so I don't have to rewrite that again. But we're not done. So we've figured out the x values that make these denominators equal to zero. But remember, we're also dividing by this entire expression here. So anything that would make the entire expression equal to zero would also be a problem because you can't divide by zero. So anything that would make this numerator equal to zero, which was this numerator right over here, would make us also divide by zero."}, {"video_title": "Dividing rational expressions Precalculus Khan Academy.mp3", "Sentence": "So we've figured out the x values that make these denominators equal to zero. But remember, we're also dividing by this entire expression here. So anything that would make the entire expression equal to zero would also be a problem because you can't divide by zero. So anything that would make this numerator equal to zero, which was this numerator right over here, would make us also divide by zero. So we have to constrain there. Not this numerator here. That one's fine."}, {"video_title": "Dividing rational expressions Precalculus Khan Academy.mp3", "Sentence": "So anything that would make this numerator equal to zero, which was this numerator right over here, would make us also divide by zero. So we have to constrain there. Not this numerator here. That one's fine. That one could be equal to zero. You can divide zero by other things. So let's see."}, {"video_title": "Dividing rational expressions Precalculus Khan Academy.mp3", "Sentence": "That one's fine. That one could be equal to zero. You can divide zero by other things. So let's see. We can see that x cannot be equal to negative four. And actually, x cannot be equal to positive four. So now we've fully constrained our domain, and now we can proceed."}, {"video_title": "Dividing rational expressions Precalculus Khan Academy.mp3", "Sentence": "So let's see. We can see that x cannot be equal to negative four. And actually, x cannot be equal to positive four. So now we've fully constrained our domain, and now we can proceed. So let me box this off right over here. And then we continue. I can rewrite all of this."}, {"video_title": "Dividing rational expressions Precalculus Khan Academy.mp3", "Sentence": "So now we've fully constrained our domain, and now we can proceed. So let me box this off right over here. And then we continue. I can rewrite all of this. So we're going to have x minus four times x plus one, all of that over negative three times x plus five. And now I'm just going to, instead of divide by this, I'm gonna multiply by the reciprocal. So this is going to be times, and I'm just gonna take the reciprocal, x minus six times x plus five."}, {"video_title": "Dividing rational expressions Precalculus Khan Academy.mp3", "Sentence": "I can rewrite all of this. So we're going to have x minus four times x plus one, all of that over negative three times x plus five. And now I'm just going to, instead of divide by this, I'm gonna multiply by the reciprocal. So this is going to be times, and I'm just gonna take the reciprocal, x minus six times x plus five. All of that over, we have x plus four times x minus four. And once again, our domain is constrained in this way. But we see we have an x minus four in the numerator now, x minus four in the denominator, x plus five in the denominator, x plus five in the numerator."}, {"video_title": "Dividing rational expressions Precalculus Khan Academy.mp3", "Sentence": "So this is going to be times, and I'm just gonna take the reciprocal, x minus six times x plus five. All of that over, we have x plus four times x minus four. And once again, our domain is constrained in this way. But we see we have an x minus four in the numerator now, x minus four in the denominator, x plus five in the denominator, x plus five in the numerator. And now we can say that this is going to be equal to x plus one times x minus six, all of that over negative three, negative three times x plus four, x plus four. So the way it's written now, we would still, it's clear that x cannot be equal to negative four. So this information, you can say it's already there in this expression now that we have reduced it to lowest terms."}, {"video_title": "Dividing rational expressions Precalculus Khan Academy.mp3", "Sentence": "But we see we have an x minus four in the numerator now, x minus four in the denominator, x plus five in the denominator, x plus five in the numerator. And now we can say that this is going to be equal to x plus one times x minus six, all of that over negative three, negative three times x plus four, x plus four. So the way it's written now, we would still, it's clear that x cannot be equal to negative four. So this information, you can say it's already there in this expression now that we have reduced it to lowest terms. But this other information right over here, this has been lost. So if you want this expression to truly be equivalent to this expression up here, you would also have to say, comma, x cannot be equal to negative five, six, or positive four. You could throw the negative four in there if you like, but that one's already in the expression, so to speak."}, {"video_title": "Evaluating composite functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "And then we're asked, what is h of g of negative six? And the way it's written might look a little strange to you. This little circle that we have in between the h and the g, that's our function composition symbol. So function composition symbol. And one way to rewrite this, it might make a little bit more sense. So this h of g of negative six, you could rewrite this as, this is going to be the same thing as g of negative six and then h of that. So h of g of negative six."}, {"video_title": "Evaluating composite functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So function composition symbol. And one way to rewrite this, it might make a little bit more sense. So this h of g of negative six, you could rewrite this as, this is going to be the same thing as g of negative six and then h of that. So h of g of negative six. Notice I spoke this out the same way that I said this. This is h of g of negative six. This is h of g of negative six."}, {"video_title": "Evaluating composite functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So h of g of negative six. Notice I spoke this out the same way that I said this. This is h of g of negative six. This is h of g of negative six. I find the second notation far more intuitive, but it's good to become familiar with this function composition notation, this little circle, because you might see that sometime and you shouldn't stress. It's just the same thing as what we have right over here. Now what is h of g of negative six?"}, {"video_title": "Evaluating composite functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "This is h of g of negative six. I find the second notation far more intuitive, but it's good to become familiar with this function composition notation, this little circle, because you might see that sometime and you shouldn't stress. It's just the same thing as what we have right over here. Now what is h of g of negative six? We just have to remind ourselves that this means that we're going to take the number negative six, we're going to input it into our function g, and then that will output g of negative six, whatever that number is, we'll figure it out in a second, and then we're going to input that into our function h. We're going to input that into our function h, and then what we output is going to be h of g of negative six which is what we want to figure out. g, h of g of negative six. We just have to do it one step at a time."}, {"video_title": "Evaluating composite functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Now what is h of g of negative six? We just have to remind ourselves that this means that we're going to take the number negative six, we're going to input it into our function g, and then that will output g of negative six, whatever that number is, we'll figure it out in a second, and then we're going to input that into our function h. We're going to input that into our function h, and then what we output is going to be h of g of negative six which is what we want to figure out. g, h of g of negative six. We just have to do it one step at a time. A lot of times when you first start looking at these function composition, it seems really convoluted and confusing, but you just have to, I want you to take a breath and take it one step at a time. Well, let's figure out what g of negative six is. It's going to evaluate to a number in this case, and then we input that number into the function h, and then we'll figure out another, and that's going to map to another number."}, {"video_title": "Evaluating composite functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "We just have to do it one step at a time. A lot of times when you first start looking at these function composition, it seems really convoluted and confusing, but you just have to, I want you to take a breath and take it one step at a time. Well, let's figure out what g of negative six is. It's going to evaluate to a number in this case, and then we input that number into the function h, and then we'll figure out another, and that's going to map to another number. So g of negative six, let's figure that out. G of negative six is equal to negative six squared plus five times negative six minus three which is equal to positive 36 minus 30 minus three. So that's equal to what?"}, {"video_title": "Evaluating composite functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "It's going to evaluate to a number in this case, and then we input that number into the function h, and then we'll figure out another, and that's going to map to another number. So g of negative six, let's figure that out. G of negative six is equal to negative six squared plus five times negative six minus three which is equal to positive 36 minus 30 minus three. So that's equal to what? 36 minus 33 which is equal to three. So g of negative six is equal to three. When you input negative six into g, it outputs three."}, {"video_title": "Evaluating composite functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So that's equal to what? 36 minus 33 which is equal to three. So g of negative six is equal to three. When you input negative six into g, it outputs three. And so h of g of negative six has now simplified to just h of three, because g of negative six is three. So let's figure out what h of three is. H of three."}, {"video_title": "Evaluating composite functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "When you input negative six into g, it outputs three. And so h of g of negative six has now simplified to just h of three, because g of negative six is three. So let's figure out what h of three is. H of three. Notice, whatever we outputted from g, we're inputting that now into h. So that's the number three. So h of three is going to be three times three minus one, three minus one squared minus five, which is equal to three times two squared, this is two right over here, minus five, which is equal to three times four minus five, which is equal to 12 minus five, which is equal to seven. And we're done."}, {"video_title": "Evaluating composite functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "H of three. Notice, whatever we outputted from g, we're inputting that now into h. So that's the number three. So h of three is going to be three times three minus one, three minus one squared minus five, which is equal to three times two squared, this is two right over here, minus five, which is equal to three times four minus five, which is equal to 12 minus five, which is equal to seven. And we're done. So you input negative six into g, you get three, and then you take that output from g and you put it into h, and you get seven. So this right over here is seven. All of this has come out to be equal to seven."}, {"video_title": "Evaluating composite functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "And we're done. So you input negative six into g, you get three, and then you take that output from g and you put it into h, and you get seven. So this right over here is seven. All of this has come out to be equal to seven. So h of g of negative six is equal to seven. H of g of negative six is equal to seven. Input negative six into g, take that output and input it into h, and you're going to get seven."}, {"video_title": "Example 2 Adding polynomials Algebra I Khan Academy.mp3", "Sentence": "And the first thing we can do, we can actually get rid of these parentheses right here, because we have this whole expression and then we're adding it to this whole expression. The parentheses really don't change our order of operations here. So let me just rewrite it once without the parentheses. So we have 5x squared plus 8x minus 3 plus 2x squared. If this was a minus, then we'd have to distribute the negative sign, but it's not. So plus 2x squared minus 7x plus 13x. Now let's just look at the different terms that have different degrees of x."}, {"video_title": "Example 2 Adding polynomials Algebra I Khan Academy.mp3", "Sentence": "So we have 5x squared plus 8x minus 3 plus 2x squared. If this was a minus, then we'd have to distribute the negative sign, but it's not. So plus 2x squared minus 7x plus 13x. Now let's just look at the different terms that have different degrees of x. Let's start with the x squared terms. So you have a 5x squared term here and you have a 2x squared term right there. So 5 of something plus 2 of that same something is going to be 7 of that something."}, {"video_title": "Example 2 Adding polynomials Algebra I Khan Academy.mp3", "Sentence": "Now let's just look at the different terms that have different degrees of x. Let's start with the x squared terms. So you have a 5x squared term here and you have a 2x squared term right there. So 5 of something plus 2 of that same something is going to be 7 of that something. So that's going to be 7x squared. And then let's look at the x terms here. So we have an 8x right there."}, {"video_title": "Example 2 Adding polynomials Algebra I Khan Academy.mp3", "Sentence": "So 5 of something plus 2 of that same something is going to be 7 of that something. So that's going to be 7x squared. And then let's look at the x terms here. So we have an 8x right there. We have a minus 7x and then we have a plus 13x. So if you have 8 of something minus 7 of something, you're just going to have 1 of that something. And if you add 14 of that something more, you're going to have 15."}, {"video_title": "Example 2 Adding polynomials Algebra I Khan Academy.mp3", "Sentence": "So we have an 8x right there. We have a minus 7x and then we have a plus 13x. So if you have 8 of something minus 7 of something, you're just going to have 1 of that something. And if you add 14 of that something more, you're going to have 15. So this is going to be plus 15x. 8x minus 7x. Oh, sorry, you're going to have 14x."}, {"video_title": "Example 2 Adding polynomials Algebra I Khan Academy.mp3", "Sentence": "And if you add 14 of that something more, you're going to have 15. So this is going to be plus 15x. 8x minus 7x. Oh, sorry, you're going to have 14x. 8 minus 7 is 1 plus 13 is 14. Plus 14x. That's these three terms."}, {"video_title": "Example 2 Adding polynomials Algebra I Khan Academy.mp3", "Sentence": "Oh, sorry, you're going to have 14x. 8 minus 7 is 1 plus 13 is 14. Plus 14x. That's these three terms. 8x minus 7x plus 13x. And then finally, you have a negative 3 or minus 3, depending on how you want to view it. And that's the only constant term."}, {"video_title": "Example 2 Adding polynomials Algebra I Khan Academy.mp3", "Sentence": "That's these three terms. 8x minus 7x plus 13x. And then finally, you have a negative 3 or minus 3, depending on how you want to view it. And that's the only constant term. You can imagine it's, you could say it's x times x to the 0. But it's a constant term. It's not being multiplied by x."}, {"video_title": "Example 2 Adding polynomials Algebra I Khan Academy.mp3", "Sentence": "And that's the only constant term. You can imagine it's, you could say it's x times x to the 0. But it's a constant term. It's not being multiplied by x. And that's the only one there. So minus 3. And we've simplified it as far as we can go."}, {"video_title": "Compound inequalities 4 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "5x minus 3 is less than 12, and 4x plus 1 is greater than 25. So let's just solve for x in each of these constraints, and just keep in mind that any x has to satisfy both of them, because it's an and over here. So first we have this 5x minus 3 is less than 12. So if we want to isolate the x, we can get rid of this negative 3 here by adding 3 to both sides. So let's add 3 to both sides of this inequality. The left hand side, we're just left with a 5x. The minus 3 and the plus 3 cancel out."}, {"video_title": "Compound inequalities 4 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "So if we want to isolate the x, we can get rid of this negative 3 here by adding 3 to both sides. So let's add 3 to both sides of this inequality. The left hand side, we're just left with a 5x. The minus 3 and the plus 3 cancel out. 5x is less than 12, plus 3 is 15. Now we can divide both sides by positive 5. That won't swap the inequality, since 5 is positive."}, {"video_title": "Compound inequalities 4 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "The minus 3 and the plus 3 cancel out. 5x is less than 12, plus 3 is 15. Now we can divide both sides by positive 5. That won't swap the inequality, since 5 is positive. So we divide both sides by positive 5, and we are left with, just from this constraint, that x is less than 15 over 5, which is 3. So that's that constraint over here. But we have the second constraint as well."}, {"video_title": "Compound inequalities 4 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "That won't swap the inequality, since 5 is positive. So we divide both sides by positive 5, and we are left with, just from this constraint, that x is less than 15 over 5, which is 3. So that's that constraint over here. But we have the second constraint as well. We have this one. We have 4x plus 1 is greater than 25. So very similarly, we can subtract 1 from both sides to get rid of that 1 on the left hand side."}, {"video_title": "Compound inequalities 4 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "But we have the second constraint as well. We have this one. We have 4x plus 1 is greater than 25. So very similarly, we can subtract 1 from both sides to get rid of that 1 on the left hand side. And we get 4x, the 1's cancel out, is greater than 25 minus 1 is 24. Divide both sides by positive 4. Don't have to do anything to the inequality, since it's a positive number."}, {"video_title": "Compound inequalities 4 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "So very similarly, we can subtract 1 from both sides to get rid of that 1 on the left hand side. And we get 4x, the 1's cancel out, is greater than 25 minus 1 is 24. Divide both sides by positive 4. Don't have to do anything to the inequality, since it's a positive number. And we get x is greater than 24 over 4 is 6. And remember, there was that AND over here. We have this AND."}, {"video_title": "Compound inequalities 4 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "Don't have to do anything to the inequality, since it's a positive number. And we get x is greater than 24 over 4 is 6. And remember, there was that AND over here. We have this AND. So x has to be less than 3, and x has to be greater than 6. So already your brain might be realizing that this is a little bit strange. This first constraint says that x needs to be less than 3."}, {"video_title": "Compound inequalities 4 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "We have this AND. So x has to be less than 3, and x has to be greater than 6. So already your brain might be realizing that this is a little bit strange. This first constraint says that x needs to be less than 3. So this is 3 on the number line. We're saying x has to be less than 3. So it has to be in this shaded area right over there."}, {"video_title": "Compound inequalities 4 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "This first constraint says that x needs to be less than 3. So this is 3 on the number line. We're saying x has to be less than 3. So it has to be in this shaded area right over there. The second constraint says that x has to be greater than 6. So this is 6 over here. It says that x has to be greater than 6."}, {"video_title": "Compound inequalities 4 Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "So it has to be in this shaded area right over there. The second constraint says that x has to be greater than 6. So this is 6 over here. It says that x has to be greater than 6. It can't even include 6. And since we have this AND here, the only x's that are a solution for this compound inequality are the ones that satisfy both, the ones that are in the overlap of their solution set. But when you look at it right over here, it's clear that there is no overlap."}, {"video_title": "Parallel lines from equation (example 3) Mathematics I High School Math Khan Academy.mp3", "Sentence": "We're asked which of these lines are parallel. So they give us three equations of three different lines. And if they're parallel, then they have to have the same slope. So all we have to do over here is figure out the slopes of each of these lines. And if any of them are equal, they're parallel. So let's do line A. Line A, it's 2y is equal to 12x plus 10."}, {"video_title": "Parallel lines from equation (example 3) Mathematics I High School Math Khan Academy.mp3", "Sentence": "So all we have to do over here is figure out the slopes of each of these lines. And if any of them are equal, they're parallel. So let's do line A. Line A, it's 2y is equal to 12x plus 10. We're almost in slope-intercept form. We can just divide both sides of this equation by 2. Let's divide all the terms by 2."}, {"video_title": "Parallel lines from equation (example 3) Mathematics I High School Math Khan Academy.mp3", "Sentence": "Line A, it's 2y is equal to 12x plus 10. We're almost in slope-intercept form. We can just divide both sides of this equation by 2. Let's divide all the terms by 2. We get y is equal to 6x. 12 divided by 2. 6x plus 5."}, {"video_title": "Parallel lines from equation (example 3) Mathematics I High School Math Khan Academy.mp3", "Sentence": "Let's divide all the terms by 2. We get y is equal to 6x. 12 divided by 2. 6x plus 5. So our slope in this case, we have it in slope-intercept form, our slope in this case is equal to 6. Let's try line B. Line B is y is equal to 6."}, {"video_title": "Parallel lines from equation (example 3) Mathematics I High School Math Khan Academy.mp3", "Sentence": "6x plus 5. So our slope in this case, we have it in slope-intercept form, our slope in this case is equal to 6. Let's try line B. Line B is y is equal to 6. Now this might be, you might say, hey, this is a bizarre character. How do I get this into slope-intercept form? Where is the x?"}, {"video_title": "Parallel lines from equation (example 3) Mathematics I High School Math Khan Academy.mp3", "Sentence": "Line B is y is equal to 6. Now this might be, you might say, hey, this is a bizarre character. How do I get this into slope-intercept form? Where is the x? And my answer to you is that it already is in slope-intercept form. I could just rewrite it as y is equal to 0x plus 6. The x term is being multiplied by 0, because the slope here is 0. y is going to be equal to 6 no matter how much you change x."}, {"video_title": "Parallel lines from equation (example 3) Mathematics I High School Math Khan Academy.mp3", "Sentence": "Where is the x? And my answer to you is that it already is in slope-intercept form. I could just rewrite it as y is equal to 0x plus 6. The x term is being multiplied by 0, because the slope here is 0. y is going to be equal to 6 no matter how much you change x. Change in y is always going to be 0. It's always going to be 6. So here, our slope is 0."}, {"video_title": "Parallel lines from equation (example 3) Mathematics I High School Math Khan Academy.mp3", "Sentence": "The x term is being multiplied by 0, because the slope here is 0. y is going to be equal to 6 no matter how much you change x. Change in y is always going to be 0. It's always going to be 6. So here, our slope is 0. Our slope is equal to 0. So these two lines are definitely not parallel. They have different slopes."}, {"video_title": "Parallel lines from equation (example 3) Mathematics I High School Math Khan Academy.mp3", "Sentence": "So here, our slope is 0. Our slope is equal to 0. So these two lines are definitely not parallel. They have different slopes. Let's try line C. I'll do it down here. So that's y minus 2 is equal to 6 times x plus 2. And this is actually in point-slope form, where the point x is equal to negative 2, y is equal to 2."}, {"video_title": "Parallel lines from equation (example 3) Mathematics I High School Math Khan Academy.mp3", "Sentence": "They have different slopes. Let's try line C. I'll do it down here. So that's y minus 2 is equal to 6 times x plus 2. And this is actually in point-slope form, where the point x is equal to negative 2, y is equal to 2. So it's a point negative 2, 2 is being represented here, because you're subtracting the points. And the slope is 6. So we already know that the slope is equal to 6."}, {"video_title": "Parallel lines from equation (example 3) Mathematics I High School Math Khan Academy.mp3", "Sentence": "And this is actually in point-slope form, where the point x is equal to negative 2, y is equal to 2. So it's a point negative 2, 2 is being represented here, because you're subtracting the points. And the slope is 6. So we already know that the slope is equal to 6. And sometimes people are more comfortable with slope-intercept form. So let's put it in slope-intercept form just to confirm that if we put it in this form, the slope will still be equal to 6. So if we distribute this 6, we get y minus 2 is equal to 6 times x, 6x plus 6 times 2 is 12."}, {"video_title": "Parallel lines from equation (example 3) Mathematics I High School Math Khan Academy.mp3", "Sentence": "So we already know that the slope is equal to 6. And sometimes people are more comfortable with slope-intercept form. So let's put it in slope-intercept form just to confirm that if we put it in this form, the slope will still be equal to 6. So if we distribute this 6, we get y minus 2 is equal to 6 times x, 6x plus 6 times 2 is 12. And then if you add this 2, if you add 2 to both sides of the equation, you get y, because these guys cancel out, is equal to 6x plus 14. So you see once again, the slope is 6. So line A and line C have the same slope."}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy (2).mp3", "Sentence": "Let's have a little bit of a review of what a function is before we talk about what the domain of a function means. So a function we can view as something, so I'll put a function in this box here, and it takes inputs, and for a given input, it's going to produce an output, which we call f of x. So for example, let's say that we have the function, let's say we have the function f of x is equal to two over x. So in this case, if, so let me see, so if that's my function f, if I were to input the number three, well, f of three that we're going to output, we know how to figure that out. We've defined it right over here. It's going to be equal to two over three. It's going to be equal to two over three."}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy (2).mp3", "Sentence": "So in this case, if, so let me see, so if that's my function f, if I were to input the number three, well, f of three that we're going to output, we know how to figure that out. We've defined it right over here. It's going to be equal to two over three. It's going to be equal to two over three. So we were able, for that input, we were able to find an output. If our input was pi, then we input into our function, and then f of pi, when x is pi, we're going to output f of pi, which is equal to two over pi. So we could write this as two over pi."}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy (2).mp3", "Sentence": "It's going to be equal to two over three. So we were able, for that input, we were able to find an output. If our input was pi, then we input into our function, and then f of pi, when x is pi, we're going to output f of pi, which is equal to two over pi. So we could write this as two over pi. So we were able to find the output pretty easily. But now let's do something interesting. Let's attempt to input zero into the function."}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy (2).mp3", "Sentence": "So we could write this as two over pi. So we were able to find the output pretty easily. But now let's do something interesting. Let's attempt to input zero into the function. If we input zero, does the function tell us what we need to output? Does this definition tell us what we need to output? So if I attempt to put x equals zero, then this definition would say, f of zero would be two over zero, but two over zero is undefined."}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy (2).mp3", "Sentence": "Let's attempt to input zero into the function. If we input zero, does the function tell us what we need to output? Does this definition tell us what we need to output? So if I attempt to put x equals zero, then this definition would say, f of zero would be two over zero, but two over zero is undefined. Let me write this, two over zero. This is undefined. This function definition does not tell us what to actually do with zero."}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy (2).mp3", "Sentence": "So if I attempt to put x equals zero, then this definition would say, f of zero would be two over zero, but two over zero is undefined. Let me write this, two over zero. This is undefined. This function definition does not tell us what to actually do with zero. It gives us an undefined answer. So this function is not defined here. It gives a question mark."}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy (2).mp3", "Sentence": "This function definition does not tell us what to actually do with zero. It gives us an undefined answer. So this function is not defined here. It gives a question mark. So this gets to the essence of what domain is. Domain is the set of all inputs over which the function is defined. So the domain of this function f would be all real numbers except for x equals zero."}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy (2).mp3", "Sentence": "It gives a question mark. So this gets to the essence of what domain is. Domain is the set of all inputs over which the function is defined. So the domain of this function f would be all real numbers except for x equals zero. So let me write down these big ideas. This is the domain, a domain of a function. Actually, let me write that out."}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy (2).mp3", "Sentence": "So the domain of this function f would be all real numbers except for x equals zero. So let me write down these big ideas. This is the domain, a domain of a function. Actually, let me write that out. Domain of a function, a domain of a function is the set of all inputs, inputs over which the function is defined, over which the function is defined, or the function has defined outputs, over which the function has defined outputs. So the domain for this f in particular, so the domain for this one, if I wanted to say its domain, I could say, look, it's going to be the set, and this curly brackets, these are kind of typical mathy set notation. I could say, okay, it's going to be the set of, and I'm going to put curly brackets like that."}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy (2).mp3", "Sentence": "Actually, let me write that out. Domain of a function, a domain of a function is the set of all inputs, inputs over which the function is defined, over which the function is defined, or the function has defined outputs, over which the function has defined outputs. So the domain for this f in particular, so the domain for this one, if I wanted to say its domain, I could say, look, it's going to be the set, and this curly brackets, these are kind of typical mathy set notation. I could say, okay, it's going to be the set of, and I'm going to put curly brackets like that. Well, x can be a member, this little symbol means a member of the real numbers, but it can't just be any real number. It can be most of the real numbers except it cannot be zero because we don't know this definition. It's undefined when you put an input of zero."}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy (2).mp3", "Sentence": "I could say, okay, it's going to be the set of, and I'm going to put curly brackets like that. Well, x can be a member, this little symbol means a member of the real numbers, but it can't just be any real number. It can be most of the real numbers except it cannot be zero because we don't know this definition. It's undefined when you put an input of zero. So x is a member of the real numbers, and when we write real numbers, we write it with this kind of double stroke right over here. That's the set of all real numbers such that, but we have to put the exception, zero is not, x equals zero is not a member of that domain, such that x does not equal zero. Now let's make this a little bit more concrete by doing some more examples."}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy (2).mp3", "Sentence": "It's undefined when you put an input of zero. So x is a member of the real numbers, and when we write real numbers, we write it with this kind of double stroke right over here. That's the set of all real numbers such that, but we have to put the exception, zero is not, x equals zero is not a member of that domain, such that x does not equal zero. Now let's make this a little bit more concrete by doing some more examples. The more examples we do, hopefully the clearer this will become. So let's say we have another function. Just to be clear, we don't always have to use f's and x's."}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy (2).mp3", "Sentence": "Now let's make this a little bit more concrete by doing some more examples. The more examples we do, hopefully the clearer this will become. So let's say we have another function. Just to be clear, we don't always have to use f's and x's. We could say, let's say we have g of y is equal to the square root of y minus six. So what's the domain here? What is the set of all inputs over which this function g is defined?"}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy (2).mp3", "Sentence": "Just to be clear, we don't always have to use f's and x's. We could say, let's say we have g of y is equal to the square root of y minus six. So what's the domain here? What is the set of all inputs over which this function g is defined? So here we are inputting a y into function g, and we're going to output g of y. Well, it's going to be defined as long as whatever we have under the radical right over here is non-negative. If this becomes negative, our traditional principal root operator here is not defined."}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy (2).mp3", "Sentence": "What is the set of all inputs over which this function g is defined? So here we are inputting a y into function g, and we're going to output g of y. Well, it's going to be defined as long as whatever we have under the radical right over here is non-negative. If this becomes negative, our traditional principal root operator here is not defined. We need something that, if this ended up being a negative number, hey, how do you take the principal root of a negative number? And we're just saying this is kind of the traditional principal root operator. So y minus six needs to be greater than or equal to zero in order for g to be defined for that input y, or you could say, add six to both sides, y needs to be greater than or equal to six."}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy (2).mp3", "Sentence": "If this becomes negative, our traditional principal root operator here is not defined. We need something that, if this ended up being a negative number, hey, how do you take the principal root of a negative number? And we're just saying this is kind of the traditional principal root operator. So y minus six needs to be greater than or equal to zero in order for g to be defined for that input y, or you could say, add six to both sides, y needs to be greater than or equal to six. Or you could say g is defined for any inputs y that are greater than or equal to six. So we could say the domain here, we could say that the domain here is the set of all y's that are a member of the real numbers such that y, such that they're also greater than or equal, such that they're also greater than or equal to six. So hopefully this is starting to make some sense."}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy (2).mp3", "Sentence": "So y minus six needs to be greater than or equal to zero in order for g to be defined for that input y, or you could say, add six to both sides, y needs to be greater than or equal to six. Or you could say g is defined for any inputs y that are greater than or equal to six. So we could say the domain here, we could say that the domain here is the set of all y's that are a member of the real numbers such that y, such that they're also greater than or equal, such that they're also greater than or equal to six. So hopefully this is starting to make some sense. And we're always used to functions defined this way, but you could even see functions that are defined in fairly exotic ways. You could see a function, let me say h of x. h of x could be defined as, it literally could be defined as, well, h of x is gonna be one if x is equal to pi, and it's equal to zero if x is equal to three. Now what's the domain here?"}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy (2).mp3", "Sentence": "So hopefully this is starting to make some sense. And we're always used to functions defined this way, but you could even see functions that are defined in fairly exotic ways. You could see a function, let me say h of x. h of x could be defined as, it literally could be defined as, well, h of x is gonna be one if x is equal to pi, and it's equal to zero if x is equal to three. Now what's the domain here? And I encourage you to pause the video and think about it. Well, this function is actually only defined for two inputs. If you, we know h of pi, if you input pi into it, we know you're gonna output one."}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy (2).mp3", "Sentence": "Now what's the domain here? And I encourage you to pause the video and think about it. Well, this function is actually only defined for two inputs. If you, we know h of pi, if you input pi into it, we know you're gonna output one. And we know that if you input three into it, h of three, when x equals three, you're going to, let me put some commas here, you're gonna get a zero. But if you input anything else, what's h of four going to be? Well, it hasn't defined it, it's undefined."}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy (2).mp3", "Sentence": "If you, we know h of pi, if you input pi into it, we know you're gonna output one. And we know that if you input three into it, h of three, when x equals three, you're going to, let me put some commas here, you're gonna get a zero. But if you input anything else, what's h of four going to be? Well, it hasn't defined it, it's undefined. What's h of negative one going to be? It hasn't defined it. So the domain here, the domain of h is literally, it's just literally going to be the two valid inputs that x can be are three and pi."}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy (2).mp3", "Sentence": "Well, it hasn't defined it, it's undefined. What's h of negative one going to be? It hasn't defined it. So the domain here, the domain of h is literally, it's just literally going to be the two valid inputs that x can be are three and pi. Three and pi. These are the only valid inputs. These are the only two numbers over which this function is actually defined."}, {"video_title": "What is the domain of a function Functions Algebra I Khan Academy (2).mp3", "Sentence": "So the domain here, the domain of h is literally, it's just literally going to be the two valid inputs that x can be are three and pi. Three and pi. These are the only valid inputs. These are the only two numbers over which this function is actually defined. So this hopefully starts to give you a flavor of why we care about domains. Not all functions are defined over all real numbers. Some are defined for only a small subset of real numbers, or for some other thing, or only whole numbers, or natural numbers, or positive numbers, or negative numbers, or they have exceptions."}, {"video_title": "Example 1 Multiplying a binomial by a binomial Algebra I Khan Academy.mp3", "Sentence": "So we're multiplying two binomials. I'm actually going to show you two really equivalent ways of doing this, one that you might hear in a classroom, and it's kind of more of a mechanical, memorizing way of doing it, which might be faster, but you really don't know what you're doing. And then there's the one where you're essentially just applying something that you already know in kind of a logical way. So I'll first do the memorizing way that you might be exposed to. And they'll use something called FOIL. Let me write this down here. So you could immediately see that whenever someone gives you a mnemonic to memorize, that you're doing something pretty mechanical."}, {"video_title": "Example 1 Multiplying a binomial by a binomial Algebra I Khan Academy.mp3", "Sentence": "So I'll first do the memorizing way that you might be exposed to. And they'll use something called FOIL. Let me write this down here. So you could immediately see that whenever someone gives you a mnemonic to memorize, that you're doing something pretty mechanical. So FOIL literally stands for First Outside. Let me write it this way. Let me write FOIL."}, {"video_title": "Example 1 Multiplying a binomial by a binomial Algebra I Khan Academy.mp3", "Sentence": "So you could immediately see that whenever someone gives you a mnemonic to memorize, that you're doing something pretty mechanical. So FOIL literally stands for First Outside. Let me write it this way. Let me write FOIL. Where the F in FOIL stands for First. The O in FOIL stands for Outside. The I stands for Inside."}, {"video_title": "Example 1 Multiplying a binomial by a binomial Algebra I Khan Academy.mp3", "Sentence": "Let me write FOIL. Where the F in FOIL stands for First. The O in FOIL stands for Outside. The I stands for Inside. And then the L stands for Last. And the reason why I don't like these things is when you're 35 years old, you're not going to remember what FOIL stood for, and then you're not going to remember how to multiply this binomial. But let's just apply FOIL."}, {"video_title": "Example 1 Multiplying a binomial by a binomial Algebra I Khan Academy.mp3", "Sentence": "The I stands for Inside. And then the L stands for Last. And the reason why I don't like these things is when you're 35 years old, you're not going to remember what FOIL stood for, and then you're not going to remember how to multiply this binomial. But let's just apply FOIL. So FIRST says just multiply the first terms in each of these binomials. So just multiply the 3x times the 5x. So 3x times the 5x."}, {"video_title": "Example 1 Multiplying a binomial by a binomial Algebra I Khan Academy.mp3", "Sentence": "But let's just apply FOIL. So FIRST says just multiply the first terms in each of these binomials. So just multiply the 3x times the 5x. So 3x times the 5x. The Outside part tells us to multiply the outside terms. So in this case, you have 3x on the outside, and you have negative 7 on the outside. So that is plus 3x times negative 7."}, {"video_title": "Example 1 Multiplying a binomial by a binomial Algebra I Khan Academy.mp3", "Sentence": "So 3x times the 5x. The Outside part tells us to multiply the outside terms. So in this case, you have 3x on the outside, and you have negative 7 on the outside. So that is plus 3x times negative 7. The inside. Well, the inside terms here are 2 and 5x. So plus 2 times 5x."}, {"video_title": "Example 1 Multiplying a binomial by a binomial Algebra I Khan Academy.mp3", "Sentence": "So that is plus 3x times negative 7. The inside. Well, the inside terms here are 2 and 5x. So plus 2 times 5x. And then finally, you have the last terms. You have the 2 and the negative 7. So the last term is 2 times negative 7."}, {"video_title": "Example 1 Multiplying a binomial by a binomial Algebra I Khan Academy.mp3", "Sentence": "So plus 2 times 5x. And then finally, you have the last terms. You have the 2 and the negative 7. So the last term is 2 times negative 7. 2 times negative 7. So what you're essentially doing is just making sure that you're multiplying each term by every other term here. What we're essentially doing is multiplying, doing the distributive property twice."}, {"video_title": "Example 1 Multiplying a binomial by a binomial Algebra I Khan Academy.mp3", "Sentence": "So the last term is 2 times negative 7. 2 times negative 7. So what you're essentially doing is just making sure that you're multiplying each term by every other term here. What we're essentially doing is multiplying, doing the distributive property twice. We're multiplying the 3x times 5x minus 7. So 3x times 5x minus 7 is 3x times 5x plus 3x minus 7. And we're multiplying the 2 times 5x minus 7 to give us these terms."}, {"video_title": "Example 1 Multiplying a binomial by a binomial Algebra I Khan Academy.mp3", "Sentence": "What we're essentially doing is multiplying, doing the distributive property twice. We're multiplying the 3x times 5x minus 7. So 3x times 5x minus 7 is 3x times 5x plus 3x minus 7. And we're multiplying the 2 times 5x minus 7 to give us these terms. But anyway, let's just multiply this out just to get our answer. 3x times 5x, the same thing as 3 times 5 times x times x, which is the same thing as 15x squared. You can use x to the first times x to the first."}, {"video_title": "Example 1 Multiplying a binomial by a binomial Algebra I Khan Academy.mp3", "Sentence": "And we're multiplying the 2 times 5x minus 7 to give us these terms. But anyway, let's just multiply this out just to get our answer. 3x times 5x, the same thing as 3 times 5 times x times x, which is the same thing as 15x squared. You can use x to the first times x to the first. You multiply the x's, you get x squared. 3 times 5 is 15. This term right here, 3 times negative 7 is negative 21."}, {"video_title": "Example 1 Multiplying a binomial by a binomial Algebra I Khan Academy.mp3", "Sentence": "You can use x to the first times x to the first. You multiply the x's, you get x squared. 3 times 5 is 15. This term right here, 3 times negative 7 is negative 21. And then you have your x right over here. And then you have this term, which is 2 times 5, which is 10 times x, so plus 10x. And then finally, you have this term here in blue."}, {"video_title": "Example 1 Multiplying a binomial by a binomial Algebra I Khan Academy.mp3", "Sentence": "This term right here, 3 times negative 7 is negative 21. And then you have your x right over here. And then you have this term, which is 2 times 5, which is 10 times x, so plus 10x. And then finally, you have this term here in blue. 2 times negative 7 is negative 14. And we aren't done yet. We can simplify this a little bit."}, {"video_title": "Example 1 Multiplying a binomial by a binomial Algebra I Khan Academy.mp3", "Sentence": "And then finally, you have this term here in blue. 2 times negative 7 is negative 14. And we aren't done yet. We can simplify this a little bit. We have two like terms here. We have this. Let me find a new color."}, {"video_title": "Example 1 Multiplying a binomial by a binomial Algebra I Khan Academy.mp3", "Sentence": "We can simplify this a little bit. We have two like terms here. We have this. Let me find a new color. We have two terms with an x to the first power, just an x term right over here. So if we have negative 21 of something and you add 10, or another way, if you have 10 of something and you subtract 21 of them, you're going to have negative 11 of that something. And we put the other terms here."}, {"video_title": "Example 1 Multiplying a binomial by a binomial Algebra I Khan Academy.mp3", "Sentence": "Let me find a new color. We have two terms with an x to the first power, just an x term right over here. So if we have negative 21 of something and you add 10, or another way, if you have 10 of something and you subtract 21 of them, you're going to have negative 11 of that something. And we put the other terms here. You have 15x squared, and then you have your minus 14. And we are done. Now I said I would show you another way to do it."}, {"video_title": "Example 1 Multiplying a binomial by a binomial Algebra I Khan Academy.mp3", "Sentence": "And we put the other terms here. You have 15x squared, and then you have your minus 14. And we are done. Now I said I would show you another way to do it. So I want to show you why the distributive property can get us here without having to memorize FOIL. So the distributive property tells us that if we're multiplying something times an expression, you just have to multiply it times every term in the expression. So we can distribute the 5x onto the 3x minus 7, this whole thing, onto the 3x plus 2."}, {"video_title": "Example 1 Multiplying a binomial by a binomial Algebra I Khan Academy.mp3", "Sentence": "Now I said I would show you another way to do it. So I want to show you why the distributive property can get us here without having to memorize FOIL. So the distributive property tells us that if we're multiplying something times an expression, you just have to multiply it times every term in the expression. So we can distribute the 5x onto the 3x minus 7, this whole thing, onto the 3x plus 2. Let me just change the order, since we're used to distributing something from the left. So this is the same thing as 5x minus 7 times 3x plus 2. I just swapped the two expressions."}, {"video_title": "Example 1 Multiplying a binomial by a binomial Algebra I Khan Academy.mp3", "Sentence": "So we can distribute the 5x onto the 3x minus 7, this whole thing, onto the 3x plus 2. Let me just change the order, since we're used to distributing something from the left. So this is the same thing as 5x minus 7 times 3x plus 2. I just swapped the two expressions. And we can distribute this whole thing times each of these terms. Now what happens if I take 5x minus 7 times 3x? Well, that's just going to be 3x times 5x minus 7."}, {"video_title": "Example 1 Multiplying a binomial by a binomial Algebra I Khan Academy.mp3", "Sentence": "I just swapped the two expressions. And we can distribute this whole thing times each of these terms. Now what happens if I take 5x minus 7 times 3x? Well, that's just going to be 3x times 5x minus 7. So I've just distributed the 5x minus 7 times 3x. And to that, I'm going to add 2 times 5x minus 7. I've just distributed the 5x minus 7 onto the 2."}, {"video_title": "Example 1 Multiplying a binomial by a binomial Algebra I Khan Academy.mp3", "Sentence": "Well, that's just going to be 3x times 5x minus 7. So I've just distributed the 5x minus 7 times 3x. And to that, I'm going to add 2 times 5x minus 7. I've just distributed the 5x minus 7 onto the 2. Now we can do distributive property again. We can distribute the 3x onto the 5x. And we can distribute the 3x onto the negative 7."}, {"video_title": "Example 1 Multiplying a binomial by a binomial Algebra I Khan Academy.mp3", "Sentence": "I've just distributed the 5x minus 7 onto the 2. Now we can do distributive property again. We can distribute the 3x onto the 5x. And we can distribute the 3x onto the negative 7. We can distribute the 2 onto the 5x over here. And we can distribute the 2 on that negative 7. Now if we do it like this, what do we get?"}, {"video_title": "Example 1 Multiplying a binomial by a binomial Algebra I Khan Academy.mp3", "Sentence": "And we can distribute the 3x onto the negative 7. We can distribute the 2 onto the 5x over here. And we can distribute the 2 on that negative 7. Now if we do it like this, what do we get? 3x times 5x. That's this right over here. If we do 3x times negative 7, that's this term right over here."}, {"video_title": "Example 1 Multiplying a binomial by a binomial Algebra I Khan Academy.mp3", "Sentence": "Now if we do it like this, what do we get? 3x times 5x. That's this right over here. If we do 3x times negative 7, that's this term right over here. If you do 2 times 5x, that's this term right over here. If you do 2 times negative 7, that is this term right over here. So we got the exact same result that we got with FOIL."}, {"video_title": "Example 1 Multiplying a binomial by a binomial Algebra I Khan Academy.mp3", "Sentence": "If we do 3x times negative 7, that's this term right over here. If you do 2 times 5x, that's this term right over here. If you do 2 times negative 7, that is this term right over here. So we got the exact same result that we got with FOIL. Now FOIL can be faster. If you just want to do it, you kind of can skip to this step. I think it's important that you know that this is how it actually works."}, {"video_title": "Adding and subtracting rational expressions 3 Algebra II Khan Academy.mp3", "Sentence": "So we have two rational expressions and we're subtracting one from the other. And just like when we first learned to subtract fractions or add fractions, we have to find a common denominator. And the best way to find a common denominator if we're just dealing with regular numbers or with algebraic expressions is to factor them out and make sure that our common denominator has all of the factors in it. That'll ensure that it's divisible by the two denominators here. So this guy right here is completely factored, he's just a plus 2. This one over here, let's see if we can factor it. a squared plus 4a plus 4."}, {"video_title": "Adding and subtracting rational expressions 3 Algebra II Khan Academy.mp3", "Sentence": "That'll ensure that it's divisible by the two denominators here. So this guy right here is completely factored, he's just a plus 2. This one over here, let's see if we can factor it. a squared plus 4a plus 4. Well, you see the pattern, 4 is 2 squared, 4 is 2 times 2, so a squared plus 4a plus 4 is a plus 2 times a plus 2, or a plus 2 squared. So we say it's a plus 2 times a plus 2. That's what a squared plus 4a plus 4 is."}, {"video_title": "Adding and subtracting rational expressions 3 Algebra II Khan Academy.mp3", "Sentence": "a squared plus 4a plus 4. Well, you see the pattern, 4 is 2 squared, 4 is 2 times 2, so a squared plus 4a plus 4 is a plus 2 times a plus 2, or a plus 2 squared. So we say it's a plus 2 times a plus 2. That's what a squared plus 4a plus 4 is. So this is obviously divisible by itself, everything is divisible by itself, except I guess for 0, is divisible by itself. And it's also divisible by a plus 2. So this is the least common multiple of this expression and that expression, and it could be a good common denominator."}, {"video_title": "Adding and subtracting rational expressions 3 Algebra II Khan Academy.mp3", "Sentence": "That's what a squared plus 4a plus 4 is. So this is obviously divisible by itself, everything is divisible by itself, except I guess for 0, is divisible by itself. And it's also divisible by a plus 2. So this is the least common multiple of this expression and that expression, and it could be a good common denominator. So let's set that up. This will be the same thing as being equal to this first term right here, a minus 2 over a plus 2, but we want the denominator now to be a plus 2 times a plus 2. We want it to be a plus 2 squared."}, {"video_title": "Adding and subtracting rational expressions 3 Algebra II Khan Academy.mp3", "Sentence": "So this is the least common multiple of this expression and that expression, and it could be a good common denominator. So let's set that up. This will be the same thing as being equal to this first term right here, a minus 2 over a plus 2, but we want the denominator now to be a plus 2 times a plus 2. We want it to be a plus 2 squared. So let's multiply this numerator and denominator by a plus 2, so its denominator is the same thing as this. So let's multiply both the numerator and the denominator by a plus 2. And we're going to assume that a is not equal to negative 2, that would have made this undefined, it would have also made this undefined."}, {"video_title": "Adding and subtracting rational expressions 3 Algebra II Khan Academy.mp3", "Sentence": "We want it to be a plus 2 squared. So let's multiply this numerator and denominator by a plus 2, so its denominator is the same thing as this. So let's multiply both the numerator and the denominator by a plus 2. And we're going to assume that a is not equal to negative 2, that would have made this undefined, it would have also made this undefined. So throughout this whole thing, we're going to assume that a cannot be equal to negative 2. The domain is all real numbers, a can be any real number except for negative 2. So the first term is that, extend the line a little bit, and then the second term doesn't change because its denominator is already the common denominator."}, {"video_title": "Adding and subtracting rational expressions 3 Algebra II Khan Academy.mp3", "Sentence": "And we're going to assume that a is not equal to negative 2, that would have made this undefined, it would have also made this undefined. So throughout this whole thing, we're going to assume that a cannot be equal to negative 2. The domain is all real numbers, a can be any real number except for negative 2. So the first term is that, extend the line a little bit, and then the second term doesn't change because its denominator is already the common denominator. So minus a minus 3 over, and we could write it either as a plus 2 times a plus 2, or as this thing over here, let's write it in the factored form, it'll make it easier to simplify later on, a plus 2 times a plus 2. And now, before we, let's set this up like this, now before we add the numerators, it'll probably be a good idea to multiply this out right there, but let me write the denominator, we know what that is. It is a plus 2 times a plus 2."}, {"video_title": "Adding and subtracting rational expressions 3 Algebra II Khan Academy.mp3", "Sentence": "So the first term is that, extend the line a little bit, and then the second term doesn't change because its denominator is already the common denominator. So minus a minus 3 over, and we could write it either as a plus 2 times a plus 2, or as this thing over here, let's write it in the factored form, it'll make it easier to simplify later on, a plus 2 times a plus 2. And now, before we, let's set this up like this, now before we add the numerators, it'll probably be a good idea to multiply this out right there, but let me write the denominator, we know what that is. It is a plus 2 times a plus 2. Now this numerator, if we have a minus 2 times a plus 2, we've seen that pattern before, we can multiply it out if you like, but we've seen it enough, hopefully, to recognize that this is going to be a squared minus 2 squared. This is going to be a squared minus 4. You can multiply it out, the middle terms cancel out, the negative 2 times a cancels out with the a times 2, and you're just left with a squared minus 4, that's that over there."}, {"video_title": "Adding and subtracting rational expressions 3 Algebra II Khan Academy.mp3", "Sentence": "It is a plus 2 times a plus 2. Now this numerator, if we have a minus 2 times a plus 2, we've seen that pattern before, we can multiply it out if you like, but we've seen it enough, hopefully, to recognize that this is going to be a squared minus 2 squared. This is going to be a squared minus 4. You can multiply it out, the middle terms cancel out, the negative 2 times a cancels out with the a times 2, and you're just left with a squared minus 4, that's that over there. And then you have this, you have minus a minus 3, so let's be very careful here. You're subtracting a minus 3, so you want to distribute the negative sign, or multiply both of these terms times negative 1. So you could put a minus a here, and then negative 3 is plus 3."}, {"video_title": "Adding and subtracting rational expressions 3 Algebra II Khan Academy.mp3", "Sentence": "You can multiply it out, the middle terms cancel out, the negative 2 times a cancels out with the a times 2, and you're just left with a squared minus 4, that's that over there. And then you have this, you have minus a minus 3, so let's be very careful here. You're subtracting a minus 3, so you want to distribute the negative sign, or multiply both of these terms times negative 1. So you could put a minus a here, and then negative 3 is plus 3. So what does this simplify to? You have a squared minus a plus, let's see, negative 4 plus 3 is negative 1. All of that over a plus 2 times a plus 2, or we could just rewrite that as a plus 2 squared."}, {"video_title": "Adding and subtracting rational expressions 3 Algebra II Khan Academy.mp3", "Sentence": "So you could put a minus a here, and then negative 3 is plus 3. So what does this simplify to? You have a squared minus a plus, let's see, negative 4 plus 3 is negative 1. All of that over a plus 2 times a plus 2, or we could just rewrite that as a plus 2 squared. Now, we might want to factor this numerator out more, just to make sure it doesn't contain a common factor with the denominator. The denominator is just 2 a plus 2s multiplied by themselves. And you can see from inspection, a plus 2 will not be a factor in this top expression."}, {"video_title": "Adding and subtracting rational expressions 3 Algebra II Khan Academy.mp3", "Sentence": "All of that over a plus 2 times a plus 2, or we could just rewrite that as a plus 2 squared. Now, we might want to factor this numerator out more, just to make sure it doesn't contain a common factor with the denominator. The denominator is just 2 a plus 2s multiplied by themselves. And you can see from inspection, a plus 2 will not be a factor in this top expression. I mean, if it was, this number right here would be divisible by 2. It's not divisible by 2, so a plus 2 is not one of the factors here. So there's not going to be any more simplification, even if we were able to factor this thing in the numerator out."}, {"video_title": "Adding and subtracting rational expressions 3 Algebra II Khan Academy.mp3", "Sentence": "And you can see from inspection, a plus 2 will not be a factor in this top expression. I mean, if it was, this number right here would be divisible by 2. It's not divisible by 2, so a plus 2 is not one of the factors here. So there's not going to be any more simplification, even if we were able to factor this thing in the numerator out. So we're done. We have simplified the rational expression, and the domain is for all a's except for a cannot, or all a's given that a does not equal negative 2. So all a's except for negative 2."}, {"video_title": "Solving quadratics by factoring leading coefficient \u00c3\u00a2\u00c2 \u00c2\u00a0 1 High School Math Khan Academy.mp3", "Sentence": "Like always, pause this video and see if you can solve for x if you can find the x values that satisfy this equation. Alright, let's work through this together. So this, the numbers here don't seem like outlandish numbers they seem like something that I might be able to deal with and I might be able to factor, so let's try to do that. So the first thing I like to do is see if I can get a coefficient of one on the second degree term on the x squared term and it looks like actually all of these terms are divisible by six. So if we divide both sides of this equation by six I'm still going to have nice integer coefficients. So let's do that, let's divide both sides by six. So if we divide the left side by six, divide by six, divide by six, divide by six and I divide the right side by six."}, {"video_title": "Solving quadratics by factoring leading coefficient \u00c3\u00a2\u00c2 \u00c2\u00a0 1 High School Math Khan Academy.mp3", "Sentence": "So the first thing I like to do is see if I can get a coefficient of one on the second degree term on the x squared term and it looks like actually all of these terms are divisible by six. So if we divide both sides of this equation by six I'm still going to have nice integer coefficients. So let's do that, let's divide both sides by six. So if we divide the left side by six, divide by six, divide by six, divide by six and I divide the right side by six. So if I do that and clearly if I do the same thing to both sides of the equation then the equality still holds. On the left hand side I am going to be left with x squared and then negative 120 divided by six, that is, let's see, 120 divided by six is 20 so that's minus 20 x and then 600 divided by six is 100 so plus 100 is equal to zero divided by six, is equal to zero. So let's see if we can factor, if we can express this quadratic as a product of two expressions."}, {"video_title": "Solving quadratics by factoring leading coefficient \u00c3\u00a2\u00c2 \u00c2\u00a0 1 High School Math Khan Academy.mp3", "Sentence": "So if we divide the left side by six, divide by six, divide by six, divide by six and I divide the right side by six. So if I do that and clearly if I do the same thing to both sides of the equation then the equality still holds. On the left hand side I am going to be left with x squared and then negative 120 divided by six, that is, let's see, 120 divided by six is 20 so that's minus 20 x and then 600 divided by six is 100 so plus 100 is equal to zero divided by six, is equal to zero. So let's see if we can factor, if we can express this quadratic as a product of two expressions. And the way we think about this, and we've done it multiple times, if we have something, if we have x plus a times x plus b and this is hopefully a review for you, if you multiply that out that is going to be equal to, that equals to x squared plus a plus b x plus ab and so what we want to do is see if we can factor this into an x plus a and an x plus b and so a plus b needs to be equal to negative 20, that needs to be a plus b and then a times b right over here, that needs to be equal to the constant term, that needs to be a times b right over there. So can we think of two numbers that if we take their product we get positive 100 and if we take their sum we get negative 20? Well since their product is positive we know that they have the same sign, so they're both going to have the same sign, so they're either both going to be positive or they're both going to be negative since we know that we have a positive product and since their sum is negative, well they both must both be negative, although you can't add up two positive numbers and get a negative, so they both must be negative."}, {"video_title": "Solving quadratics by factoring leading coefficient \u00c3\u00a2\u00c2 \u00c2\u00a0 1 High School Math Khan Academy.mp3", "Sentence": "So let's see if we can factor, if we can express this quadratic as a product of two expressions. And the way we think about this, and we've done it multiple times, if we have something, if we have x plus a times x plus b and this is hopefully a review for you, if you multiply that out that is going to be equal to, that equals to x squared plus a plus b x plus ab and so what we want to do is see if we can factor this into an x plus a and an x plus b and so a plus b needs to be equal to negative 20, that needs to be a plus b and then a times b right over here, that needs to be equal to the constant term, that needs to be a times b right over there. So can we think of two numbers that if we take their product we get positive 100 and if we take their sum we get negative 20? Well since their product is positive we know that they have the same sign, so they're both going to have the same sign, so they're either both going to be positive or they're both going to be negative since we know that we have a positive product and since their sum is negative, well they both must both be negative, although you can't add up two positive numbers and get a negative, so they both must be negative. So let's think about it a little bit. What negative numbers, when I add them together I get negative 20 and when I multiply I get 100? Well you could try to factor 100, you could say well negative two times negative 50 or negative four times negative 25, but the one that might jump out at you is this is negative 10 times, I'll write it this way, negative 10 times negative 10 and this is negative 10 plus negative 10."}, {"video_title": "Solving quadratics by factoring leading coefficient \u00c3\u00a2\u00c2 \u00c2\u00a0 1 High School Math Khan Academy.mp3", "Sentence": "Well since their product is positive we know that they have the same sign, so they're both going to have the same sign, so they're either both going to be positive or they're both going to be negative since we know that we have a positive product and since their sum is negative, well they both must both be negative, although you can't add up two positive numbers and get a negative, so they both must be negative. So let's think about it a little bit. What negative numbers, when I add them together I get negative 20 and when I multiply I get 100? Well you could try to factor 100, you could say well negative two times negative 50 or negative four times negative 25, but the one that might jump out at you is this is negative 10 times, I'll write it this way, negative 10 times negative 10 and this is negative 10 plus negative 10. So in that case both our A and our B would be negative 10 and so we can rewrite the left side of this equation as, we can rewrite it as X, and I'll write it this way at first, X plus negative 10 times, times X plus negative 10 again, X plus negative 10 and that is going to be equal to zero. So all I've done is I've factored this quadratic. Or another way, these are both the same thing as X minus 10, I could rewrite this as X minus 10 squared is equal to zero and so the only way that the left hand side is going to be equal to zero is if X minus 10 is equal to zero."}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "So, we already know how to take exponents. If I were to say 2 to the 4th power, what does that mean? Well, that means 2 times 2 times 2 times 2. 2 multiplied, or repeatedly multiplied, 4 times. And so this is going to be 2 times 2 is 4, times 2 is 8, times 2 is 16. But what if we think about things in another way? What if we're essentially, we know that we get to 16 when we raise 2 to some power, and we want to know what that power is."}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "2 multiplied, or repeatedly multiplied, 4 times. And so this is going to be 2 times 2 is 4, times 2 is 8, times 2 is 16. But what if we think about things in another way? What if we're essentially, we know that we get to 16 when we raise 2 to some power, and we want to know what that power is. So, for example, let's say that I start with 2, and I say I'm raising it to some power. What does that power have to be to get 16? Well, we just figure that out."}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "What if we're essentially, we know that we get to 16 when we raise 2 to some power, and we want to know what that power is. So, for example, let's say that I start with 2, and I say I'm raising it to some power. What does that power have to be to get 16? Well, we just figure that out. X would have to be 4. And this is what logarithms are fundamentally about, figuring out what power you have to raise to to get another number. Now, the way that we would denote this with logarithm notation is we would say log base, actually let me make it a little bit more colorful."}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, we just figure that out. X would have to be 4. And this is what logarithms are fundamentally about, figuring out what power you have to raise to to get another number. Now, the way that we would denote this with logarithm notation is we would say log base, actually let me make it a little bit more colorful. Log base 2, so I'll do this 2 in blue, log base 2 of 16 is equal to what? Or is equal in this case, since we have the x there, is equal to x. This and this are completely equivalent statements."}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "Now, the way that we would denote this with logarithm notation is we would say log base, actually let me make it a little bit more colorful. Log base 2, so I'll do this 2 in blue, log base 2 of 16 is equal to what? Or is equal in this case, since we have the x there, is equal to x. This and this are completely equivalent statements. This is saying, hey, well, if I take 2 to some x power, I get 16. This is saying, what power do I need to raise 2 to to get 16, and I'm going to set that to be equal to x. And you would say, well, you've got to raise it to the fourth power."}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "This and this are completely equivalent statements. This is saying, hey, well, if I take 2 to some x power, I get 16. This is saying, what power do I need to raise 2 to to get 16, and I'm going to set that to be equal to x. And you would say, well, you've got to raise it to the fourth power. Once again, x is equal to 4. So with that out of the way, let's try more examples of evaluating logarithmic expressions. So let's say you had log base 3 of 81."}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "And you would say, well, you've got to raise it to the fourth power. Once again, x is equal to 4. So with that out of the way, let's try more examples of evaluating logarithmic expressions. So let's say you had log base 3 of 81. What would this evaluate to? Well, just as a reminder, this evaluates to the power we have to raise 3 to to get to 81. So if you want to, you could set this to be equal to an x, set that to be equal to an x, and you can restate this equation as 3 to the x power is equal to 81."}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "So let's say you had log base 3 of 81. What would this evaluate to? Well, just as a reminder, this evaluates to the power we have to raise 3 to to get to 81. So if you want to, you could set this to be equal to an x, set that to be equal to an x, and you can restate this equation as 3 to the x power is equal to 81. Why is a logarithm useful? And you'll see that it has very interesting properties later on. But you didn't necessarily have to use algebra to do it this way, to say that the x is the power that you raise 3 to to get to 81."}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "So if you want to, you could set this to be equal to an x, set that to be equal to an x, and you can restate this equation as 3 to the x power is equal to 81. Why is a logarithm useful? And you'll see that it has very interesting properties later on. But you didn't necessarily have to use algebra to do it this way, to say that the x is the power that you raise 3 to to get to 81. You had to use algebra here. Well, with just a straight-up logarithmic expression, you didn't really have to use any algebra. We didn't have to set it equal to x."}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "But you didn't necessarily have to use algebra to do it this way, to say that the x is the power that you raise 3 to to get to 81. You had to use algebra here. Well, with just a straight-up logarithmic expression, you didn't really have to use any algebra. We didn't have to set it equal to x. We could just say this evaluates to the power I need to raise 3 to to get to 81. Well, what power do you have to raise 3 to to get to 81? Well, let's experiment a little bit."}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "We didn't have to set it equal to x. We could just say this evaluates to the power I need to raise 3 to to get to 81. Well, what power do you have to raise 3 to to get to 81? Well, let's experiment a little bit. So 3 to the first power is just 3. 3 to the second power is 9. 3 to the third power is 27."}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, let's experiment a little bit. So 3 to the first power is just 3. 3 to the second power is 9. 3 to the third power is 27. 3 to the fourth power, 27 times 3, is equal to 81. 3 to the fourth power is equal to 81. x is equal to 4. So we could say log base 3 of 81 is equal to 4."}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "3 to the third power is 27. 3 to the fourth power, 27 times 3, is equal to 81. 3 to the fourth power is equal to 81. x is equal to 4. So we could say log base 3 of 81 is equal to 4. Let's do several more of these examples. And I really encourage you to give a shot on your own, and you'll hopefully get the hang of it. So let's try a little larger number."}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "So we could say log base 3 of 81 is equal to 4. Let's do several more of these examples. And I really encourage you to give a shot on your own, and you'll hopefully get the hang of it. So let's try a little larger number. Let's say we want to take log base 6 of 216. What will this evaluate to? Well, we're asking ourselves, what power do we have to raise 6 to to get to 216?"}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "So let's try a little larger number. Let's say we want to take log base 6 of 216. What will this evaluate to? Well, we're asking ourselves, what power do we have to raise 6 to to get to 216? 6 to the first power is 6. 6 to the second power is 36. 36 times 6 is 216."}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, we're asking ourselves, what power do we have to raise 6 to to get to 216? 6 to the first power is 6. 6 to the second power is 36. 36 times 6 is 216. This is equal to 216. So this is 6 to the third power is equal to 216. So if someone says, what power do I have to raise 6 to, this base here, to get to 216?"}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "36 times 6 is 216. This is equal to 216. So this is 6 to the third power is equal to 216. So if someone says, what power do I have to raise 6 to, this base here, to get to 216? Well, that's just going to be equal to 3. 6 to the third power is equal to 216. Let's try another one."}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "So if someone says, what power do I have to raise 6 to, this base here, to get to 216? Well, that's just going to be equal to 3. 6 to the third power is equal to 216. Let's try another one. Let's say I had log base 2 of 64. So what does this evaluate to? Well, once again, we're asking ourselves, or this will evaluate to the exponent that I have to raise this base to."}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "Let's try another one. Let's say I had log base 2 of 64. So what does this evaluate to? Well, once again, we're asking ourselves, or this will evaluate to the exponent that I have to raise this base to. And you do this as a little subscript right here. The exponent that I have to raise 2 to to get to 64. So 2 to the first power is 2."}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, once again, we're asking ourselves, or this will evaluate to the exponent that I have to raise this base to. And you do this as a little subscript right here. The exponent that I have to raise 2 to to get to 64. So 2 to the first power is 2. 2 to the second power is 4. 8, 16, 32, 64. So this right over here is 2 to the sixth power is equal to 64."}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "So 2 to the first power is 2. 2 to the second power is 4. 8, 16, 32, 64. So this right over here is 2 to the sixth power is equal to 64. So when you evaluate this expression, you say, what power do I have to raise 2 to to get to 64? Well, I have to raise it to the sixth power. Let's do a slightly more straightforward one."}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "So this right over here is 2 to the sixth power is equal to 64. So when you evaluate this expression, you say, what power do I have to raise 2 to to get to 64? Well, I have to raise it to the sixth power. Let's do a slightly more straightforward one. Or maybe this will be less straightforward, depending on how you view it. What is log base 100 of 1? Let me think about that for a second."}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "Let's do a slightly more straightforward one. Or maybe this will be less straightforward, depending on how you view it. What is log base 100 of 1? Let me think about that for a second. So the 100 is a subscript, and then it's log base 100 of 1. That's one way to think about it. I could put a parentheses around the 1."}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "Let me think about that for a second. So the 100 is a subscript, and then it's log base 100 of 1. That's one way to think about it. I could put a parentheses around the 1. What does this evaluate to? Well, this is asking ourselves, or we would evaluate this as, what power do I have to raise 102 to get to 1? So let me write this down as an equation."}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "I could put a parentheses around the 1. What does this evaluate to? Well, this is asking ourselves, or we would evaluate this as, what power do I have to raise 102 to get to 1? So let me write this down as an equation. So if I set this to be equal to x, this is literally saying 100 to what power is equal to 1? Well, anything to the zeroth power is equal to 1. So in this case, x is equal to 0."}, {"video_title": "Interpret quadratic models Factored form Algebra I Khan Academy.mp3", "Sentence": "We're told that Rodrigo watches a helicopter take off from a platform. The height of the helicopter in meters above the ground, t minutes after takeoff, is modeled by, and we see this function right over here, Rodrigo wants to know when the helicopter will land on the ground. So pause this video and see if you can figure that out. All right, now let's think about this together. So let's just imagine actually what the graph of this function looks like. And it'll also help us imagine what's going on with the helicopter. So our horizontal axis, this is t, time in minutes, and then our vertical axis is height, so height as a function of time, and maybe I just write it like this."}, {"video_title": "Interpret quadratic models Factored form Algebra I Khan Academy.mp3", "Sentence": "All right, now let's think about this together. So let's just imagine actually what the graph of this function looks like. And it'll also help us imagine what's going on with the helicopter. So our horizontal axis, this is t, time in minutes, and then our vertical axis is height, so height as a function of time, and maybe I just write it like this. I'll just write height, and this is given in meters above the ground. Now, I don't know exactly what the graph looks like, but given that I have a negative coefficient on my quadratic term, I know that it is a downward-opening parabola like that. And it says that the helicopter takes off of a platform."}, {"video_title": "Interpret quadratic models Factored form Algebra I Khan Academy.mp3", "Sentence": "So our horizontal axis, this is t, time in minutes, and then our vertical axis is height, so height as a function of time, and maybe I just write it like this. I'll just write height, and this is given in meters above the ground. Now, I don't know exactly what the graph looks like, but given that I have a negative coefficient on my quadratic term, I know that it is a downward-opening parabola like that. And it says that the helicopter takes off of a platform. So however high the platform is, then it takes off, and it's going to do something like this. I don't know exactly what the graph looks like, but probably something like this. Now, if they asked us what is the highest point of the helicopter and at what time does it happen, then we'd wanna figure out what the vertex is of this parabola."}, {"video_title": "Interpret quadratic models Factored form Algebra I Khan Academy.mp3", "Sentence": "And it says that the helicopter takes off of a platform. So however high the platform is, then it takes off, and it's going to do something like this. I don't know exactly what the graph looks like, but probably something like this. Now, if they asked us what is the highest point of the helicopter and at what time does it happen, then we'd wanna figure out what the vertex is of this parabola. But that's not what they're asking. They're asking when does a helicopter land on the ground? That's this time right over here."}, {"video_title": "Interpret quadratic models Factored form Algebra I Khan Academy.mp3", "Sentence": "Now, if they asked us what is the highest point of the helicopter and at what time does it happen, then we'd wanna figure out what the vertex is of this parabola. But that's not what they're asking. They're asking when does a helicopter land on the ground? That's this time right over here. So if we wanted to find the vertex, we would wanna put this into vertex form. But here, we wanna figure out when does that function equal zero? We want to find a zero of this quadratic right over here."}, {"video_title": "Interpret quadratic models Factored form Algebra I Khan Academy.mp3", "Sentence": "That's this time right over here. So if we wanted to find the vertex, we would wanna put this into vertex form. But here, we wanna figure out when does that function equal zero? We want to find a zero of this quadratic right over here. So the best way that I can think about doing it is try to factor it. Try to set this thing equal to zero, and then factor it, and then see what t values make that equal to zero. So let me do that."}, {"video_title": "Interpret quadratic models Factored form Algebra I Khan Academy.mp3", "Sentence": "We want to find a zero of this quadratic right over here. So the best way that I can think about doing it is try to factor it. Try to set this thing equal to zero, and then factor it, and then see what t values make that equal to zero. So let me do that. So I say negative three t squared plus 24t plus 60. Remember, we care when our height is equal to zero, equals zero. So let's see."}, {"video_title": "Interpret quadratic models Factored form Algebra I Khan Academy.mp3", "Sentence": "So let me do that. So I say negative three t squared plus 24t plus 60. Remember, we care when our height is equal to zero, equals zero. So let's see. Maybe the first thing I would do, just to simplify this second degree term a little bit, let's just divide both sides by negative three. If we did that, this would become t squared, 24 divided by negative three is negative eight, negative eight t. 60 divided by negative three is negative 20. And then zero divided by negative three is of course still zero."}, {"video_title": "Interpret quadratic models Factored form Algebra I Khan Academy.mp3", "Sentence": "So let's see. Maybe the first thing I would do, just to simplify this second degree term a little bit, let's just divide both sides by negative three. If we did that, this would become t squared, 24 divided by negative three is negative eight, negative eight t. 60 divided by negative three is negative 20. And then zero divided by negative three is of course still zero. And now can I think of two numbers whose product is negative 20, so they would have to have different signs in order to get a negative product, and whose sum is negative eight? So let's see. What about negative 10 and two?"}, {"video_title": "Interpret quadratic models Factored form Algebra I Khan Academy.mp3", "Sentence": "And then zero divided by negative three is of course still zero. And now can I think of two numbers whose product is negative 20, so they would have to have different signs in order to get a negative product, and whose sum is negative eight? So let's see. What about negative 10 and two? That seems to work. So I could write this as t minus 10 times t plus two is equal to zero. And so in order to make this entire expression equal to zero, either one of these could be equal to zero."}, {"video_title": "Interpret quadratic models Factored form Algebra I Khan Academy.mp3", "Sentence": "What about negative 10 and two? That seems to work. So I could write this as t minus 10 times t plus two is equal to zero. And so in order to make this entire expression equal to zero, either one of these could be equal to zero. So either t minus 10 is equal to zero, or t plus two is equal to zero. And of course on the left here, I can add 10 to both sides. So either t equals 10, or I could subtract two from both sides here, t is equal to negative two."}, {"video_title": "Interpret quadratic models Factored form Algebra I Khan Academy.mp3", "Sentence": "And so in order to make this entire expression equal to zero, either one of these could be equal to zero. So either t minus 10 is equal to zero, or t plus two is equal to zero. And of course on the left here, I can add 10 to both sides. So either t equals 10, or I could subtract two from both sides here, t is equal to negative two. So there's two places where the function is equal to zero. One at time t equals negative two, and one at time t is equal to 10. Now we're assuming we're dealing with positive time here."}, {"video_title": "Interpret quadratic models Factored form Algebra I Khan Academy.mp3", "Sentence": "So either t equals 10, or I could subtract two from both sides here, t is equal to negative two. So there's two places where the function is equal to zero. One at time t equals negative two, and one at time t is equal to 10. Now we're assuming we're dealing with positive time here. We don't know what the helicopter was doing before the takeoff. So we wouldn't really think about this. So what we really care about is that t is equal to 10 minutes."}, {"video_title": "Interpret quadratic models Factored form Algebra I Khan Academy.mp3", "Sentence": "Now we're assuming we're dealing with positive time here. We don't know what the helicopter was doing before the takeoff. So we wouldn't really think about this. So what we really care about is that t is equal to 10 minutes. That's when the helicopter is right over there. And actually we know at t equals zero, these two terms become zero. We know it takes off at 60 meters."}, {"video_title": "Comparing linear rates example.mp3", "Sentence": "We're told that a griffin flew east over a castle at 50 kilometers per hour. Then, 42 minutes later, a dragon also flew east over the castle. The dragon flew 225 kilometers per hour. Assume both the griffin and the dragon continue flying east at the same speeds. How many minutes will the dragon have flown since passing the castle when it catches up to the griffin? And they also ask us, how many kilometers east of the castle will they be at that time? So pause this video and see if you can figure this out before we do this together."}, {"video_title": "Comparing linear rates example.mp3", "Sentence": "Assume both the griffin and the dragon continue flying east at the same speeds. How many minutes will the dragon have flown since passing the castle when it catches up to the griffin? And they also ask us, how many kilometers east of the castle will they be at that time? So pause this video and see if you can figure this out before we do this together. All right, so the question is, how many minutes will the dragon have flown since passing the castle when it catches up to the griffin? So let's set that variable to be equal to t, the number of minutes that the dragon has flown. Dragon flown since castle, since castle, and catches up."}, {"video_title": "Comparing linear rates example.mp3", "Sentence": "So pause this video and see if you can figure this out before we do this together. All right, so the question is, how many minutes will the dragon have flown since passing the castle when it catches up to the griffin? So let's set that variable to be equal to t, the number of minutes that the dragon has flown. Dragon flown since castle, since castle, and catches up. Catches up. So let's think about the distance that the dragon would have flown in that t minutes. Well, the dragon's flying at 225 kilometers per hour, so the distance is going to be the rate, 225 kilometers per hour times the time, so times t minutes, but we have to be careful."}, {"video_title": "Comparing linear rates example.mp3", "Sentence": "Dragon flown since castle, since castle, and catches up. Catches up. So let's think about the distance that the dragon would have flown in that t minutes. Well, the dragon's flying at 225 kilometers per hour, so the distance is going to be the rate, 225 kilometers per hour times the time, so times t minutes, but we have to be careful. This is in minutes while the rate is given in kilometers per hour. So we have to make sure that our units work out. And so for every one hour, we have 60 minutes."}, {"video_title": "Comparing linear rates example.mp3", "Sentence": "Well, the dragon's flying at 225 kilometers per hour, so the distance is going to be the rate, 225 kilometers per hour times the time, so times t minutes, but we have to be careful. This is in minutes while the rate is given in kilometers per hour. So we have to make sure that our units work out. And so for every one hour, we have 60 minutes. And we can see here that the units indeed do work out. This hour cancels with that hour in the numerator and the denominator, and this minute cancels out with this minute. And so the distance that the dragon would have flown after t minutes is going to be 225t over 60 kilometers."}, {"video_title": "Comparing linear rates example.mp3", "Sentence": "And so for every one hour, we have 60 minutes. And we can see here that the units indeed do work out. This hour cancels with that hour in the numerator and the denominator, and this minute cancels out with this minute. And so the distance that the dragon would have flown after t minutes is going to be 225t over 60 kilometers. So let me write it this way. 225 over 60t kilometers. Now we could try to simplify this, but I'll leave it like this for now."}, {"video_title": "Comparing linear rates example.mp3", "Sentence": "And so the distance that the dragon would have flown after t minutes is going to be 225t over 60 kilometers. So let me write it this way. 225 over 60t kilometers. Now we could try to simplify this, but I'll leave it like this for now. Maybe I'll simplify it a little bit later. Now let's think about how far the griffin would have flown. So they tell us that the griffin is flying at 50 kilometers per hour."}, {"video_title": "Comparing linear rates example.mp3", "Sentence": "Now we could try to simplify this, but I'll leave it like this for now. Maybe I'll simplify it a little bit later. Now let's think about how far the griffin would have flown. So they tell us that the griffin is flying at 50 kilometers per hour. So 50 kilometers per hour. And how long would the griffin have flown by that point? Well, the griffin passed the castle 42 minutes before the dragon passed it."}, {"video_title": "Comparing linear rates example.mp3", "Sentence": "So they tell us that the griffin is flying at 50 kilometers per hour. So 50 kilometers per hour. And how long would the griffin have flown by that point? Well, the griffin passed the castle 42 minutes before the dragon passed it. So if t is how many minutes that the dragon has been flying east of the castle, well, then the griffin is going to be t plus 42 minutes. So t plus 42 minutes is how long that the griffin has been traveling east of the castle. And then once again, we have to make sure that our units work out."}, {"video_title": "Comparing linear rates example.mp3", "Sentence": "Well, the griffin passed the castle 42 minutes before the dragon passed it. So if t is how many minutes that the dragon has been flying east of the castle, well, then the griffin is going to be t plus 42 minutes. So t plus 42 minutes is how long that the griffin has been traveling east of the castle. And then once again, we have to make sure that our units work out. So we're gonna say one hour for every 60 minutes. The minutes cancel out, the hours cancel out. And so we are going to be left with 50 over 60, or I could write 5 6ths times t plus 42 kilometers."}, {"video_title": "Comparing linear rates example.mp3", "Sentence": "And then once again, we have to make sure that our units work out. So we're gonna say one hour for every 60 minutes. The minutes cancel out, the hours cancel out. And so we are going to be left with 50 over 60, or I could write 5 6ths times t plus 42 kilometers. Or if we wanna simplify this even more, this is going to be 5 6ths t plus, let's see, 5 6ths of 42. 42 divided by six is seven times five is 35, plus 35 kilometers. So we know that they would have flown the exact same distance, because we're talking about when the dragon catches up with the griffin."}, {"video_title": "Comparing linear rates example.mp3", "Sentence": "And so we are going to be left with 50 over 60, or I could write 5 6ths times t plus 42 kilometers. Or if we wanna simplify this even more, this is going to be 5 6ths t plus, let's see, 5 6ths of 42. 42 divided by six is seven times five is 35, plus 35 kilometers. So we know that they would have flown the exact same distance, because we're talking about when the dragon catches up with the griffin. So these two things need to be equal to each other, and then we can just solve for t. So let's do that. We get 225 over 60 t, and we know that both sides are in kilometers, so just for the sake of simplicity, I won't write the units here. So this is going to be equal to 5 6ths t plus 35."}, {"video_title": "Comparing linear rates example.mp3", "Sentence": "So we know that they would have flown the exact same distance, because we're talking about when the dragon catches up with the griffin. So these two things need to be equal to each other, and then we can just solve for t. So let's do that. We get 225 over 60 t, and we know that both sides are in kilometers, so just for the sake of simplicity, I won't write the units here. So this is going to be equal to 5 6ths t plus 35. And now let us solve for t. We can subtract 5 6ths t from both sides, or actually, since I already have 60 as a denominator, I could subtract 50 over 60 t from both sides, which is the same thing as 5 6ths t. So I'm going to have 225 over 60 minus 50 over 60, and then all of that times t is equal to 35. And so let me get myself a little bit more real estate. So this is going to be simplified as 175 over 60 t is equal to 35, or that if I just multiply both sides by 60 over 175, I'll get that t is equal to 35 times 60 over 175."}, {"video_title": "Comparing linear rates example.mp3", "Sentence": "So this is going to be equal to 5 6ths t plus 35. And now let us solve for t. We can subtract 5 6ths t from both sides, or actually, since I already have 60 as a denominator, I could subtract 50 over 60 t from both sides, which is the same thing as 5 6ths t. So I'm going to have 225 over 60 minus 50 over 60, and then all of that times t is equal to 35. And so let me get myself a little bit more real estate. So this is going to be simplified as 175 over 60 t is equal to 35, or that if I just multiply both sides by 60 over 175, I'll get that t is equal to 35 times 60 over 175. And you might recognize that 35 is the same thing as five times seven, and 175 is the same thing as 25 times seven. So these sevens cancel out, and then if we divide both this and this by five, this becomes a one, this becomes a five, and then 60 divided by five is equal to 12. And so remember, t was in minutes, so the answer to the first part of the question is 12 minutes."}, {"video_title": "Comparing linear rates example.mp3", "Sentence": "So this is going to be simplified as 175 over 60 t is equal to 35, or that if I just multiply both sides by 60 over 175, I'll get that t is equal to 35 times 60 over 175. And you might recognize that 35 is the same thing as five times seven, and 175 is the same thing as 25 times seven. So these sevens cancel out, and then if we divide both this and this by five, this becomes a one, this becomes a five, and then 60 divided by five is equal to 12. And so remember, t was in minutes, so the answer to the first part of the question is 12 minutes. So let's go back up to what they were asking us. How many minutes will the dragon have flown since passing the castle when it catches up to the griffin? Well, we define that as t, and then we got 12 minutes."}, {"video_title": "Comparing linear rates example.mp3", "Sentence": "And so remember, t was in minutes, so the answer to the first part of the question is 12 minutes. So let's go back up to what they were asking us. How many minutes will the dragon have flown since passing the castle when it catches up to the griffin? Well, we define that as t, and then we got 12 minutes. Now, the next part of the question is how many kilometers east of the castle will they be at that time? So to figure out how many kilometers east of the castle, we just have to calculate this expression or this expression when t is equal to 12. So I'm going to use this first one."}, {"video_title": "Comparing linear rates example.mp3", "Sentence": "Well, we define that as t, and then we got 12 minutes. Now, the next part of the question is how many kilometers east of the castle will they be at that time? So to figure out how many kilometers east of the castle, we just have to calculate this expression or this expression when t is equal to 12. So I'm going to use this first one. So you're going to have 225 over 60 times 12 is going to give us, let's see, 60 and 12 are both divisible by 12, so you get that one over five. And if you divide 225 by five, that is going to give us 45, and the units all work out to kilometers. So we answered the first two parts of the question."}, {"video_title": "Comparing linear rates example.mp3", "Sentence": "So I'm going to use this first one. So you're going to have 225 over 60 times 12 is going to give us, let's see, 60 and 12 are both divisible by 12, so you get that one over five. And if you divide 225 by five, that is going to give us 45, and the units all work out to kilometers. So we answered the first two parts of the question. And the second part of this question, they tell us that Latanya and Jair both wrote correct inequalities for the times in minutes when the dragon is farther east of the castle than the griffin is. Latanya wrote t is greater than 12, and Jair wrote t is greater than 54. How did Latanya and Jair define their variables?"}, {"video_title": "Comparing linear rates example.mp3", "Sentence": "So we answered the first two parts of the question. And the second part of this question, they tell us that Latanya and Jair both wrote correct inequalities for the times in minutes when the dragon is farther east of the castle than the griffin is. Latanya wrote t is greater than 12, and Jair wrote t is greater than 54. How did Latanya and Jair define their variables? Well, Latanya defined her variables the exact same way that I define mine, because we got t is equal to 12 when the dragon passes up the griffin. So for t is greater than 12, the dragon is farther east of the castle than the griffin. So we did the exact same thing as Latanya."}, {"video_title": "Comparing linear rates example.mp3", "Sentence": "How did Latanya and Jair define their variables? Well, Latanya defined her variables the exact same way that I define mine, because we got t is equal to 12 when the dragon passes up the griffin. So for t is greater than 12, the dragon is farther east of the castle than the griffin. So we did the exact same thing as Latanya. But what Jair did, if he got t is greater than 54, is he must have defined t as being equal to the number of minutes since the griffin, the slower, the first but slower creature, passed the castle. And so he would have gotten t is greater than 54. And then if you wanted to know how many minutes since the dragon passed the castle, because the dragon got there 42 minutes later, he would have subtracted 42 from that."}, {"video_title": "Comparing linear rates example.mp3", "Sentence": "So we did the exact same thing as Latanya. But what Jair did, if he got t is greater than 54, is he must have defined t as being equal to the number of minutes since the griffin, the slower, the first but slower creature, passed the castle. And so he would have gotten t is greater than 54. And then if you wanted to know how many minutes since the dragon passed the castle, because the dragon got there 42 minutes later, he would have subtracted 42 from that. And that's how you connect these two numbers over here. But the important thing to realize is there's multiple ways to solve the same problem. What matters is to be very clear how you are defining that variable, and use it consistently throughout, and then interpret it correctly when you're answering the questions."}, {"video_title": "Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "So let's find all of the x's that satisfy this. So the first thing I'd like to do is get rid of this 3. So let's subtract 3 from both sides of this equation. So the left-hand side is just going to end up being 4x. These 3's cancel out. That just ends up with a 0. No reason to change the inequality just yet."}, {"video_title": "Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "So the left-hand side is just going to end up being 4x. These 3's cancel out. That just ends up with a 0. No reason to change the inequality just yet. We're just adding and subtracting from both sides. In this case, subtracting. That doesn't change the inequality as long as we're subtracting the same value."}, {"video_title": "Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "No reason to change the inequality just yet. We're just adding and subtracting from both sides. In this case, subtracting. That doesn't change the inequality as long as we're subtracting the same value. We have negative 1 minus 3. That is negative 4. Negative 1 minus 3 is negative 4."}, {"video_title": "Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "That doesn't change the inequality as long as we're subtracting the same value. We have negative 1 minus 3. That is negative 4. Negative 1 minus 3 is negative 4. And then we'll want to divide both sides of this equation by 4. Let's divide both sides of this equation by 4. Once again, when you multiply or divide both sides of an inequality by a positive number, it doesn't change the inequality."}, {"video_title": "Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "Negative 1 minus 3 is negative 4. And then we'll want to divide both sides of this equation by 4. Let's divide both sides of this equation by 4. Once again, when you multiply or divide both sides of an inequality by a positive number, it doesn't change the inequality. So the left-hand side is just x. x is less than negative 4 divided by 4 is negative 1. x is less than negative 1. Or we could write this in interval notation. All of the x's from negative infinity to negative 1, but not including negative 1."}, {"video_title": "Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "Once again, when you multiply or divide both sides of an inequality by a positive number, it doesn't change the inequality. So the left-hand side is just x. x is less than negative 4 divided by 4 is negative 1. x is less than negative 1. Or we could write this in interval notation. All of the x's from negative infinity to negative 1, but not including negative 1. So we put a parenthesis right there. Let's do a slightly harder one. Let's say we have 5x is greater than 8x plus 27."}, {"video_title": "Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "All of the x's from negative infinity to negative 1, but not including negative 1. So we put a parenthesis right there. Let's do a slightly harder one. Let's say we have 5x is greater than 8x plus 27. Let's get all our x's on the left-hand side. The best way to do that is subtract 8x from both sides. You subtract 8x from both sides."}, {"video_title": "Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "Let's say we have 5x is greater than 8x plus 27. Let's get all our x's on the left-hand side. The best way to do that is subtract 8x from both sides. You subtract 8x from both sides. The left-hand side becomes 5x minus 8x. That's negative 3x. We still have a greater than sign."}, {"video_title": "Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "You subtract 8x from both sides. The left-hand side becomes 5x minus 8x. That's negative 3x. We still have a greater than sign. We're just adding or subtracting the same quantity to both sides. These 8x's cancel out and you're just left with a 27. So you have negative 3x is greater than 27."}, {"video_title": "Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "We still have a greater than sign. We're just adding or subtracting the same quantity to both sides. These 8x's cancel out and you're just left with a 27. So you have negative 3x is greater than 27. Now, to just turn this into an x, we want to divide both sides by negative 3. But remember, when you multiply or divide both sides of an inequality by a negative number, you swap the inequality. So if we divide both sides of this by negative 3, we have to swap this inequality."}, {"video_title": "Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "So you have negative 3x is greater than 27. Now, to just turn this into an x, we want to divide both sides by negative 3. But remember, when you multiply or divide both sides of an inequality by a negative number, you swap the inequality. So if we divide both sides of this by negative 3, we have to swap this inequality. It will go from being a greater than sign to a less than sign. And this is a bit of a way that I remember greater than. The left-hand side just looks bigger."}, {"video_title": "Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "So if we divide both sides of this by negative 3, we have to swap this inequality. It will go from being a greater than sign to a less than sign. And this is a bit of a way that I remember greater than. The left-hand side just looks bigger. This is greater than. If you just imagine this height, that height is greater than that height right there, which is just a point. I don't know if that confuses you or not."}, {"video_title": "Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "The left-hand side just looks bigger. This is greater than. If you just imagine this height, that height is greater than that height right there, which is just a point. I don't know if that confuses you or not. This is less than. This little point is less than the distance of that big opening. That's how I remember it."}, {"video_title": "Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "I don't know if that confuses you or not. This is less than. This little point is less than the distance of that big opening. That's how I remember it. But anyway, 3x over negative 3. So anyway, now that we divided both sides by a negative number, by negative 3, we swapped the inequality from greater than to less than. In the left-hand side, the negative 3's cancel out."}, {"video_title": "Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "That's how I remember it. But anyway, 3x over negative 3. So anyway, now that we divided both sides by a negative number, by negative 3, we swapped the inequality from greater than to less than. In the left-hand side, the negative 3's cancel out. You get x is less than 27 over negative 3, which is negative 9. Or in interval notation, it would be everything from negative infinity to negative 9, not including negative 9. If you wanted to do it as a number line, it would look like this."}, {"video_title": "Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "In the left-hand side, the negative 3's cancel out. You get x is less than 27 over negative 3, which is negative 9. Or in interval notation, it would be everything from negative infinity to negative 9, not including negative 9. If you wanted to do it as a number line, it would look like this. This would be negative 9. Maybe this would be negative 8. Maybe this would be negative 10."}, {"video_title": "Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "If you wanted to do it as a number line, it would look like this. This would be negative 9. Maybe this would be negative 8. Maybe this would be negative 10. You would start at negative 9, not include it because we don't have an equal sign here. You go all the way, everything less than that, all the way down as we see to negative infinity. Let's do a nice, hairy problem."}, {"video_title": "Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "Maybe this would be negative 10. You would start at negative 9, not include it because we don't have an equal sign here. You go all the way, everything less than that, all the way down as we see to negative infinity. Let's do a nice, hairy problem. Let's say we have 8x minus 5 times 4x plus 1 is greater than or equal to negative 1 plus 2 times 4x minus 3. This might seem very daunting, but if we just simplify it step by step, you'll see it's no harder than any of the other problems we've tackled. Let's just simplify this."}, {"video_title": "Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "Let's do a nice, hairy problem. Let's say we have 8x minus 5 times 4x plus 1 is greater than or equal to negative 1 plus 2 times 4x minus 3. This might seem very daunting, but if we just simplify it step by step, you'll see it's no harder than any of the other problems we've tackled. Let's just simplify this. You get 8x minus, let's distribute this negative 5. Let me say 8x and then distribute the negative 5. Negative 5 times 4x is negative 20x."}, {"video_title": "Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "Let's just simplify this. You get 8x minus, let's distribute this negative 5. Let me say 8x and then distribute the negative 5. Negative 5 times 4x is negative 20x. Negative 5, when I say negative 5, I'm talking about this whole thing. Negative 5 times 1 is negative 5. Then that's going to be greater than or equal to negative 1 plus 2 times 4x is 8x."}, {"video_title": "Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "Negative 5 times 4x is negative 20x. Negative 5, when I say negative 5, I'm talking about this whole thing. Negative 5 times 1 is negative 5. Then that's going to be greater than or equal to negative 1 plus 2 times 4x is 8x. 2 times negative 3 is negative 6. Now we can merge these two terms. 8x minus 20x is negative 12x minus 5 is greater than or equal to, we can merge these constant terms, negative 1 minus 6, that's negative 7."}, {"video_title": "Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "Then that's going to be greater than or equal to negative 1 plus 2 times 4x is 8x. 2 times negative 3 is negative 6. Now we can merge these two terms. 8x minus 20x is negative 12x minus 5 is greater than or equal to, we can merge these constant terms, negative 1 minus 6, that's negative 7. Then we have this plus 8x left over. I like to get all my x terms on the left-hand side, so let's subtract 8x from both sides of this equation. Let's subtract 8x from both sides of this equation."}, {"video_title": "Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "8x minus 20x is negative 12x minus 5 is greater than or equal to, we can merge these constant terms, negative 1 minus 6, that's negative 7. Then we have this plus 8x left over. I like to get all my x terms on the left-hand side, so let's subtract 8x from both sides of this equation. Let's subtract 8x from both sides of this equation. From both sides, I'm subtracting 8x. This left-hand side, negative 12 minus 8, that's negative 20. Negative 20x minus 5."}, {"video_title": "Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "Let's subtract 8x from both sides of this equation. From both sides, I'm subtracting 8x. This left-hand side, negative 12 minus 8, that's negative 20. Negative 20x minus 5. Once again, no reason to change the inequality just yet. All we're doing is simplifying the sides or adding and subtracting from them. The right-hand side becomes, this thing cancels out, 8x minus 8x, that's 0."}, {"video_title": "Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "Negative 20x minus 5. Once again, no reason to change the inequality just yet. All we're doing is simplifying the sides or adding and subtracting from them. The right-hand side becomes, this thing cancels out, 8x minus 8x, that's 0. You're just left with a negative 7. Now I want to get rid of this negative 5, so let's add 5 to both sides of this equation. The left-hand side, you're just left with a negative 20x."}, {"video_title": "Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "The right-hand side becomes, this thing cancels out, 8x minus 8x, that's 0. You're just left with a negative 7. Now I want to get rid of this negative 5, so let's add 5 to both sides of this equation. The left-hand side, you're just left with a negative 20x. The 5 and these 5s cancel out. No reason to change the inequality just yet. Negative 7 plus 5, that's negative 2."}, {"video_title": "Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "The left-hand side, you're just left with a negative 20x. The 5 and these 5s cancel out. No reason to change the inequality just yet. Negative 7 plus 5, that's negative 2. Now we're at an interesting point. We have negative 20x is greater than or equal to negative 2. If this was an equation or really any type of even inequality, we want to divide both sides by negative 20, but we have to remember, when you multiply or divide both sides of an inequality by a negative number, you have to swap the inequality, so let's remember that."}, {"video_title": "Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "Negative 7 plus 5, that's negative 2. Now we're at an interesting point. We have negative 20x is greater than or equal to negative 2. If this was an equation or really any type of even inequality, we want to divide both sides by negative 20, but we have to remember, when you multiply or divide both sides of an inequality by a negative number, you have to swap the inequality, so let's remember that. If we divide this side by negative 20 and we divide this side by negative 20, all I did is took both of these sides, divided by negative 20. We have to swap the inequality. The greater than or equal to has to become a less than or equal sign."}, {"video_title": "Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "If this was an equation or really any type of even inequality, we want to divide both sides by negative 20, but we have to remember, when you multiply or divide both sides of an inequality by a negative number, you have to swap the inequality, so let's remember that. If we divide this side by negative 20 and we divide this side by negative 20, all I did is took both of these sides, divided by negative 20. We have to swap the inequality. The greater than or equal to has to become a less than or equal sign. Of course, these cancel out. You get x is less than or equal to, the negatives cancel out. 2 over 20 is 1 over 10."}, {"video_title": "Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "The greater than or equal to has to become a less than or equal sign. Of course, these cancel out. You get x is less than or equal to, the negatives cancel out. 2 over 20 is 1 over 10. If we were writing it in interval notation, the upper bound would be 1 over 10. Notice we're including it because we have an equal sign, less than or equal, so we're including 1 over 10, and we're going to go all the way down to negative infinity, everything less than or equal to 1 over 10. This is just another way of writing that."}, {"video_title": "Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "2 over 20 is 1 over 10. If we were writing it in interval notation, the upper bound would be 1 over 10. Notice we're including it because we have an equal sign, less than or equal, so we're including 1 over 10, and we're going to go all the way down to negative infinity, everything less than or equal to 1 over 10. This is just another way of writing that. Just for fun, let's draw the number line right here. This is maybe 0. That is 1."}, {"video_title": "Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3", "Sentence": "This is just another way of writing that. Just for fun, let's draw the number line right here. This is maybe 0. That is 1. 1 over 10 might be over here. Everything less than or equal to 1 over 10. We're going to include the 1 tenth and everything less than that is included in the solutions."}, {"video_title": "Finding features of quadratic functions Mathematics II High School Math Khan Academy.mp3", "Sentence": "I know they're all called f, but we'll just assume they are different functions. And for each of these, I want to do three things. I want to find the zeros. And so the zeros are the input values that make the value of the function equal to zero. So here would be the t values that make f of t equal zero. Here would be the x values that make the function equal zero. So I want to find the zeros."}, {"video_title": "Finding features of quadratic functions Mathematics II High School Math Khan Academy.mp3", "Sentence": "And so the zeros are the input values that make the value of the function equal to zero. So here would be the t values that make f of t equal zero. Here would be the x values that make the function equal zero. So I want to find the zeros. I also want to find the coordinates of the vertex. And finally, I want to find the equation of the line of symmetry. Line of symmetry."}, {"video_title": "Finding features of quadratic functions Mathematics II High School Math Khan Academy.mp3", "Sentence": "So I want to find the zeros. I also want to find the coordinates of the vertex. And finally, I want to find the equation of the line of symmetry. Line of symmetry. And in particular, to make it a little bit more specific, the vertical line of symmetry, which will actually be the only line of symmetry for these three. So pause the video and see if you can figure out the zeros, the vertex, and the line of symmetry. And I'm assuming you just did that."}, {"video_title": "Finding features of quadratic functions Mathematics II High School Math Khan Academy.mp3", "Sentence": "Line of symmetry. And in particular, to make it a little bit more specific, the vertical line of symmetry, which will actually be the only line of symmetry for these three. So pause the video and see if you can figure out the zeros, the vertex, and the line of symmetry. And I'm assuming you just did that. And now I'm going to attempt to do it. And if at any point you get inspired, pause the video again and keep on working on it. The best way to learn this stuff is to do it yourself."}, {"video_title": "Finding features of quadratic functions Mathematics II High School Math Khan Academy.mp3", "Sentence": "And I'm assuming you just did that. And now I'm going to attempt to do it. And if at any point you get inspired, pause the video again and keep on working on it. The best way to learn this stuff is to do it yourself. So let's see. So let's first find the zeros. So to find the zeros, we can set t minus five squared minus nine equal to zero."}, {"video_title": "Finding features of quadratic functions Mathematics II High School Math Khan Academy.mp3", "Sentence": "The best way to learn this stuff is to do it yourself. So let's see. So let's first find the zeros. So to find the zeros, we can set t minus five squared minus nine equal to zero. So we could say t, for what t's does t minus five squared minus nine equal zero? Let's see, to solve this, we could add nine to both sides. And so we could say, if we add nine to both sides, the left-hand side's just t minus five squared."}, {"video_title": "Finding features of quadratic functions Mathematics II High School Math Khan Academy.mp3", "Sentence": "So to find the zeros, we can set t minus five squared minus nine equal to zero. So we could say t, for what t's does t minus five squared minus nine equal zero? Let's see, to solve this, we could add nine to both sides. And so we could say, if we add nine to both sides, the left-hand side's just t minus five squared. The right-hand side is going to be nine. And so if t minus five squared is nine, that means that t minus five could be equal to the positive square root of nine, or t minus five could equal the negative square root of nine. And to solve for t, we could add five to both sides, so we get t is equal to eight, or t is equal to, if we add five to both sides here, t is equal to two."}, {"video_title": "Finding features of quadratic functions Mathematics II High School Math Khan Academy.mp3", "Sentence": "And so we could say, if we add nine to both sides, the left-hand side's just t minus five squared. The right-hand side is going to be nine. And so if t minus five squared is nine, that means that t minus five could be equal to the positive square root of nine, or t minus five could equal the negative square root of nine. And to solve for t, we could add five to both sides, so we get t is equal to eight, or t is equal to, if we add five to both sides here, t is equal to two. And just like that, we have found the zeros for this function, because if t is equal to eight or two, the function is going to be zero. F of eight is zero, and f of two is going to be zero. Now let's find the vertex, the coordinates of the vertex."}, {"video_title": "Finding features of quadratic functions Mathematics II High School Math Khan Academy.mp3", "Sentence": "And to solve for t, we could add five to both sides, so we get t is equal to eight, or t is equal to, if we add five to both sides here, t is equal to two. And just like that, we have found the zeros for this function, because if t is equal to eight or two, the function is going to be zero. F of eight is zero, and f of two is going to be zero. Now let's find the vertex, the coordinates of the vertex. The coordinates of the vertex. So the x-coordinate of the vertex, or sorry, I should say the t-coordinate of the vertex, since the input variable here is t. The t-coordinate of the vertex is going to be halfway in between the zeros. It's going to be halfway in between where the parabola in this case is going to intersect the x-axis, or the t-axis."}, {"video_title": "Finding features of quadratic functions Mathematics II High School Math Khan Academy.mp3", "Sentence": "Now let's find the vertex, the coordinates of the vertex. The coordinates of the vertex. So the x-coordinate of the vertex, or sorry, I should say the t-coordinate of the vertex, since the input variable here is t. The t-coordinate of the vertex is going to be halfway in between the zeros. It's going to be halfway in between where the parabola in this case is going to intersect the x-axis, or the t-axis. I keep saying x-axis, the t-axis for this case. So halfway between eight and two, well it's going to be the average of them, eight plus two over two, that's 10 over two, that's five. So the t-coordinate is five, and five is three away from eight, and three away from two."}, {"video_title": "Finding features of quadratic functions Mathematics II High School Math Khan Academy.mp3", "Sentence": "It's going to be halfway in between where the parabola in this case is going to intersect the x-axis, or the t-axis. I keep saying x-axis, the t-axis for this case. So halfway between eight and two, well it's going to be the average of them, eight plus two over two, that's 10 over two, that's five. So the t-coordinate is five, and five is three away from eight, and three away from two. And when t is equal to five, what is f of t, or what is f of five? Well when t is equal to five, five minus five squared is just zero, and then f of, so f of five is just going to be negative nine. And this form of a function, this is actually called vertex form, because it's very easy to pick out the vertex."}, {"video_title": "Finding features of quadratic functions Mathematics II High School Math Khan Academy.mp3", "Sentence": "So the t-coordinate is five, and five is three away from eight, and three away from two. And when t is equal to five, what is f of t, or what is f of five? Well when t is equal to five, five minus five squared is just zero, and then f of, so f of five is just going to be negative nine. And this form of a function, this is actually called vertex form, because it's very easy to pick out the vertex. It's very easy to realize, like, okay look, for this particular one, we're going to hit a minimum point when this part of the expression is equal to zero, because this thing, the lowest value it can take on is zero, because you're squaring it, it can never take on a negative value, and it takes on zero when t is equal to five, and if this part is zero, then the f of five is going to be negative nine. So just like that, we have established the vertex. Now we actually have a lot of information if we wanted to draw it."}, {"video_title": "Finding features of quadratic functions Mathematics II High School Math Khan Academy.mp3", "Sentence": "And this form of a function, this is actually called vertex form, because it's very easy to pick out the vertex. It's very easy to realize, like, okay look, for this particular one, we're going to hit a minimum point when this part of the expression is equal to zero, because this thing, the lowest value it can take on is zero, because you're squaring it, it can never take on a negative value, and it takes on zero when t is equal to five, and if this part is zero, then the f of five is going to be negative nine. So just like that, we have established the vertex. Now we actually have a lot of information if we wanted to draw it. So if we want to draw this function, I'll just do a very quick sketch of it. Whoops. So a very quick sketch of it."}, {"video_title": "Finding features of quadratic functions Mathematics II High School Math Khan Academy.mp3", "Sentence": "Now we actually have a lot of information if we wanted to draw it. So if we want to draw this function, I'll just do a very quick sketch of it. Whoops. So a very quick sketch of it. So that is our t-axis, not our x-axis, after I keep reminding myself. And that is my, let's call that my y-axis, and we're going to graph y is equal to f of t. Well, we know the vertex is at the point five, negative nine. So this is t is equal to five, and y is equal to negative nine."}, {"video_title": "Finding features of quadratic functions Mathematics II High School Math Khan Academy.mp3", "Sentence": "So a very quick sketch of it. So that is our t-axis, not our x-axis, after I keep reminding myself. And that is my, let's call that my y-axis, and we're going to graph y is equal to f of t. Well, we know the vertex is at the point five, negative nine. So this is t is equal to five, and y is equal to negative nine. So that's the vertex, right over there. And then we know we have zeros at t equals eight and t equals two. So t equals eight and t equals two."}, {"video_title": "Finding features of quadratic functions Mathematics II High School Math Khan Academy.mp3", "Sentence": "So this is t is equal to five, and y is equal to negative nine. So that's the vertex, right over there. And then we know we have zeros at t equals eight and t equals two. So t equals eight and t equals two. Let me make that a little bit, and t equals two. Those are the two zeros, so eight and two. And so just like that, we can graph f of t, or we can graph y is equal to f of t. So y is equal to f of t is going to look something like, let me draw something like that."}, {"video_title": "Finding features of quadratic functions Mathematics II High School Math Khan Academy.mp3", "Sentence": "So t equals eight and t equals two. Let me make that a little bit, and t equals two. Those are the two zeros, so eight and two. And so just like that, we can graph f of t, or we can graph y is equal to f of t. So y is equal to f of t is going to look something like, let me draw something like that. That's the graph of y is equal to f of t. Now the last thing that I said is the line of symmetry. Well, the line of symmetry is going to be the vertical line that goes through the vertex. So the equation of that line of symmetry is going to be t is equal to five."}, {"video_title": "Finding features of quadratic functions Mathematics II High School Math Khan Academy.mp3", "Sentence": "And so just like that, we can graph f of t, or we can graph y is equal to f of t. So y is equal to f of t is going to look something like, let me draw something like that. That's the graph of y is equal to f of t. Now the last thing that I said is the line of symmetry. Well, the line of symmetry is going to be the vertical line that goes through the vertex. So the equation of that line of symmetry is going to be t is equal to five. And it's really just the t coordinate of the vertex that defines the line of symmetry. Let's do the other two right over here. So what are the zeros?"}, {"video_title": "Finding features of quadratic functions Mathematics II High School Math Khan Academy.mp3", "Sentence": "So the equation of that line of symmetry is going to be t is equal to five. And it's really just the t coordinate of the vertex that defines the line of symmetry. Let's do the other two right over here. So what are the zeros? Well, if you set this equal to zero, if we say x plus two times x plus four is equal to zero, well that's going to happen if x plus two is equal to zero, or x plus four is equal to zero. This is going to happen if we subtract two from both sides when x is negative two, and if we subtract four from both sides, or when x is equal to negative four. As we said, the vertex, the x coordinate of the vertex is going to be halfway in between these."}, {"video_title": "Finding features of quadratic functions Mathematics II High School Math Khan Academy.mp3", "Sentence": "So what are the zeros? Well, if you set this equal to zero, if we say x plus two times x plus four is equal to zero, well that's going to happen if x plus two is equal to zero, or x plus four is equal to zero. This is going to happen if we subtract two from both sides when x is negative two, and if we subtract four from both sides, or when x is equal to negative four. As we said, the vertex, the x coordinate of the vertex is going to be halfway in between these. So it's going to be negative two plus negative four over two. So that would be negative six over two, which is just negative three. Negative three."}, {"video_title": "Finding features of quadratic functions Mathematics II High School Math Khan Academy.mp3", "Sentence": "As we said, the vertex, the x coordinate of the vertex is going to be halfway in between these. So it's going to be negative two plus negative four over two. So that would be negative six over two, which is just negative three. Negative three. And when x is negative three, f of x is going to be, let's see, it's going to be negative one times one. Right? Negative three plus two is negative one, and so times one."}, {"video_title": "Finding features of quadratic functions Mathematics II High School Math Khan Academy.mp3", "Sentence": "Negative three. And when x is negative three, f of x is going to be, let's see, it's going to be negative one times one. Right? Negative three plus two is negative one, and so times one. So it's just going to be negative one. There you have it. And the line of symmetry is going to be the vertical line x is equal to negative three."}, {"video_title": "Finding features of quadratic functions Mathematics II High School Math Khan Academy.mp3", "Sentence": "Negative three plus two is negative one, and so times one. So it's just going to be negative one. There you have it. And the line of symmetry is going to be the vertical line x is equal to negative three. And once again, we can graph that really fast. So let me, this is my y-axis. See, everything is happening for negative x's, so I'll draw it a little bit more skewed this way."}, {"video_title": "Finding features of quadratic functions Mathematics II High School Math Khan Academy.mp3", "Sentence": "And the line of symmetry is going to be the vertical line x is equal to negative three. And once again, we can graph that really fast. So let me, this is my y-axis. See, everything is happening for negative x's, so I'll draw it a little bit more skewed this way. This is my x-axis, and we see that we have zeros at x equals negative two and x is equal to negative four. So negative one, two, three, four. So we have zeros, we have zeros there."}, {"video_title": "Finding features of quadratic functions Mathematics II High School Math Khan Academy.mp3", "Sentence": "See, everything is happening for negative x's, so I'll draw it a little bit more skewed this way. This is my x-axis, and we see that we have zeros at x equals negative two and x is equal to negative four. So negative one, two, three, four. So we have zeros, we have zeros there. Negative two, let me be careful. Negative two and negative four. And the vertex is at negative three comma negative one."}, {"video_title": "Finding features of quadratic functions Mathematics II High School Math Khan Academy.mp3", "Sentence": "So we have zeros, we have zeros there. Negative two, let me be careful. Negative two and negative four. And the vertex is at negative three comma negative one. So negative three comma negative one. Let me make sure we see that. So this is negative one right over here."}, {"video_title": "Finding features of quadratic functions Mathematics II High School Math Khan Academy.mp3", "Sentence": "And the vertex is at negative three comma negative one. So negative three comma negative one. Let me make sure we see that. So this is negative one right over here. Negative one, this is negative two, this is negative four. And so we can sketch out what the graph of y is equal to f of x is going to look like. It's going to look something like that."}, {"video_title": "Finding features of quadratic functions Mathematics II High School Math Khan Academy.mp3", "Sentence": "So this is negative one right over here. Negative one, this is negative two, this is negative four. And so we can sketch out what the graph of y is equal to f of x is going to look like. It's going to look something like that. That is y is equal to f of x. Let's do one more. Hopefully we're getting the hang of this."}, {"video_title": "Finding features of quadratic functions Mathematics II High School Math Khan Academy.mp3", "Sentence": "It's going to look something like that. That is y is equal to f of x. Let's do one more. Hopefully we're getting the hang of this. So here to solve x squared plus six x plus eight is equal to zero, it will be useful to factor this. And so this can be written as, and if you have trouble doing this, I encourage you to watch videos on factoring polynomials. What adds up to six and when you take their product is eight?"}, {"video_title": "Finding features of quadratic functions Mathematics II High School Math Khan Academy.mp3", "Sentence": "Hopefully we're getting the hang of this. So here to solve x squared plus six x plus eight is equal to zero, it will be useful to factor this. And so this can be written as, and if you have trouble doing this, I encourage you to watch videos on factoring polynomials. What adds up to six and when you take their product is eight? Well, four and two. Four plus two is six, and four times two is eight. So is equal to zero."}, {"video_title": "Finding features of quadratic functions Mathematics II High School Math Khan Academy.mp3", "Sentence": "What adds up to six and when you take their product is eight? Well, four and two. Four plus two is six, and four times two is eight. So is equal to zero. And then this is actually the exact same thing as what we have in blue right over here. These are actually the exact same function. They're just written in different forms."}, {"video_title": "Finding features of quadratic functions Mathematics II High School Math Khan Academy.mp3", "Sentence": "So is equal to zero. And then this is actually the exact same thing as what we have in blue right over here. These are actually the exact same function. They're just written in different forms. And so the solutions are going to be the exact solutions that we just saw right over here, and the graph is going to be the same thing that we have right over there. So same vertex, same line of symmetry, same zeros. These functions were just written in different ways."}, {"video_title": "Introduction to increasing, decreasing, positive or negative intervals Algebra I Khan Academy.mp3", "Sentence": "Well, positive means that the value of the function is greater than zero. It means that the value of the function, this means that the function is sitting above the x-axis. So it's sitting above the x-axis in this place right over here that I am highlighting in yellow, and it is also sitting above the x-axis over here. And if we wanted to, if we wanted to write those intervals mathematically, well let's see, let's say that this point, let's say that this point right over here is x equals a, let's say that this right over here is x equals b, and this right over here is x equals c, then it's positive, it's positive as long as x is between a and b. At x equals a or at x equals b, the value of our function is zero, but it's positive when x is between a and b, a and b, or if x is greater than c. X is, we could write it there, c is less than x, or we could write that x is greater than c. These are the intervals when our function is positive. Let me write this, f of x, f of x positive when x is in this interval or this interval, or that interval. So when is f of x negative?"}, {"video_title": "Introduction to increasing, decreasing, positive or negative intervals Algebra I Khan Academy.mp3", "Sentence": "And if we wanted to, if we wanted to write those intervals mathematically, well let's see, let's say that this point, let's say that this point right over here is x equals a, let's say that this right over here is x equals b, and this right over here is x equals c, then it's positive, it's positive as long as x is between a and b. At x equals a or at x equals b, the value of our function is zero, but it's positive when x is between a and b, a and b, or if x is greater than c. X is, we could write it there, c is less than x, or we could write that x is greater than c. These are the intervals when our function is positive. Let me write this, f of x, f of x positive when x is in this interval or this interval, or that interval. So when is f of x negative? Let me do this in another color. F of x is going to be negative. Well, it's going to be negative if x is less than a."}, {"video_title": "Introduction to increasing, decreasing, positive or negative intervals Algebra I Khan Academy.mp3", "Sentence": "So when is f of x negative? Let me do this in another color. F of x is going to be negative. Well, it's going to be negative if x is less than a. So this is if x is less than a, or if x is between b and c, then we see that f of x is below the x axis. F of x is down here. So this is where it's negative."}, {"video_title": "Introduction to increasing, decreasing, positive or negative intervals Algebra I Khan Academy.mp3", "Sentence": "Well, it's going to be negative if x is less than a. So this is if x is less than a, or if x is between b and c, then we see that f of x is below the x axis. F of x is down here. So this is where it's negative. So here, or, or x is between b or c. x is between b and c, and I'm not saying less than or equal to, because at b or c, the value of the function, f of b is zero. f of c is zero. That's where we are actually intersecting the x axis."}, {"video_title": "Introduction to increasing, decreasing, positive or negative intervals Algebra I Khan Academy.mp3", "Sentence": "So this is where it's negative. So here, or, or x is between b or c. x is between b and c, and I'm not saying less than or equal to, because at b or c, the value of the function, f of b is zero. f of c is zero. That's where we are actually intersecting the x axis. So that was reasonably straightforward. Now let's ask ourselves a different question. When is the function increasing or decreasing?"}, {"video_title": "Introduction to increasing, decreasing, positive or negative intervals Algebra I Khan Academy.mp3", "Sentence": "That's where we are actually intersecting the x axis. So that was reasonably straightforward. Now let's ask ourselves a different question. When is the function increasing or decreasing? So when is f of x, f of x in, increasing? Well, increasing, one way to think about it is every time that x is increasing, then y should be increasing. Or another way to think about it, you have a positive rate of change of y with respect to x."}, {"video_title": "Introduction to increasing, decreasing, positive or negative intervals Algebra I Khan Academy.mp3", "Sentence": "When is the function increasing or decreasing? So when is f of x, f of x in, increasing? Well, increasing, one way to think about it is every time that x is increasing, then y should be increasing. Or another way to think about it, you have a positive rate of change of y with respect to x. Or you could even think about it as, imagine if you had a tangent line at any of these points. If you had a tangent line at any of these points, the slope of that tangent line is going to be positive. But the easiest way for me to think about it is, as you increase x, you're going to be increasing y."}, {"video_title": "Introduction to increasing, decreasing, positive or negative intervals Algebra I Khan Academy.mp3", "Sentence": "Or another way to think about it, you have a positive rate of change of y with respect to x. Or you could even think about it as, imagine if you had a tangent line at any of these points. If you had a tangent line at any of these points, the slope of that tangent line is going to be positive. But the easiest way for me to think about it is, as you increase x, you're going to be increasing y. So where is the function increasing? Well, I'm doing it in blue. So it's increasing right until we get to this, this point right over here."}, {"video_title": "Introduction to increasing, decreasing, positive or negative intervals Algebra I Khan Academy.mp3", "Sentence": "But the easiest way for me to think about it is, as you increase x, you're going to be increasing y. So where is the function increasing? Well, I'm doing it in blue. So it's increasing right until we get to this, this point right over here. Right until we get to that point over there. Then it starts decreasing until we get to this point right over here. And then it starts increasing again."}, {"video_title": "Introduction to increasing, decreasing, positive or negative intervals Algebra I Khan Academy.mp3", "Sentence": "So it's increasing right until we get to this, this point right over here. Right until we get to that point over there. Then it starts decreasing until we get to this point right over here. And then it starts increasing again. It starts, it starts increasing again. So let me make some more, let me make some more labels here. So let's say that this is x equals d, and that this right over here, actually let me do that green color."}, {"video_title": "Introduction to increasing, decreasing, positive or negative intervals Algebra I Khan Academy.mp3", "Sentence": "And then it starts increasing again. It starts, it starts increasing again. So let me make some more, let me make some more labels here. So let's say that this is x equals d, and that this right over here, actually let me do that green color. So let's say this is x equals d. Now it's not a, d, b, but you get the picture. And let's say that this is x is equal to, x is equal to, let me do it a little bit. x is equal to e. x is equal to e. So when is this function increasing?"}, {"video_title": "Introduction to increasing, decreasing, positive or negative intervals Algebra I Khan Academy.mp3", "Sentence": "So let's say that this is x equals d, and that this right over here, actually let me do that green color. So let's say this is x equals d. Now it's not a, d, b, but you get the picture. And let's say that this is x is equal to, x is equal to, let me do it a little bit. x is equal to e. x is equal to e. So when is this function increasing? Well, it's increasing if x is less than d. x is less than d. And I'm not gonna say less than or equal to, because right at x equals d, it looks like, just for that moment, the slope of the tangent line looks like it would be, it would be constant. We're going from increasing to decreasing. So right at d, we're neither increasing or decreasing."}, {"video_title": "Introduction to increasing, decreasing, positive or negative intervals Algebra I Khan Academy.mp3", "Sentence": "x is equal to e. x is equal to e. So when is this function increasing? Well, it's increasing if x is less than d. x is less than d. And I'm not gonna say less than or equal to, because right at x equals d, it looks like, just for that moment, the slope of the tangent line looks like it would be, it would be constant. We're going from increasing to decreasing. So right at d, we're neither increasing or decreasing. But then we're also increasing. So if x is less than d, or x is greater than e, or x is greater than e. And where is f of x decreasing? So f of x, let me do this in a different color."}, {"video_title": "Introduction to increasing, decreasing, positive or negative intervals Algebra I Khan Academy.mp3", "Sentence": "So right at d, we're neither increasing or decreasing. But then we're also increasing. So if x is less than d, or x is greater than e, or x is greater than e. And where is f of x decreasing? So f of x, let me do this in a different color. When is, let me pick a move. So f of x decreasing, decreasing, well it's going to be right over here. It's gonna be right between d and e, between x equals d and x equals e, but not exactly at those points, because at both of those points, you're neither increasing nor decreasing."}, {"video_title": "Introduction to increasing, decreasing, positive or negative intervals Algebra I Khan Academy.mp3", "Sentence": "So f of x, let me do this in a different color. When is, let me pick a move. So f of x decreasing, decreasing, well it's going to be right over here. It's gonna be right between d and e, between x equals d and x equals e, but not exactly at those points, because at both of those points, you're neither increasing nor decreasing. But you see right over here, as x increases, as you increase your x, what's happening to your y? If you go from this point and you increase your x, what happened to your y? Your y has decreased."}, {"video_title": "Introduction to increasing, decreasing, positive or negative intervals Algebra I Khan Academy.mp3", "Sentence": "It's gonna be right between d and e, between x equals d and x equals e, but not exactly at those points, because at both of those points, you're neither increasing nor decreasing. But you see right over here, as x increases, as you increase your x, what's happening to your y? If you go from this point and you increase your x, what happened to your y? Your y has decreased. You increase your x, your y has decreased. You increase your x, y has decreased. Increase x, y has decreased, all the way until this point over here."}, {"video_title": "Introduction to increasing, decreasing, positive or negative intervals Algebra I Khan Academy.mp3", "Sentence": "Your y has decreased. You increase your x, your y has decreased. You increase your x, y has decreased. Increase x, y has decreased, all the way until this point over here. So f of x is decreasing for x between d and e. So hopefully that gives you a sense of things. Notice these aren't the same intervals. The intervals where we're positive or negative don't perfectly coincide with when we are increasing or decreasing, so it's very important to think about these separately, even though they kind of sound the same."}, {"video_title": "Graphing shifted functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "It's shown in the grid below. Graph the function g of x is equal to x minus two squared minus four in the interactive graph. And this is from the shifting functions exercise on Khan Academy. And we can see we can change, we can change the, we can change the graph of g of x. But let's see, we want to graph it properly. So let's see how they relate. Well, let's think about a few things."}, {"video_title": "Graphing shifted functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "And we can see we can change, we can change the, we can change the graph of g of x. But let's see, we want to graph it properly. So let's see how they relate. Well, let's think about a few things. Let's first just make g of x completely overlap. Well, actually, that's completely easier said than, okay, there you go. Now they're completely overlapping."}, {"video_title": "Graphing shifted functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Well, let's think about a few things. Let's first just make g of x completely overlap. Well, actually, that's completely easier said than, okay, there you go. Now they're completely overlapping. And let's see how they're different. Well, g of x, if you look at what's going on here, instead of having an x squared, we have an x minus two squared. So one way to think about it is, when x is zero, you have zero squared is equal to zero."}, {"video_title": "Graphing shifted functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Now they're completely overlapping. And let's see how they're different. Well, g of x, if you look at what's going on here, instead of having an x squared, we have an x minus two squared. So one way to think about it is, when x is zero, you have zero squared is equal to zero. But how do you get zero here? Well, x has got to be equal to two. Two minus two squared is zero squared."}, {"video_title": "Graphing shifted functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So one way to think about it is, when x is zero, you have zero squared is equal to zero. But how do you get zero here? Well, x has got to be equal to two. Two minus two squared is zero squared. If we don't look at the negative four just yet. And so we would want to shift this graph over two to the right. This is essentially how much do we shift to the right."}, {"video_title": "Graphing shifted functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Two minus two squared is zero squared. If we don't look at the negative four just yet. And so we would want to shift this graph over two to the right. This is essentially how much do we shift to the right. It's sometimes a little bit counterintuitive that we have a negative there, because you might say, well, negative, that makes me think that I want to shift to the left. But you have to remind yourself is like, well, okay, for the original graph, when it was just x squared, to get the zero squared, I just had to put x equals zero. Now to get a zero squared, I have to put in a two."}, {"video_title": "Graphing shifted functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "This is essentially how much do we shift to the right. It's sometimes a little bit counterintuitive that we have a negative there, because you might say, well, negative, that makes me think that I want to shift to the left. But you have to remind yourself is like, well, okay, for the original graph, when it was just x squared, to get the zero squared, I just had to put x equals zero. Now to get a zero squared, I have to put in a two. So this is actually shifting the graph to the right. And so what do we do with this negative four? Well, this is a little bit more intuitive, or at least for me when I first learned it."}, {"video_title": "Graphing shifted functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Now to get a zero squared, I have to put in a two. So this is actually shifting the graph to the right. And so what do we do with this negative four? Well, this is a little bit more intuitive, or at least for me when I first learned it. This literally will just shift the graph down. Whatever your value is of x minus two squared, it's gonna shift it down by four. So what we want to do is just shift both of these points down by four."}, {"video_title": "Graphing shifted functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Well, this is a little bit more intuitive, or at least for me when I first learned it. This literally will just shift the graph down. Whatever your value is of x minus two squared, it's gonna shift it down by four. So what we want to do is just shift both of these points down by four. So this is gonna go from nine, this is gonna go from the coordinate five comma nine to five comma, if we go down four, five comma five. And this is gonna go from two comma zero to two comma negative four. Two comma negative four."}, {"video_title": "Graphing shifted functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So what we want to do is just shift both of these points down by four. So this is gonna go from nine, this is gonna go from the coordinate five comma nine to five comma, if we go down four, five comma five. And this is gonna go from two comma zero to two comma negative four. Two comma negative four. Did I do that right? I think that is, I think that's right. What essentially what we have going on is g of x is f of x shifted two to the right and four down, two to the right and four down."}, {"video_title": "Graphing shifted functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Two comma negative four. Did I do that right? I think that is, I think that's right. What essentially what we have going on is g of x is f of x shifted two to the right and four down, two to the right and four down. And notice, if you look at the vertex here, we shifted two to the right and four down. And I shifted this one also, this one also I shifted two to the right and four down. And there you have it."}, {"video_title": "Introduction to factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "We recognized that an expression like x squared could be written as x times x. We also recognized that a polynomial like three x squared plus four x, that in this situation, both terms had the common factor of x, and you could factor that out. And so you could rewrite this as x times three x plus four. And we also learned to do fancier things. We learned to factor things like x squared plus seven x plus 12. We were able to say, hey, what two numbers would add up to seven, and if I were to multiply them, I'd get 12, and in those early videos, we show why that works, and say, well, three and four, so maybe this can be factored as x plus three times x plus four. If this is unfamiliar to you, I encourage you to go review that in some of the introductory factoring quadratics on Khan Academy."}, {"video_title": "Introduction to factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "And we also learned to do fancier things. We learned to factor things like x squared plus seven x plus 12. We were able to say, hey, what two numbers would add up to seven, and if I were to multiply them, I'd get 12, and in those early videos, we show why that works, and say, well, three and four, so maybe this can be factored as x plus three times x plus four. If this is unfamiliar to you, I encourage you to go review that in some of the introductory factoring quadratics on Khan Academy. It should be a review at this point in your journey. We also looked at things like differences of squares, x squared minus nine. Say, hey, that's x squared minus three squared, so we could factor that as x plus three times x minus three, and we looked at other types of quadratics."}, {"video_title": "Introduction to factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "If this is unfamiliar to you, I encourage you to go review that in some of the introductory factoring quadratics on Khan Academy. It should be a review at this point in your journey. We also looked at things like differences of squares, x squared minus nine. Say, hey, that's x squared minus three squared, so we could factor that as x plus three times x minus three, and we looked at other types of quadratics. Now, as we go deeper into our algebra journeys, we're going to build on this to factor higher degree polynomials, third degree, fourth degree, fifth degree, which will be very useful in your mathematical careers, but we're going to start doing it by really looking at some of the structure, some of the patterns that we've seen in introductory algebra. For example, let's say someone walks up to you on the street and says, can you factor x to the third plus seven x squared plus 12x? Well, at first you might say, oh, this is a third degree polynomial, that seems kind of intimidating, until you realize, hey, all of these terms have the common factor x, so if I factor that out, then it becomes x times x squared plus seven x plus 12, and then this is exactly what we saw over here, so we could rewrite all of this as x times x plus three times x plus four, so we're going to see that we might be able to do some simple factoring like this and even factoring multiple times."}, {"video_title": "Introduction to factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "Say, hey, that's x squared minus three squared, so we could factor that as x plus three times x minus three, and we looked at other types of quadratics. Now, as we go deeper into our algebra journeys, we're going to build on this to factor higher degree polynomials, third degree, fourth degree, fifth degree, which will be very useful in your mathematical careers, but we're going to start doing it by really looking at some of the structure, some of the patterns that we've seen in introductory algebra. For example, let's say someone walks up to you on the street and says, can you factor x to the third plus seven x squared plus 12x? Well, at first you might say, oh, this is a third degree polynomial, that seems kind of intimidating, until you realize, hey, all of these terms have the common factor x, so if I factor that out, then it becomes x times x squared plus seven x plus 12, and then this is exactly what we saw over here, so we could rewrite all of this as x times x plus three times x plus four, so we're going to see that we might be able to do some simple factoring like this and even factoring multiple times. We might also start to appreciate structure that brings us back to some of what we saw in our introductory algebra, so for example, you might see something like this, where once again, someone walks up to you on the street and says, hey, you factor this, a to the fourth power plus seven a squared plus 12, and at first, you're like, wow, there's a fourth power here, what do I do, until you say, well, what if I were to rewrite this as a squared squared plus seven a squared plus 12, and now this a squared is looking an awful lot like this x over here. If this were an x, then this would be x squared. If this were an x, then this would just be an x, and then these expressions would be the same, so when I factor it, everywhere I see an x, I could replace with an a squared, so I could factor this out, really looking at the same structure we have here as a squared plus three times a squared plus four."}, {"video_title": "Introduction to factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "Well, at first you might say, oh, this is a third degree polynomial, that seems kind of intimidating, until you realize, hey, all of these terms have the common factor x, so if I factor that out, then it becomes x times x squared plus seven x plus 12, and then this is exactly what we saw over here, so we could rewrite all of this as x times x plus three times x plus four, so we're going to see that we might be able to do some simple factoring like this and even factoring multiple times. We might also start to appreciate structure that brings us back to some of what we saw in our introductory algebra, so for example, you might see something like this, where once again, someone walks up to you on the street and says, hey, you factor this, a to the fourth power plus seven a squared plus 12, and at first, you're like, wow, there's a fourth power here, what do I do, until you say, well, what if I were to rewrite this as a squared squared plus seven a squared plus 12, and now this a squared is looking an awful lot like this x over here. If this were an x, then this would be x squared. If this were an x, then this would just be an x, and then these expressions would be the same, so when I factor it, everywhere I see an x, I could replace with an a squared, so I could factor this out, really looking at the same structure we have here as a squared plus three times a squared plus four. Now, I'm going really fast through this. This is really the introductory video, the overview video. Don't worry if this is a little bit much too fast."}, {"video_title": "Introduction to factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "If this were an x, then this would just be an x, and then these expressions would be the same, so when I factor it, everywhere I see an x, I could replace with an a squared, so I could factor this out, really looking at the same structure we have here as a squared plus three times a squared plus four. Now, I'm going really fast through this. This is really the introductory video, the overview video. Don't worry if this is a little bit much too fast. This is really just to give you a sense of things. Later in this unit, we're going to dig deeper into each of these cases, but just to give you a sense of where we're going, I'll give you another example that builds off of what you likely saw in your introductory algebra learning, so building off of the structure here, if someone were to walk up to you, again, a lot of people are walking up to you, and say factor four x to the sixth minus nine y to the fourth Well, at first, this looks quite intimidating until you realize that, hey, I could write both of these as squares. I could write this first one as two x to the third squared minus, and I could write this second term as three y squared squared, and now, this is just a difference of squares, so it'd be two x to the third plus three y squared times two x to the third minus three y squared."}, {"video_title": "Introduction to factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "Don't worry if this is a little bit much too fast. This is really just to give you a sense of things. Later in this unit, we're going to dig deeper into each of these cases, but just to give you a sense of where we're going, I'll give you another example that builds off of what you likely saw in your introductory algebra learning, so building off of the structure here, if someone were to walk up to you, again, a lot of people are walking up to you, and say factor four x to the sixth minus nine y to the fourth Well, at first, this looks quite intimidating until you realize that, hey, I could write both of these as squares. I could write this first one as two x to the third squared minus, and I could write this second term as three y squared squared, and now, this is just a difference of squares, so it'd be two x to the third plus three y squared times two x to the third minus three y squared. We'll also see things like this where we're going to be factoring multiple times, so once again, someone walks up to you on the street, and they say you're a very popular person. Someone walks up to you and says, up to you on the street and says, factor x to the fourth minus y to the fourth. Well, based on what we just saw, you could realize that this is the same thing as x squared squared minus y squared squared, and you say, okay, this is a difference of squares just like this was a difference of squares, so it's going to be the sum of x squared and y squared, x squared plus y squared, times the difference of them, x squared minus y squared."}, {"video_title": "Introduction to factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "I could write this first one as two x to the third squared minus, and I could write this second term as three y squared squared, and now, this is just a difference of squares, so it'd be two x to the third plus three y squared times two x to the third minus three y squared. We'll also see things like this where we're going to be factoring multiple times, so once again, someone walks up to you on the street, and they say you're a very popular person. Someone walks up to you and says, up to you on the street and says, factor x to the fourth minus y to the fourth. Well, based on what we just saw, you could realize that this is the same thing as x squared squared minus y squared squared, and you say, okay, this is a difference of squares just like this was a difference of squares, so it's going to be the sum of x squared and y squared, x squared plus y squared, times the difference of them, x squared minus y squared. Now, this is fun because this is two a difference of squares, so we can rewrite this whole thing as I'll rewrite this first part, x squared, x squared plus y squared, and then we could factor this as a difference of squares just as we factored this up here, and we get x plus y times x minus y, so I'll leave you there. I've just bombarded you with a bunch of information, but this is really just to get you warmed up. Don't stress about it because we're gonna go deep into each of these, and there's gonna be plenty of chances to practice it on Khan Academy to make sure you understand where all of this is coming from."}, {"video_title": "Evaluating quotient of fractional exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "And like always, pause the video and see if you can figure this out. All right, let's work through this together. And at first you might find this kind of daunting, especially when you see something like two to the 4 7ths power. Is that even, that's not going to be a whole number. How do I do this, especially without a calculator? And I should have said do this without a calculator. But then the key is to see that we can use our exponent properties to simplify this a little bit so that we can do this on paper."}, {"video_title": "Evaluating quotient of fractional exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "Is that even, that's not going to be a whole number. How do I do this, especially without a calculator? And I should have said do this without a calculator. But then the key is to see that we can use our exponent properties to simplify this a little bit so that we can do this on paper. And the main property that might jump out at you is if I have something, if I have x to the a power over y to the a power, this is the same thing as x over y to the a power. And our situation right over here, 256 would be x, two would be y, and then a is 4 7ths. So we can rewrite this."}, {"video_title": "Evaluating quotient of fractional exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "But then the key is to see that we can use our exponent properties to simplify this a little bit so that we can do this on paper. And the main property that might jump out at you is if I have something, if I have x to the a power over y to the a power, this is the same thing as x over y to the a power. And our situation right over here, 256 would be x, two would be y, and then a is 4 7ths. So we can rewrite this. This is going to be equal to, this is equal to 256 over two to the 4 7ths power. And so this is nice, we're already able to simplify this because we know 256 divided by two is 128. So this is 128 to the 4 7ths power."}, {"video_title": "Evaluating quotient of fractional exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "So we can rewrite this. This is going to be equal to, this is equal to 256 over two to the 4 7ths power. And so this is nice, we're already able to simplify this because we know 256 divided by two is 128. So this is 128 to the 4 7ths power. Now this might also seem a little bit difficult. How do I raise 128 to a fractional power? But we just have to remind ourselves this is the same thing."}, {"video_title": "Evaluating quotient of fractional exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "So this is 128 to the 4 7ths power. Now this might also seem a little bit difficult. How do I raise 128 to a fractional power? But we just have to remind ourselves this is the same thing. This is the same thing as 128 to the 1 7th power then raised to the 4th power. We could also view it the other way around. We could say that this is also 128 to the 4th power and then raise that to the 1 7th."}, {"video_title": "Evaluating quotient of fractional exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "But we just have to remind ourselves this is the same thing. This is the same thing as 128 to the 1 7th power then raised to the 4th power. We could also view it the other way around. We could say that this is also 128 to the 4th power and then raise that to the 1 7th. But multiplying 128 four times, that's going to be very computationally intensive and then you have to find the 7th root of that. That seems pretty difficult so we don't want to go in that way. But if we can get the smaller number first, what is 128 to the 1 7th power, then that might be easier to raise to the 4th power."}, {"video_title": "Evaluating quotient of fractional exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "We could say that this is also 128 to the 4th power and then raise that to the 1 7th. But multiplying 128 four times, that's going to be very computationally intensive and then you have to find the 7th root of that. That seems pretty difficult so we don't want to go in that way. But if we can get the smaller number first, what is 128 to the 1 7th power, then that might be easier to raise to the 4th power. Now when you look at this, and knowing that probably the question writer in this case, I'm the person who presented it with you, is telling you that you're not going to use a calculator, it's a pretty good clue that all right, this is probably going to be something that I can figure out on my own. And you might recognize 128 as a power of two and maybe two to the 7th is 128 and we can verify that. So let's see, two to the 1st is two, four, eight, 16, 32, 64, 128."}, {"video_title": "Evaluating quotient of fractional exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "But if we can get the smaller number first, what is 128 to the 1 7th power, then that might be easier to raise to the 4th power. Now when you look at this, and knowing that probably the question writer in this case, I'm the person who presented it with you, is telling you that you're not going to use a calculator, it's a pretty good clue that all right, this is probably going to be something that I can figure out on my own. And you might recognize 128 as a power of two and maybe two to the 7th is 128 and we can verify that. So let's see, two to the 1st is two, four, eight, 16, 32, 64, 128. Two times two is four, times two is eight, times two is 16, times two is 32, times two is 64, times two is 128. So two to the 7th power is equal to 128. Or another way of saying this exact same thing is that 128, 128 is equal to, or 128 to the 1 7th power is equal to two."}, {"video_title": "Evaluating quotient of fractional exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "So let's see, two to the 1st is two, four, eight, 16, 32, 64, 128. Two times two is four, times two is eight, times two is 16, times two is 32, times two is 64, times two is 128. So two to the 7th power is equal to 128. Or another way of saying this exact same thing is that 128, 128 is equal to, or 128 to the 1 7th power is equal to two. Or you could even say that the 7th root, the 7th root of 128 is equal to two. So we can simplify this, this is two. So our whole expression is now just two to the 4th power."}, {"video_title": "Evaluating quotient of fractional exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "Or another way of saying this exact same thing is that 128, 128 is equal to, or 128 to the 1 7th power is equal to two. Or you could even say that the 7th root, the 7th root of 128 is equal to two. So we can simplify this, this is two. So our whole expression is now just two to the 4th power. Well that's just two times two, times two times two. So that's two to the 4th power, two to the 4th power which is just going to be equal to 16. That's two times two, times two, times two, right over there and so we're done."}, {"video_title": "How to find the domain and the range of a function given its graph (example) Khan Academy.mp3", "Sentence": "What is its domain? So the way it's graphed right over here, we could assume that this is the entire function definition for f of x. So for example, if we say, well, what does f of x equal when x is equal to negative 9? Well, we go up here. We don't see it's graphed here. It's not defined for x equals negative 9, or x equals negative 8 and 1 half, or x equals negative 8. It's not defined for any of these values."}, {"video_title": "How to find the domain and the range of a function given its graph (example) Khan Academy.mp3", "Sentence": "Well, we go up here. We don't see it's graphed here. It's not defined for x equals negative 9, or x equals negative 8 and 1 half, or x equals negative 8. It's not defined for any of these values. It only starts getting defined at x equals negative 6. At x equals negative 6, f of x is equal to 5. And then it keeps getting defined."}, {"video_title": "How to find the domain and the range of a function given its graph (example) Khan Academy.mp3", "Sentence": "It's not defined for any of these values. It only starts getting defined at x equals negative 6. At x equals negative 6, f of x is equal to 5. And then it keeps getting defined. f of x is defined for x all the way from x equals negative 6 all the way to x equals 7. When x equals 7, f of x is equal to 5. You can take any x value between negative 6, including negative 6, and positive 7, including positive 7, and you just have to move up above that number wherever you are to find out what the value of the function is at that point."}, {"video_title": "How to find the domain and the range of a function given its graph (example) Khan Academy.mp3", "Sentence": "And then it keeps getting defined. f of x is defined for x all the way from x equals negative 6 all the way to x equals 7. When x equals 7, f of x is equal to 5. You can take any x value between negative 6, including negative 6, and positive 7, including positive 7, and you just have to move up above that number wherever you are to find out what the value of the function is at that point. So the domain of this function definition, well, f of x is defined for any x that is greater than or equal to negative 6. Or we could say negative 6 is less than or equal to x, which is less than or equal to 7. If x satisfies this condition right over here, the function is defined."}, {"video_title": "How to find the domain and the range of a function given its graph (example) Khan Academy.mp3", "Sentence": "You can take any x value between negative 6, including negative 6, and positive 7, including positive 7, and you just have to move up above that number wherever you are to find out what the value of the function is at that point. So the domain of this function definition, well, f of x is defined for any x that is greater than or equal to negative 6. Or we could say negative 6 is less than or equal to x, which is less than or equal to 7. If x satisfies this condition right over here, the function is defined. So that's its domain. So let's check our answer. Let's do a few more of these."}, {"video_title": "How to find the domain and the range of a function given its graph (example) Khan Academy.mp3", "Sentence": "If x satisfies this condition right over here, the function is defined. So that's its domain. So let's check our answer. Let's do a few more of these. The function f of x is graphed. What is its domain? Well, exact similar argument."}, {"video_title": "How to find the domain and the range of a function given its graph (example) Khan Academy.mp3", "Sentence": "Let's do a few more of these. The function f of x is graphed. What is its domain? Well, exact similar argument. This function is not defined for x's negative 9, negative 8, all the way up, I should say, to negative 1. At negative 1, it starts getting defined. f of negative 1 is negative 5."}, {"video_title": "How to find the domain and the range of a function given its graph (example) Khan Academy.mp3", "Sentence": "Well, exact similar argument. This function is not defined for x's negative 9, negative 8, all the way up, I should say, to negative 1. At negative 1, it starts getting defined. f of negative 1 is negative 5. So it's defined for negative 1 is less than or equal to x. And it's defined all the way up to x equals 7, including x equals 7. So this right over here, negative 1 is less than or equal to x is less than or equal to 7."}, {"video_title": "How to find the domain and the range of a function given its graph (example) Khan Academy.mp3", "Sentence": "f of negative 1 is negative 5. So it's defined for negative 1 is less than or equal to x. And it's defined all the way up to x equals 7, including x equals 7. So this right over here, negative 1 is less than or equal to x is less than or equal to 7. The function is defined for any x that satisfies this double inequality right over here. Let's do a few more. The function f of x is graphed."}, {"video_title": "How to find the domain and the range of a function given its graph (example) Khan Academy.mp3", "Sentence": "So this right over here, negative 1 is less than or equal to x is less than or equal to 7. The function is defined for any x that satisfies this double inequality right over here. Let's do a few more. The function f of x is graphed. What is its range? So now we're not thinking about the x's for which this function is defined. We're thinking about the set of y values."}, {"video_title": "How to find the domain and the range of a function given its graph (example) Khan Academy.mp3", "Sentence": "The function f of x is graphed. What is its range? So now we're not thinking about the x's for which this function is defined. We're thinking about the set of y values. Where do all of the y values fall into? Well, let's see. The lowest possible y value, or the lowest possible value of f of x that we get here, looks like it's 0."}, {"video_title": "How to find the domain and the range of a function given its graph (example) Khan Academy.mp3", "Sentence": "We're thinking about the set of y values. Where do all of the y values fall into? Well, let's see. The lowest possible y value, or the lowest possible value of f of x that we get here, looks like it's 0. The function never goes below 0. So f of x, so 0 is less than or equal to f of x. It does equal 0 right over here."}, {"video_title": "How to find the domain and the range of a function given its graph (example) Khan Academy.mp3", "Sentence": "The lowest possible y value, or the lowest possible value of f of x that we get here, looks like it's 0. The function never goes below 0. So f of x, so 0 is less than or equal to f of x. It does equal 0 right over here. f of negative 4 is 0. And then the highest y value, or the highest value that f of x obtains in this function definition is 8. f of 7 is 8. It never gets above 8, but it does equal 8 right over here when x is equal to 7."}, {"video_title": "How to find the domain and the range of a function given its graph (example) Khan Academy.mp3", "Sentence": "It does equal 0 right over here. f of negative 4 is 0. And then the highest y value, or the highest value that f of x obtains in this function definition is 8. f of 7 is 8. It never gets above 8, but it does equal 8 right over here when x is equal to 7. So 0 is less than f of x, which is less than or equal to 8. So that's its range. Let's do a few more."}, {"video_title": "How to find the domain and the range of a function given its graph (example) Khan Academy.mp3", "Sentence": "It never gets above 8, but it does equal 8 right over here when x is equal to 7. So 0 is less than f of x, which is less than or equal to 8. So that's its range. Let's do a few more. This is kind of fun. The function f of x is graphed. What is its domain?"}, {"video_title": "How to find the domain and the range of a function given its graph (example) Khan Academy.mp3", "Sentence": "Let's do a few more. This is kind of fun. The function f of x is graphed. What is its domain? So once again, this function is defined for negative 2 is less than or equal to x, which is less than or equal to 5. If you give me an x anywhere in between negative 2 and 5, I can look at this graph to see where the function is defined. f of negative 2 is negative 4. f of negative 1 is negative 3, so on and so forth."}, {"video_title": "Polynomial subtraction Mathematics II High School Math Khan Academy.mp3", "Sentence": "we're asked to subtract negative two x squared plus four x minus one from six x squared plus three x minus nine. And like always, I encourage you to pause the video and see if you can give it a go. All right, now let's work through this together. So I could rewrite this as six x squared plus three x minus nine minus, minus this expression right over here. So I'll put that in parentheses, minus negative two x squared, negative two x squared, plus four x minus one. Now what can we do from here? Well, we can distribute this negative sign."}, {"video_title": "Polynomial subtraction Mathematics II High School Math Khan Academy.mp3", "Sentence": "So I could rewrite this as six x squared plus three x minus nine minus, minus this expression right over here. So I'll put that in parentheses, minus negative two x squared, negative two x squared, plus four x minus one. Now what can we do from here? Well, we can distribute this negative sign. We can distribute this negative sign. And then if we did that, we would get the six x squared plus three x minus nine won't change, so we still have that. Six x squared plus three x minus nine."}, {"video_title": "Polynomial subtraction Mathematics II High School Math Khan Academy.mp3", "Sentence": "Well, we can distribute this negative sign. We can distribute this negative sign. And then if we did that, we would get the six x squared plus three x minus nine won't change, so we still have that. Six x squared plus three x minus nine. But then if I distribute the negative sign, the negative of negative two x squared is positive two x squared. So that's going to be positive two, good, give it a little more space, positive two x squared. And then subtract, and then the negative of positive four x is, I'm going to subtract four x now."}, {"video_title": "Polynomial subtraction Mathematics II High School Math Khan Academy.mp3", "Sentence": "Six x squared plus three x minus nine. But then if I distribute the negative sign, the negative of negative two x squared is positive two x squared. So that's going to be positive two, good, give it a little more space, positive two x squared. And then subtract, and then the negative of positive four x is, I'm going to subtract four x now. And then the negative of negative one, or the opposite of negative one, is gonna be positive one. So I've just distributed the negative sign. And now I can add terms that have the same degree on our x, the same degree terms, I guess you could say."}, {"video_title": "Polynomial subtraction Mathematics II High School Math Khan Academy.mp3", "Sentence": "And then subtract, and then the negative of positive four x is, I'm going to subtract four x now. And then the negative of negative one, or the opposite of negative one, is gonna be positive one. So I've just distributed the negative sign. And now I can add terms that have the same degree on our x, the same degree terms, I guess you could say. So I have an x squared term. Here it's six x squared. Here I have a two x squared term."}, {"video_title": "Polynomial subtraction Mathematics II High School Math Khan Academy.mp3", "Sentence": "And now I can add terms that have the same degree on our x, the same degree terms, I guess you could say. So I have an x squared term. Here it's six x squared. Here I have a two x squared term. So I can add those two together. Six x squared plus two x squared. If I have six x squareds, and then I have another two x squareds, how many x squares am I now going to have?"}, {"video_title": "Polynomial subtraction Mathematics II High School Math Khan Academy.mp3", "Sentence": "Here I have a two x squared term. So I can add those two together. Six x squared plus two x squared. If I have six x squareds, and then I have another two x squareds, how many x squares am I now going to have? I'm now going to have eight x squareds. Eight x squareds. Or six x squared plus two x squared, we add the coefficients, the six and the two, to get eight, eight x squared."}, {"video_title": "Polynomial subtraction Mathematics II High School Math Khan Academy.mp3", "Sentence": "If I have six x squareds, and then I have another two x squareds, how many x squares am I now going to have? I'm now going to have eight x squareds. Eight x squareds. Or six x squared plus two x squared, we add the coefficients, the six and the two, to get eight, eight x squared. Then we can add the x terms. You could view these as the first degree terms. Three x, we have three x, and then we have minus four x."}, {"video_title": "Polynomial subtraction Mathematics II High School Math Khan Academy.mp3", "Sentence": "Or six x squared plus two x squared, we add the coefficients, the six and the two, to get eight, eight x squared. Then we can add the x terms. You could view these as the first degree terms. Three x, we have three x, and then we have minus four x. So three x minus four x, if I have three of something and I take away four of them, I'm now going to have negative one of that thing. So, or you could say that the coefficients, three minus four would be negative one. So I now have negative one x. I could write it as negative one x, but I might as well just write it as negative x."}, {"video_title": "Polynomial subtraction Mathematics II High School Math Khan Academy.mp3", "Sentence": "Three x, we have three x, and then we have minus four x. So three x minus four x, if I have three of something and I take away four of them, I'm now going to have negative one of that thing. So, or you could say that the coefficients, three minus four would be negative one. So I now have negative one x. I could write it as negative one x, but I might as well just write it as negative x. That's the same thing as negative one x. And then finally, I can deal with our constant terms. I have, I'm subtracting a nine, and then I'm adding a one."}, {"video_title": "Polynomial subtraction Mathematics II High School Math Khan Academy.mp3", "Sentence": "So I now have negative one x. I could write it as negative one x, but I might as well just write it as negative x. That's the same thing as negative one x. And then finally, I can deal with our constant terms. I have, I'm subtracting a nine, and then I'm adding a one. So you could say, well, what's negative nine plus one? Well, that's going to be negative eight. That's going to be negative eight."}, {"video_title": "Verifying inverse functions from tables Precalculus Khan Academy.mp3", "Sentence": "We have some x's and then they tell us what the corresponding s of x is. And then in this table, we have some x's and they tell us the corresponding t of x. It says complete the table for the composite function, s of t of x. We want to fill in these five entries here. And then they ask us our s and t inverses. So pause this video and see if you can figure this out on your own before we work through it together. Alright, now let's work through this together."}, {"video_title": "Verifying inverse functions from tables Precalculus Khan Academy.mp3", "Sentence": "We want to fill in these five entries here. And then they ask us our s and t inverses. So pause this video and see if you can figure this out on your own before we work through it together. Alright, now let's work through this together. So let's just remind ourselves what's going on with a composite function like this. So you're going to take some x value. And it looks like we're first going to put it into the function t. That is going to output t of x."}, {"video_title": "Verifying inverse functions from tables Precalculus Khan Academy.mp3", "Sentence": "Alright, now let's work through this together. So let's just remind ourselves what's going on with a composite function like this. So you're going to take some x value. And it looks like we're first going to put it into the function t. That is going to output t of x. And then we're going to take that output, take that t of x. And then it will be the input into s. So then we're going to input that into s. And then that would output s of what we inputted, which in this case is t of x. So let's go on that journey."}, {"video_title": "Verifying inverse functions from tables Precalculus Khan Academy.mp3", "Sentence": "And it looks like we're first going to put it into the function t. That is going to output t of x. And then we're going to take that output, take that t of x. And then it will be the input into s. So then we're going to input that into s. And then that would output s of what we inputted, which in this case is t of x. So let's go on that journey. So what we're going to do is first take these numbers, put them into the function t, figure out what it outputs, and then take that output and then put it into the function s. It's going to be a fun little ride. So when x is equal to 12, we're going to put it into our function t first. So when x is an input into t, the output is equal to negative 1."}, {"video_title": "Verifying inverse functions from tables Precalculus Khan Academy.mp3", "Sentence": "So let's go on that journey. So what we're going to do is first take these numbers, put them into the function t, figure out what it outputs, and then take that output and then put it into the function s. It's going to be a fun little ride. So when x is equal to 12, we're going to put it into our function t first. So when x is an input into t, the output is equal to negative 1. So that's our t of x. And then we're going to take this negative 1 and input it into s. So negative 1 here. And when you input that into s, you get as the output, s of negative 1 is 12."}, {"video_title": "Verifying inverse functions from tables Precalculus Khan Academy.mp3", "Sentence": "So when x is an input into t, the output is equal to negative 1. So that's our t of x. And then we're going to take this negative 1 and input it into s. So negative 1 here. And when you input that into s, you get as the output, s of negative 1 is 12. So s of t of x is 12. So interestingly, this is 12. Now let's do the next one."}, {"video_title": "Verifying inverse functions from tables Precalculus Khan Academy.mp3", "Sentence": "And when you input that into s, you get as the output, s of negative 1 is 12. So s of t of x is 12. So interestingly, this is 12. Now let's do the next one. So when we input 18 into t, so 18 is the input, t of x, t of 18 is 2. And then if we want to input that into s, so this is going to be the input into s, the output is 18. Very interesting."}, {"video_title": "Verifying inverse functions from tables Precalculus Khan Academy.mp3", "Sentence": "Now let's do the next one. So when we input 18 into t, so 18 is the input, t of x, t of 18 is 2. And then if we want to input that into s, so this is going to be the input into s, the output is 18. Very interesting. All right. Let's keep going. So when we input 61 into t, the output is 8."}, {"video_title": "Verifying inverse functions from tables Precalculus Khan Academy.mp3", "Sentence": "Very interesting. All right. Let's keep going. So when we input 61 into t, the output is 8. Then when we take 8 and we input it into s of x, or s of 8, I should say, it's going to be 61. All right. Things are looking good so far."}, {"video_title": "Verifying inverse functions from tables Precalculus Khan Academy.mp3", "Sentence": "So when we input 61 into t, the output is 8. Then when we take 8 and we input it into s of x, or s of 8, I should say, it's going to be 61. All right. Things are looking good so far. And I'm running out of colors. I'll do green. So when we take 70 and we input it into t, t of 70 is 7."}, {"video_title": "Verifying inverse functions from tables Precalculus Khan Academy.mp3", "Sentence": "Things are looking good so far. And I'm running out of colors. I'll do green. So when we take 70 and we input it into t, t of 70 is 7. When you take 7 and input it into s, you get 70. All right. And then one last one I will do in this blue color."}, {"video_title": "Verifying inverse functions from tables Precalculus Khan Academy.mp3", "Sentence": "So when we take 70 and we input it into t, t of 70 is 7. When you take 7 and input it into s, you get 70. All right. And then one last one I will do in this blue color. When you take 100 and put it into t, it outputs negative 5. You take negative 5, input it into s, you get 100. So in every situation that we have looked at right over here, in all of these situations, we see that s of t of x is equal to x, which inclines us to believe that they are inverses."}, {"video_title": "Verifying inverse functions from tables Precalculus Khan Academy.mp3", "Sentence": "And then one last one I will do in this blue color. When you take 100 and put it into t, it outputs negative 5. You take negative 5, input it into s, you get 100. So in every situation that we have looked at right over here, in all of these situations, we see that s of t of x is equal to x, which inclines us to believe that they are inverses. Remember, if these two are inverses of each other, this would be true. And also t of s of x is going to be equal to x. But we don't really know 100% unless we know that we have looked at every combination in the domains for each of them."}, {"video_title": "Verifying inverse functions from tables Precalculus Khan Academy.mp3", "Sentence": "So in every situation that we have looked at right over here, in all of these situations, we see that s of t of x is equal to x, which inclines us to believe that they are inverses. Remember, if these two are inverses of each other, this would be true. And also t of s of x is going to be equal to x. But we don't really know 100% unless we know that we have looked at every combination in the domains for each of them. Now, when you look at these two tables up here, and I could have done this, this is the one we looked at on our journey to get to this 12 right over here. The following table gives all of the input-output pairs for the function s and t. So this right over here is the domain for the function s, and this right over here is the domain for the function t. So because for every member of the function s, every member of the domain of the function s, the corresponding output right over there is the domain for the function t, and it takes us back to where we began. And then the opposite is true as well."}, {"video_title": "Example 3 Factoring quadratics as a perfect square of a sum (a+b)^2 Algebra I Khan Academy.mp3", "Sentence": "We need to factor negative 4t squared minus 12t minus 9. And a good place to start is to say, well, are there any common factors for all of these terms? And when you look at them, well, these first two are divisible by 4, these last two are divisible by 3, but not all of them are divisible by any one number. Well, but you could factor out a negative 1. But even if you factor out a negative 1, so you say this is the same thing as negative 1, times positive 4t squared plus 12t plus 9, you still end up with a non-1 coefficient out here and on the second degree term, on the t squared term. So you might want to immediately start grouping this. And if you did factor it by grouping, it would work."}, {"video_title": "Example 3 Factoring quadratics as a perfect square of a sum (a+b)^2 Algebra I Khan Academy.mp3", "Sentence": "Well, but you could factor out a negative 1. But even if you factor out a negative 1, so you say this is the same thing as negative 1, times positive 4t squared plus 12t plus 9, you still end up with a non-1 coefficient out here and on the second degree term, on the t squared term. So you might want to immediately start grouping this. And if you did factor it by grouping, it would work. You would get the right answer. But there is something you might be able to see, or there is something about this equation that might pop out at you that might make it a little bit simpler to solve. And to understand that, let's take a little bit of a break here on the right-hand side and just think about what happens if you take a plus b times a plus b, if you just have a binomial squared."}, {"video_title": "Example 3 Factoring quadratics as a perfect square of a sum (a+b)^2 Algebra I Khan Academy.mp3", "Sentence": "And if you did factor it by grouping, it would work. You would get the right answer. But there is something you might be able to see, or there is something about this equation that might pop out at you that might make it a little bit simpler to solve. And to understand that, let's take a little bit of a break here on the right-hand side and just think about what happens if you take a plus b times a plus b, if you just have a binomial squared. Well, you have a times a, which is a squared. Then you have a times that b, which is plus ab. Then you have b times a, which is the same thing as ab."}, {"video_title": "Example 3 Factoring quadratics as a perfect square of a sum (a+b)^2 Algebra I Khan Academy.mp3", "Sentence": "And to understand that, let's take a little bit of a break here on the right-hand side and just think about what happens if you take a plus b times a plus b, if you just have a binomial squared. Well, you have a times a, which is a squared. Then you have a times that b, which is plus ab. Then you have b times a, which is the same thing as ab. And then you have b times b, or you have b squared. And so if you add these middle two terms right here, you're left with a squared plus 2ab plus b squared. This is the square of a binomial."}, {"video_title": "Example 3 Factoring quadratics as a perfect square of a sum (a+b)^2 Algebra I Khan Academy.mp3", "Sentence": "Then you have b times a, which is the same thing as ab. And then you have b times b, or you have b squared. And so if you add these middle two terms right here, you're left with a squared plus 2ab plus b squared. This is the square of a binomial. Now, does this right here, does 4t squared plus 12t plus 9 fit this pattern? Well, if 4t squared is a squared, so if this right here is a squared, if that is a squared right there, then what does a have to be? If this is a squared, then a would be equal to the square root of this."}, {"video_title": "Example 3 Factoring quadratics as a perfect square of a sum (a+b)^2 Algebra I Khan Academy.mp3", "Sentence": "This is the square of a binomial. Now, does this right here, does 4t squared plus 12t plus 9 fit this pattern? Well, if 4t squared is a squared, so if this right here is a squared, if that is a squared right there, then what does a have to be? If this is a squared, then a would be equal to the square root of this. It would be 2t. And if this is b squared, let me do that in a different color. If this right here is b squared, if the 9 is b squared right there, then that means that b is equal to 3, equal to the positive square root of the 9."}, {"video_title": "Example 3 Factoring quadratics as a perfect square of a sum (a+b)^2 Algebra I Khan Academy.mp3", "Sentence": "If this is a squared, then a would be equal to the square root of this. It would be 2t. And if this is b squared, let me do that in a different color. If this right here is b squared, if the 9 is b squared right there, then that means that b is equal to 3, equal to the positive square root of the 9. Now, this number right here, and actually it doesn't have to just be equal to 3. It might have been negative 3 as well. It could be plus or minus 3."}, {"video_title": "Example 3 Factoring quadratics as a perfect square of a sum (a+b)^2 Algebra I Khan Academy.mp3", "Sentence": "If this right here is b squared, if the 9 is b squared right there, then that means that b is equal to 3, equal to the positive square root of the 9. Now, this number right here, and actually it doesn't have to just be equal to 3. It might have been negative 3 as well. It could be plus or minus 3. But this number here, is it 2 times ab? That's the middle term that we care about. Is it 2 times ab?"}, {"video_title": "Example 3 Factoring quadratics as a perfect square of a sum (a+b)^2 Algebra I Khan Academy.mp3", "Sentence": "It could be plus or minus 3. But this number here, is it 2 times ab? That's the middle term that we care about. Is it 2 times ab? Well, if we multiply 2t times 3, we get 6t. And then we multiply that times 2, you get 12t. This right here, 12t, is equal to 2 times 2t times 3."}, {"video_title": "Example 3 Factoring quadratics as a perfect square of a sum (a+b)^2 Algebra I Khan Academy.mp3", "Sentence": "Is it 2 times ab? Well, if we multiply 2t times 3, we get 6t. And then we multiply that times 2, you get 12t. This right here, 12t, is equal to 2 times 2t times 3. It is 2 times ab. And if this was a negative 3, we would look to see if this was a negative 12, but this does work for a positive 3. So this does fit the pattern of a perfect square."}, {"video_title": "Example 3 Factoring quadratics as a perfect square of a sum (a+b)^2 Algebra I Khan Academy.mp3", "Sentence": "This right here, 12t, is equal to 2 times 2t times 3. It is 2 times ab. And if this was a negative 3, we would look to see if this was a negative 12, but this does work for a positive 3. So this does fit the pattern of a perfect square. This is a special type of, or you could view this as a square of a binomial. So if you wanted to factor this, the stuff on the inside, you still have that negative 1 out there, the 4t squared plus 12t plus 9, you could immediately say, well that's going to be a plus b times a plus b. Or 2t plus 3 times 2t plus 3."}, {"video_title": "Example 3 Factoring quadratics as a perfect square of a sum (a+b)^2 Algebra I Khan Academy.mp3", "Sentence": "So this does fit the pattern of a perfect square. This is a special type of, or you could view this as a square of a binomial. So if you wanted to factor this, the stuff on the inside, you still have that negative 1 out there, the 4t squared plus 12t plus 9, you could immediately say, well that's going to be a plus b times a plus b. Or 2t plus 3 times 2t plus 3. Or you could just say it's 2t plus 3 squared. It fits this pattern. And of course, you can't forget about this negative 1 out here."}, {"video_title": "Example 3 Factoring quadratics as a perfect square of a sum (a+b)^2 Algebra I Khan Academy.mp3", "Sentence": "Or 2t plus 3 times 2t plus 3. Or you could just say it's 2t plus 3 squared. It fits this pattern. And of course, you can't forget about this negative 1 out here. You could have also solved it by grouping, but this might be a quicker thing to recognize. This is a number squared. That's another number squared."}, {"video_title": "How to find two function inputs with the same output given graph (example) Khan Academy.mp3", "Sentence": "The graph of the function f is shown below. What is the input value other than negative five for which f of x is equal to f of negative five? So we have our x-axis, we have our y-axis, and then in blue, they've graphed y equals f of x. So for example, when x is equal to one, f of x, when y is going to be equal to f of x, that's what this graph is, f of x is equal to one. When x is equal to seven, f of seven, we see, is equal to five. When x is equal to nine, we see that f of nine is equal to six. So what is the input value other than negative five for which f of x is equal to f of negative five?"}, {"video_title": "How to find two function inputs with the same output given graph (example) Khan Academy.mp3", "Sentence": "So for example, when x is equal to one, f of x, when y is going to be equal to f of x, that's what this graph is, f of x is equal to one. When x is equal to seven, f of seven, we see, is equal to five. When x is equal to nine, we see that f of nine is equal to six. So what is the input value other than negative five for which f of x is equal to f of negative five? So let's see. If x is equal to negative five, f of negative five is, we move up here to the graph, it is equal to four, because this, once again, this is the graph y is equal to f of x. So the y-coordinate here, this is what f of x, this is what f of negative five is equal to."}, {"video_title": "How to find two function inputs with the same output given graph (example) Khan Academy.mp3", "Sentence": "So what is the input value other than negative five for which f of x is equal to f of negative five? So let's see. If x is equal to negative five, f of negative five is, we move up here to the graph, it is equal to four, because this, once again, this is the graph y is equal to f of x. So the y-coordinate here, this is what f of x, this is what f of negative five is equal to. So f of negative five is equal to four. So where else does that happen? So let's see, let's just move horizontally to the right."}, {"video_title": "How to find two function inputs with the same output given graph (example) Khan Academy.mp3", "Sentence": "So the y-coordinate here, this is what f of x, this is what f of negative five is equal to. So f of negative five is equal to four. So where else does that happen? So let's see, let's just move horizontally to the right. That happens here as well. So what input do we have to give? What's the x-coordinate here to get y is equal to f of x, or f of x is equal to four right over here?"}, {"video_title": "How to find two function inputs with the same output given graph (example) Khan Academy.mp3", "Sentence": "So let's see, let's just move horizontally to the right. That happens here as well. So what input do we have to give? What's the x-coordinate here to get y is equal to f of x, or f of x is equal to four right over here? Well, we see the x-coordinate is also four. So this tells us that f of four is equal to four, which is the same thing as f of negative five. So when x is four, the function takes on the same value as when x is negative five."}, {"video_title": "Point-slope and slope-intercept form from two points Algebra I Khan Academy.mp3", "Sentence": "Intercept form and I encourage you like always pause the video and see if you can do it So let's first think about point slope form point point slope Point slope form and point slope form is very easy to generate if you know a point on The line or if you know a point that satisfies where the x and y coordinates satisfy the linear equation and if you were to know the slope of The of the line that represents a solution set of that linear equation Now for sure we actually had we were given two points that are solutions that represent solutions to the linear equation To fully apply point slope or to apply point slope easily We just have to figure out the slope and what we could do is we could just evaluate What's the slope between the two points that we know and we just have to remind ourselves that slope slope is equal to Slope is equal to change in y over change in x Sometimes people say rise over run and what's that going to be? well if we say if we if we say that this the second point right over here if We say if we say this is kind of our if we starting at this point And we go to that point then our change in y going from this point to that point is going to be it's going to be equal to 1 minus 1 minus 9 1 minus 9 this point right over here is the point 6 comma 1 so we started at y equals 9 we finish at y equals 1 our Change in y is going to be 1 minus 9 we have a negative 8 change in y which makes sense We've gone down 8 so this is going to be equal to this is going to be equal to negative 8 That's our change in y and what's our change in x well we go from x equals 4 to x equals 6 So we end up at x equals 6 and we started at x equals 4 We started at x equals 4 so our change in x is 6 minus 4 which is equal to 2 Which is equal to 2 and you could have even done it visually to go from this point to this point your change in y your change in y is You went down 8 so your change in let me write this so your change in y is equal to negative 8 And what was your change in x to get and get to this point? Well your change in x is positive 2 So your change in x is equal to 2 And so what's your slope change in y over change in x negative 8 over 2 is equal to negative 4 So now that we have a now that we know the slope And we know a point and we know a and we know a point we actually know two points on the line We can express this in point slope form and so let's do that And the way I like to do is I always like to just take it straight from the definition of what slope is We know that the slope between any two points on this line is going to be negative 4 So if we take an arbitrary y that sits on this line And if we find the difference between that y and let's focus on this point up here so if we find the difference that y and this y and 9 and and It's over over the difference between some x on the line and this x and 4 and 4 this is going to be the slope between any x y on this line and This point right over here and the slope between any two points on a line are going to have to be constant So this is going to be equal to the slope of the line It's going to be equal to negative 4 and we're not in point slope form or classic point slope form just yet to do that We just multiply both sides times x minus 4 so we get y minus 9 We get y minus 9 is equal to our slope negative 4 times x minus 4 times X minus 4 and this right over here is our classic this right over here is our cat classic point slope form We have the point sometimes they even put parentheses like this But we could figure out the point from this point slope form the point that sits on this line Would things that make both sides of this equation equal to 0 so it would be x equals 4 y equals 9 Which we have right up there, and then the slope is right over here. It's negative 4 Now from this can we now express this this this linear equation in y-intercept form and Y intercept form just as a bit of a reminder. It's y is equal to mx plus B where this coefficient is our slope and this constant right over here is Allows us to figure out our y-intercept and to get this and to get this in this form We just have to simplify a little bit of this algebra, so you have y minus 9 y minus 9 is Equal to let's distribute this negative 4 and I'll just switch some colors Let's distribute this negative 4 negative 4 times x is negative 4x negative 4 times negative 4 is plus 16 And now if we just want to isolate the y on the left hand side we can add 9 to both sides Let's do that. Let's add 9 Let's add 9 to both sides Let's add 9 to both sides on the left hand side. We're just left with y and on the right hand side."}, {"video_title": "How to match function input to output given the formula (example) Algebra I Khan Academy.mp3", "Sentence": "The function f is defined as follows. f of t is equal to negative two t plus five. So whatever we input into this function, we multiply it times negative two, and then we add five. So what is the input value for which f of t is equal to 13? So if f of t is equal to 13, that means that this thing over here is equal to 13 for some t, for some input. So we can just solve the equation and negative two t plus five is equal to 13. So let's do that."}, {"video_title": "How to match function input to output given the formula (example) Algebra I Khan Academy.mp3", "Sentence": "So what is the input value for which f of t is equal to 13? So if f of t is equal to 13, that means that this thing over here is equal to 13 for some t, for some input. So we can just solve the equation and negative two t plus five is equal to 13. So let's do that. Negative two t plus five is equal to 13. Well, we can subtract five from both sides. I'm just trying to isolate the t on the left-hand side."}, {"video_title": "How to match function input to output given the formula (example) Algebra I Khan Academy.mp3", "Sentence": "So let's do that. Negative two t plus five is equal to 13. Well, we can subtract five from both sides. I'm just trying to isolate the t on the left-hand side. So subtract negative five from the left. That's the whole reason why we did that, so those disappear. But we have to do it from the right as well."}, {"video_title": "How to match function input to output given the formula (example) Algebra I Khan Academy.mp3", "Sentence": "I'm just trying to isolate the t on the left-hand side. So subtract negative five from the left. That's the whole reason why we did that, so those disappear. But we have to do it from the right as well. So you have 13 minus five is eight, and on the left-hand side, you still have your negative two t. So you have negative two t is equal to eight. Now to just have a t on the left-hand side, I want to divide both sides by negative two. And I'm left with t is equal to eight divided by negative two is equal to negative four."}, {"video_title": "How to match function input to output given the formula (example) Algebra I Khan Academy.mp3", "Sentence": "But we have to do it from the right as well. So you have 13 minus five is eight, and on the left-hand side, you still have your negative two t. So you have negative two t is equal to eight. Now to just have a t on the left-hand side, I want to divide both sides by negative two. And I'm left with t is equal to eight divided by negative two is equal to negative four. So you input negative four, you input negative four into this function, into this function, and it will output 13. 13. Or we could write that f of negative four is equal to 13."}, {"video_title": "How to match function input to output given the formula (example) Algebra I Khan Academy.mp3", "Sentence": "And I'm left with t is equal to eight divided by negative two is equal to negative four. So you input negative four, you input negative four into this function, into this function, and it will output 13. 13. Or we could write that f of negative four is equal to 13. But this is what they're looking for. This is the input value. Negative four is the input value for which f of t is equal to 13."}, {"video_title": "Multiplying binomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "And we want to write that product in standard quadratic form, which is just a fancy way of saying a form where you have some coefficient on the second degree term, ax squared, plus some coefficient b on the first degree term, plus the constant term. So this right over here would be standard quadratic form. So that's the form that we want to express this product in. And I encourage you to pause the video and try to work through it on your own. All right, now let's work through this. And the key when we're multiplying two binomials like this, or actually when we're multiplying any polynomials, is just to remember the distributive property that we all by this point know quite well. So what we could view this as is we can distribute this x minus four, this entire expression, over the x and the seven."}, {"video_title": "Multiplying binomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "And I encourage you to pause the video and try to work through it on your own. All right, now let's work through this. And the key when we're multiplying two binomials like this, or actually when we're multiplying any polynomials, is just to remember the distributive property that we all by this point know quite well. So what we could view this as is we can distribute this x minus four, this entire expression, over the x and the seven. So we could say that this is the same thing as x minus four times x, plus x minus four times seven. So let's write that. So x minus four times x, or we could write this as x times x minus four."}, {"video_title": "Multiplying binomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "So what we could view this as is we can distribute this x minus four, this entire expression, over the x and the seven. So we could say that this is the same thing as x minus four times x, plus x minus four times seven. So let's write that. So x minus four times x, or we could write this as x times x minus four. That's distributing the, or multiplying the x minus four times x. That's right there. Plus seven times x minus four."}, {"video_title": "Multiplying binomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "So x minus four times x, or we could write this as x times x minus four. That's distributing the, or multiplying the x minus four times x. That's right there. Plus seven times x minus four. Times x minus four. Notice, all we did is distribute the x minus four. We took this whole thing and we multiplied it by each term over here."}, {"video_title": "Multiplying binomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "Plus seven times x minus four. Times x minus four. Notice, all we did is distribute the x minus four. We took this whole thing and we multiplied it by each term over here. We multiplied x by x minus four, and we multiplied seven by x minus four. Now we see that we have these, I guess you'd call them two separate terms, and to simplify each of them, or to multiply them out, we just have to distribute in this first, we have to distribute this blue x, and over here we have to distribute this blue seven. So let's do that."}, {"video_title": "Multiplying binomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "We took this whole thing and we multiplied it by each term over here. We multiplied x by x minus four, and we multiplied seven by x minus four. Now we see that we have these, I guess you'd call them two separate terms, and to simplify each of them, or to multiply them out, we just have to distribute in this first, we have to distribute this blue x, and over here we have to distribute this blue seven. So let's do that. So here we could say x times x is going to be x squared. X times, we have a negative here, so we could say negative four is going to be negative four x. And just like that, we get x squared minus four x."}, {"video_title": "Multiplying binomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "So let's do that. So here we could say x times x is going to be x squared. X times, we have a negative here, so we could say negative four is going to be negative four x. And just like that, we get x squared minus four x. And then over here, we have seven times x, so that's going to be plus seven x. And then we have seven times the negative four, which is negative 28. And we are almost done."}, {"video_title": "Multiplying binomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "And just like that, we get x squared minus four x. And then over here, we have seven times x, so that's going to be plus seven x. And then we have seven times the negative four, which is negative 28. And we are almost done. We can simplify it a little bit more. We have two first degree terms here. If I have negative four x's, and to that I add seven x's, what is that going to be?"}, {"video_title": "Multiplying binomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "And we are almost done. We can simplify it a little bit more. We have two first degree terms here. If I have negative four x's, and to that I add seven x's, what is that going to be? Well, those two terms together, these two terms together are going to be negative four plus seven x's. Negative four plus seven. Negative four plus seven x's."}, {"video_title": "Multiplying binomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "If I have negative four x's, and to that I add seven x's, what is that going to be? Well, those two terms together, these two terms together are going to be negative four plus seven x's. Negative four plus seven. Negative four plus seven x's. So all I'm doing here, I'm making it very clear that I'm adding these two coefficients, and then we have all the other terms. We have the x squared, x squared plus this, and then we have the minus 28. And we're at the home stretch."}, {"video_title": "Multiplying binomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "Negative four plus seven x's. So all I'm doing here, I'm making it very clear that I'm adding these two coefficients, and then we have all the other terms. We have the x squared, x squared plus this, and then we have the minus 28. And we're at the home stretch. This would simplify to x squared. Now negative four plus seven is three, so this is going to be plus three x. That's what these two middle terms simplify to, to three x, and then we have minus 28."}, {"video_title": "Multiplying binomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "And we're at the home stretch. This would simplify to x squared. Now negative four plus seven is three, so this is going to be plus three x. That's what these two middle terms simplify to, to three x, and then we have minus 28. Minus 28. And just like that, we are done. And a fun thing to think about, and notice it's in the same form."}, {"video_title": "Multiplying binomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "That's what these two middle terms simplify to, to three x, and then we have minus 28. Minus 28. And just like that, we are done. And a fun thing to think about, and notice it's in the same form. If we were to compare, a is one, b is three, and c is negative 28. But it's interesting here to look at the pattern. When we multiply these two binomials, especially these two binomials, where the coefficient on the x term was a one."}, {"video_title": "Multiplying binomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "And a fun thing to think about, and notice it's in the same form. If we were to compare, a is one, b is three, and c is negative 28. But it's interesting here to look at the pattern. When we multiply these two binomials, especially these two binomials, where the coefficient on the x term was a one. Notice, we have x times x, that's what actually forms the x squared term over here. We have negative four, let me do this in a new color. We have negative four times, that's not a new color."}, {"video_title": "Multiplying binomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "When we multiply these two binomials, especially these two binomials, where the coefficient on the x term was a one. Notice, we have x times x, that's what actually forms the x squared term over here. We have negative four, let me do this in a new color. We have negative four times, that's not a new color. We have negative four times seven, which is going to be negative 28. And then how did we get this middle term? How did we get this three x?"}, {"video_title": "Multiplying binomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "We have negative four times, that's not a new color. We have negative four times seven, which is going to be negative 28. And then how did we get this middle term? How did we get this three x? Well, you had the negative four x plus the seven x, or you had the negative four plus the seven times x. You had the negative four plus the seven times x. So hopefully you see a little bit of a pattern here."}, {"video_title": "Multiplying binomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "How did we get this three x? Well, you had the negative four x plus the seven x, or you had the negative four plus the seven times x. You had the negative four plus the seven times x. So hopefully you see a little bit of a pattern here. If you're multiplying two binomials where the coefficients on the x term are both one, it's going to be x squared, and then the last term, the constant term, is going to be the product of these two constants, negative four and seven. And then the first degree term, right over here, its coefficient is going to be the sum of these two constants, negative four and seven. Now this might, you could view this pattern, if you practice it, as just something that'll help you multiply binomials a little bit faster."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "Welcome back. Well, I'm going to show you the last two logarithm properties now. So this one, and I always found this one to be in some ways the most obvious one. But don't feel bad if it's not obvious. Maybe it will take a little bit of introspection. And I encourage you to really experiment with all these logarithm properties, because that's the only way that you'll really learn them. And the point of math isn't just to pass the next exam or to get an A on the next exam."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "But don't feel bad if it's not obvious. Maybe it will take a little bit of introspection. And I encourage you to really experiment with all these logarithm properties, because that's the only way that you'll really learn them. And the point of math isn't just to pass the next exam or to get an A on the next exam. The point of math is to understand math. And so you can actually apply it in life later on and not have to relearn everything every time. So the next logarithm property is if I have A times the logarithm base b of c. If I have A times this whole thing, that that equals logarithm base b of c to the a power."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "And the point of math isn't just to pass the next exam or to get an A on the next exam. The point of math is to understand math. And so you can actually apply it in life later on and not have to relearn everything every time. So the next logarithm property is if I have A times the logarithm base b of c. If I have A times this whole thing, that that equals logarithm base b of c to the a power. Fascinating. So let's see if this works out. So let's say if I have, I don't know, 3 times logarithm base 2 of 8."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "So the next logarithm property is if I have A times the logarithm base b of c. If I have A times this whole thing, that that equals logarithm base b of c to the a power. Fascinating. So let's see if this works out. So let's say if I have, I don't know, 3 times logarithm base 2 of 8. So this property tells us that this is going to be the same thing as logarithm base 2 of 8 to the third power. And that's the same thing as, well, we could figure it out. So let's see what this is."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "So let's say if I have, I don't know, 3 times logarithm base 2 of 8. So this property tells us that this is going to be the same thing as logarithm base 2 of 8 to the third power. And that's the same thing as, well, we could figure it out. So let's see what this is. 3 times log base, what's log base 2 of 8? The reason why I kind of hesitated a second ago is because every time I want to figure something out, I implicitly want to use log and exponential rules kind of to do it. So I'm trying to avoid that."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "So let's see what this is. 3 times log base, what's log base 2 of 8? The reason why I kind of hesitated a second ago is because every time I want to figure something out, I implicitly want to use log and exponential rules kind of to do it. So I'm trying to avoid that. But anyway, going back. So what is this? 2 to what power is 8?"}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "So I'm trying to avoid that. But anyway, going back. So what is this? 2 to what power is 8? Well, 2 to the third power is 8, right? So that's 3. And we have this 3 here."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "2 to what power is 8? Well, 2 to the third power is 8, right? So that's 3. And we have this 3 here. So 3 times 3. So this thing right here should equal 9. If this equals 9, then we know that this property works at least for this example."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "And we have this 3 here. So 3 times 3. So this thing right here should equal 9. If this equals 9, then we know that this property works at least for this example. You don't know if it works for all examples. And for that, maybe you'd want to look at the proof we have in the other videos. But that's kind of a more advanced topic."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "If this equals 9, then we know that this property works at least for this example. You don't know if it works for all examples. And for that, maybe you'd want to look at the proof we have in the other videos. But that's kind of a more advanced topic. But the important thing first is just to understand how to use it. So let's see. What is 2 to the ninth power?"}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "But that's kind of a more advanced topic. But the important thing first is just to understand how to use it. So let's see. What is 2 to the ninth power? Well, it's going to be some large number. Actually, I know what it is. It's 256, because in the last video, we figured out that 2 to the eighth was equal to 256."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "What is 2 to the ninth power? Well, it's going to be some large number. Actually, I know what it is. It's 256, because in the last video, we figured out that 2 to the eighth was equal to 256. And so 2 to the ninth should be 512. So 2 to the ninth should be 512. So if 8 to the third is also 512, then we are correct, right, because log base 2 of 512 is going to be equal to 9."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "It's 256, because in the last video, we figured out that 2 to the eighth was equal to 256. And so 2 to the ninth should be 512. So 2 to the ninth should be 512. So if 8 to the third is also 512, then we are correct, right, because log base 2 of 512 is going to be equal to 9. What's 8 to the third? It's 64 times, right? 8 times 8 squared is 64, so 8 cubed."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "So if 8 to the third is also 512, then we are correct, right, because log base 2 of 512 is going to be equal to 9. What's 8 to the third? It's 64 times, right? 8 times 8 squared is 64, so 8 cubed. So let's see. 4 times 2 is 3. 6 times 8, it looks like it's 512."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "8 times 8 squared is 64, so 8 cubed. So let's see. 4 times 2 is 3. 6 times 8, it looks like it's 512. Correct. And there's other ways you could have done it, because you could have said 8 to the third is the same thing as 2 to the ninth. How do we know that?"}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "6 times 8, it looks like it's 512. Correct. And there's other ways you could have done it, because you could have said 8 to the third is the same thing as 2 to the ninth. How do we know that? Well, 8 to the third is equal to 2 to the third to the third, right? I just rewrote 8. And we know from our exponent rules that 2 to the third to the third is the same thing as 2 to the ninth."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "How do we know that? Well, 8 to the third is equal to 2 to the third to the third, right? I just rewrote 8. And we know from our exponent rules that 2 to the third to the third is the same thing as 2 to the ninth. And actually, it's this exponent property where you can multiply. When you take something to an exponent and then take that to an exponent, and you can essentially just multiply the exponents, that's the exponent property that actually leads to this logarithm property. But I'm not going to dwell on that too much in this presentation."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "And we know from our exponent rules that 2 to the third to the third is the same thing as 2 to the ninth. And actually, it's this exponent property where you can multiply. When you take something to an exponent and then take that to an exponent, and you can essentially just multiply the exponents, that's the exponent property that actually leads to this logarithm property. But I'm not going to dwell on that too much in this presentation. There's a whole video on proving it a little bit more formally. The next logarithm property I'm going to show you, and then I'll review everything and maybe do some examples. This is probably the single most useful logarithm property if you are a calculator addict."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "But I'm not going to dwell on that too much in this presentation. There's a whole video on proving it a little bit more formally. The next logarithm property I'm going to show you, and then I'll review everything and maybe do some examples. This is probably the single most useful logarithm property if you are a calculator addict. And I'll show you why. So let's say I have log base B of A is equal to log base C of A divided by log base C of B. Now why is this a useful property if you are a calculator addict?"}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "This is probably the single most useful logarithm property if you are a calculator addict. And I'll show you why. So let's say I have log base B of A is equal to log base C of A divided by log base C of B. Now why is this a useful property if you are a calculator addict? Well, let's say you go to class and there's a quiz. The teacher says, you can use your calculator. And using your calculator, I want you to figure out the log base 17 of 357."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "Now why is this a useful property if you are a calculator addict? Well, let's say you go to class and there's a quiz. The teacher says, you can use your calculator. And using your calculator, I want you to figure out the log base 17 of 357. And you will scramble and look for the log base 17 button on your calculator and not find it. Because there is no log base 17 number, a button on your calculator. You'll probably either have a log button or you'll have an LN button."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "And using your calculator, I want you to figure out the log base 17 of 357. And you will scramble and look for the log base 17 button on your calculator and not find it. Because there is no log base 17 number, a button on your calculator. You'll probably either have a log button or you'll have an LN button. And just so you know, the log button on your calculator is probably base 10. And your LN number, your LN button on your calculator is going to be base E. For those of you who aren't familiar with E, don't worry about it. But it's 2.71 something something."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "You'll probably either have a log button or you'll have an LN button. And just so you know, the log button on your calculator is probably base 10. And your LN number, your LN button on your calculator is going to be base E. For those of you who aren't familiar with E, don't worry about it. But it's 2.71 something something. It's a number. It's nothing. It's an amazing number, but we'll talk more about that."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "But it's 2.71 something something. It's a number. It's nothing. It's an amazing number, but we'll talk more about that. In a future presentation. So there's only two bases you have on your calculator. So if you want to figure out another base logarithm, you use this property."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "It's an amazing number, but we'll talk more about that. In a future presentation. So there's only two bases you have on your calculator. So if you want to figure out another base logarithm, you use this property. So if you're given this on an exam, you can very confidently say, oh, well, that is just the same thing as you would have to switch to your yellow color in order to act with confidence. We could do either E or 10. But you could say that's the same thing as log base 10 of 357 divided by log base 10 of 17."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "So if you want to figure out another base logarithm, you use this property. So if you're given this on an exam, you can very confidently say, oh, well, that is just the same thing as you would have to switch to your yellow color in order to act with confidence. We could do either E or 10. But you could say that's the same thing as log base 10 of 357 divided by log base 10 of 17. So you literally could just say, type in 357 in your calculator and press the log button and you're going to get bam, bam, bam, bam. Then you can clear it or if you know how to use a parenthesis on your calculator, you can do that. But then you say, type 17 in your calculator, press the log button, you get bam, bam, bam, bam."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "But you could say that's the same thing as log base 10 of 357 divided by log base 10 of 17. So you literally could just say, type in 357 in your calculator and press the log button and you're going to get bam, bam, bam, bam. Then you can clear it or if you know how to use a parenthesis on your calculator, you can do that. But then you say, type 17 in your calculator, press the log button, you get bam, bam, bam, bam. And then you just divide them and you get your answer. So this is a super useful property for calculator addicts. And once again, I'm not going to go into a lot of depth of how this one to me is the most useful, but it doesn't completely fall out of, obviously, of the exponent properties, but it's hard for me to describe the intuition simply, so you probably want to watch the proof on it if you don't believe why this happens."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "But then you say, type 17 in your calculator, press the log button, you get bam, bam, bam, bam. And then you just divide them and you get your answer. So this is a super useful property for calculator addicts. And once again, I'm not going to go into a lot of depth of how this one to me is the most useful, but it doesn't completely fall out of, obviously, of the exponent properties, but it's hard for me to describe the intuition simply, so you probably want to watch the proof on it if you don't believe why this happens. But anyway, with all of those aside, and this is probably the one you're going to be using the most in everyday life. I still use this in my job, so just so you know, the logarithms are useful. Let's do some examples."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "And once again, I'm not going to go into a lot of depth of how this one to me is the most useful, but it doesn't completely fall out of, obviously, of the exponent properties, but it's hard for me to describe the intuition simply, so you probably want to watch the proof on it if you don't believe why this happens. But anyway, with all of those aside, and this is probably the one you're going to be using the most in everyday life. I still use this in my job, so just so you know, the logarithms are useful. Let's do some examples. So let's just rewrite a bunch of things in simpler forms. So if I wanted to write the log base 2 of the square root of 32 divided by the square root of 8, how can I rewrite this so it's reasonably not messy? Well, let's think about this."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "Let's do some examples. So let's just rewrite a bunch of things in simpler forms. So if I wanted to write the log base 2 of the square root of 32 divided by the square root of 8, how can I rewrite this so it's reasonably not messy? Well, let's think about this. This is the same thing. This is equal to, I don't know if I move vertically or horizontally, but I'll move vertically. This is the same thing as the log base 2 of 32 over the square root of 8 to the 1 half power, right?"}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, let's think about this. This is the same thing. This is equal to, I don't know if I move vertically or horizontally, but I'll move vertically. This is the same thing as the log base 2 of 32 over the square root of 8 to the 1 half power, right? And we know from our logarithm properties, the third one we learned, that that is the same thing as 1 half times the logarithm of 32 divided by the square root of 8, right? I just took the exponent and made that the coefficient on the entire thing, and we learned that at the beginning of this video. And now we have a little quotient here, right?"}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "This is the same thing as the log base 2 of 32 over the square root of 8 to the 1 half power, right? And we know from our logarithm properties, the third one we learned, that that is the same thing as 1 half times the logarithm of 32 divided by the square root of 8, right? I just took the exponent and made that the coefficient on the entire thing, and we learned that at the beginning of this video. And now we have a little quotient here, right? Logarithm of 32 divided by logarithm of square root of 8. Well, we can use our other logarithm. Let's keep the 1 half out."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "And now we have a little quotient here, right? Logarithm of 32 divided by logarithm of square root of 8. Well, we can use our other logarithm. Let's keep the 1 half out. That's going to equal, oops, parentheses, logarithm, oh, I forgot my base, logarithm base 2 of 32 minus, right, because this is in the quotient, minus the logarithm base 2 of the square root of 8, right? Let's see. Well, here, once again, we have a square root here, so we could say that this is equal to 1 half times log base 2 of 32 minus, this 8 to the 1 half, which is the same thing as 1 half log base 2 of 8."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "Let's keep the 1 half out. That's going to equal, oops, parentheses, logarithm, oh, I forgot my base, logarithm base 2 of 32 minus, right, because this is in the quotient, minus the logarithm base 2 of the square root of 8, right? Let's see. Well, here, once again, we have a square root here, so we could say that this is equal to 1 half times log base 2 of 32 minus, this 8 to the 1 half, which is the same thing as 1 half log base 2 of 8. We learned that property at the beginning of this presentation. And then if we want, we can distribute this original 1 half, and this equals 1 half log base 2 of 32 minus 1 half, minus 1 fourth, because we have to distribute that 1 half, minus 1 fourth log base 2 of 8. This is 5 halves minus, this is 3, 3 times 1 fourth minus 3 fourths."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, here, once again, we have a square root here, so we could say that this is equal to 1 half times log base 2 of 32 minus, this 8 to the 1 half, which is the same thing as 1 half log base 2 of 8. We learned that property at the beginning of this presentation. And then if we want, we can distribute this original 1 half, and this equals 1 half log base 2 of 32 minus 1 half, minus 1 fourth, because we have to distribute that 1 half, minus 1 fourth log base 2 of 8. This is 5 halves minus, this is 3, 3 times 1 fourth minus 3 fourths. Or 10 fourths minus 3 fourths is equal to 7 fourths. I probably made some arithmetic errors, but you get the point. See you soon."}, {"video_title": "Example 2 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Now in order to really use the quadratic equation, or to figure out what our a's, b's, and c's are, we have to have our equation in the form a x squared plus bx plus c is equal to 0. And then if we know our a's, b's, and c's, we'll say that the solutions to this equation are x is equal to negative b plus or minus the square root of b squared minus 4ac, all of that over 2a. So the first thing we have to do for this equation right here is to put it in this form. And on one side of the equation, we have a negative x squared plus 8x. So that looks like the first two terms. But our constant's on the other side. So let's get the constant on the left-hand side and get a 0 here on the right-hand side."}, {"video_title": "Example 2 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "And on one side of the equation, we have a negative x squared plus 8x. So that looks like the first two terms. But our constant's on the other side. So let's get the constant on the left-hand side and get a 0 here on the right-hand side. So let's subtract 1 from both sides of this equation. The left-hand side of the equation will become negative x squared plus 8x minus 1. And then the right-hand side, 1 minus 1 is 0."}, {"video_title": "Example 2 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So let's get the constant on the left-hand side and get a 0 here on the right-hand side. So let's subtract 1 from both sides of this equation. The left-hand side of the equation will become negative x squared plus 8x minus 1. And then the right-hand side, 1 minus 1 is 0. Now we have it in that form. We have ax squared. a is negative 1."}, {"video_title": "Example 2 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "And then the right-hand side, 1 minus 1 is 0. Now we have it in that form. We have ax squared. a is negative 1. So let me write this down. a is equal to negative 1. a is right here. a is equal to negative 1."}, {"video_title": "Example 2 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "a is negative 1. So let me write this down. a is equal to negative 1. a is right here. a is equal to negative 1. It's implicit there. You could put a 1 here if you like, a negative 1. Negative x squared is the same thing as negative 1x squared."}, {"video_title": "Example 2 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "a is equal to negative 1. It's implicit there. You could put a 1 here if you like, a negative 1. Negative x squared is the same thing as negative 1x squared. b is equal to 8. So b is equal to 8. That's the 8 right there."}, {"video_title": "Example 2 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Negative x squared is the same thing as negative 1x squared. b is equal to 8. So b is equal to 8. That's the 8 right there. And c is equal to negative 1. And c is equal to negative 1. That's the negative 1 right there."}, {"video_title": "Example 2 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "That's the 8 right there. And c is equal to negative 1. And c is equal to negative 1. That's the negative 1 right there. So now we can just apply the quadratic formula. The solutions to this equation are x is equal to negative b. So negative b plus or minus the square root of b squared, of 8 squared, minus 4ac."}, {"video_title": "Example 2 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "That's the negative 1 right there. So now we can just apply the quadratic formula. The solutions to this equation are x is equal to negative b. So negative b plus or minus the square root of b squared, of 8 squared, minus 4ac. Let me do that in that green color. Minus 4. The green's the part of the formula."}, {"video_title": "Example 2 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So negative b plus or minus the square root of b squared, of 8 squared, minus 4ac. Let me do that in that green color. Minus 4. The green's the part of the formula. The colored parts are the things that we're substituting into the formula. Minus 4 times a, which is negative 1, times c, which is also negative 1. And then all of that, let me extend the square root sign a little bit further, all of that is going to be over 2 times a."}, {"video_title": "Example 2 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "The green's the part of the formula. The colored parts are the things that we're substituting into the formula. Minus 4 times a, which is negative 1, times c, which is also negative 1. And then all of that, let me extend the square root sign a little bit further, all of that is going to be over 2 times a. In this case, a is negative 1. So let's simplify this. So this becomes negative 8."}, {"video_title": "Example 2 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "And then all of that, let me extend the square root sign a little bit further, all of that is going to be over 2 times a. In this case, a is negative 1. So let's simplify this. So this becomes negative 8. This is negative 8 plus or minus the square root of 8 squared is 64. And then you have a negative 1 times a negative 1. These just cancel out just to be a 1."}, {"video_title": "Example 2 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So this becomes negative 8. This is negative 8 plus or minus the square root of 8 squared is 64. And then you have a negative 1 times a negative 1. These just cancel out just to be a 1. So 64 minus 4, that's just that 4 over there. All of that over negative 2. So this is equal to negative 8 plus or minus the square root of 60."}, {"video_title": "Example 2 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "These just cancel out just to be a 1. So 64 minus 4, that's just that 4 over there. All of that over negative 2. So this is equal to negative 8 plus or minus the square root of 60. All of that over negative 2. Now let's see if we can simplify the radical expression here, the square root of 60. Let's see, 60 is equal to 2 times 30."}, {"video_title": "Example 2 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So this is equal to negative 8 plus or minus the square root of 60. All of that over negative 2. Now let's see if we can simplify the radical expression here, the square root of 60. Let's see, 60 is equal to 2 times 30. 30 is equal to 2 times 15. And then 15 is 3 times 5. So we do have a perfect square here."}, {"video_title": "Example 2 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Let's see, 60 is equal to 2 times 30. 30 is equal to 2 times 15. And then 15 is 3 times 5. So we do have a perfect square here. We do have a 2 times 2 in there. It is 2 times 2 times 15, or 4 times 15. So we could write the square root of 60 is equal to the square root of 4 times the square root of 15."}, {"video_title": "Example 2 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So we do have a perfect square here. We do have a 2 times 2 in there. It is 2 times 2 times 15, or 4 times 15. So we could write the square root of 60 is equal to the square root of 4 times the square root of 15. Right? Square root of 4 times the square root of 15. That's what 60 is, 4 times 15."}, {"video_title": "Example 2 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So we could write the square root of 60 is equal to the square root of 4 times the square root of 15. Right? Square root of 4 times the square root of 15. That's what 60 is, 4 times 15. And so this is equal to, square root of 4 is 2 times the square root of 15. So we can rewrite this expression right here as being equal to negative 8 plus or minus 2 times the square root of 15, all of that over negative 2. Now, we can divide both of these terms right here are divisible by either 2 or negative 2."}, {"video_title": "Example 2 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "That's what 60 is, 4 times 15. And so this is equal to, square root of 4 is 2 times the square root of 15. So we can rewrite this expression right here as being equal to negative 8 plus or minus 2 times the square root of 15, all of that over negative 2. Now, we can divide both of these terms right here are divisible by either 2 or negative 2. So let's divide it. So we have negative 8 divided by negative 2, which is positive 4. So let me write it over here."}, {"video_title": "Example 2 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Now, we can divide both of these terms right here are divisible by either 2 or negative 2. So let's divide it. So we have negative 8 divided by negative 2, which is positive 4. So let me write it over here. Negative 8 divided by negative 2 is positive 4. And then you have this weird thing, plus or minus 2 divided by negative 2. And really what we have here is two expressions."}, {"video_title": "Example 2 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So let me write it over here. Negative 8 divided by negative 2 is positive 4. And then you have this weird thing, plus or minus 2 divided by negative 2. And really what we have here is two expressions. But if we're plus 2 and we divide by negative 2, it will be negative 1. And if we take negative 2 and divide by negative 2, we're going to have positive 1. So instead of plus or minus, you could imagine it now as being minus or plus."}, {"video_title": "Example 2 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "And really what we have here is two expressions. But if we're plus 2 and we divide by negative 2, it will be negative 1. And if we take negative 2 and divide by negative 2, we're going to have positive 1. So instead of plus or minus, you could imagine it now as being minus or plus. But it's really the same thing, right? It's really now minus or plus. If it was plus, it's not going to be a minus."}, {"video_title": "Example 2 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "So instead of plus or minus, you could imagine it now as being minus or plus. But it's really the same thing, right? It's really now minus or plus. If it was plus, it's not going to be a minus. Now if it was a minus, it's not going to be a plus. Minus or plus 2 times the square root of 15. Or another way to view it is that the two solutions here are 4 minus 2 roots of 15 and 4 plus 2 roots of 15."}, {"video_title": "Example 2 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "If it was plus, it's not going to be a minus. Now if it was a minus, it's not going to be a plus. Minus or plus 2 times the square root of 15. Or another way to view it is that the two solutions here are 4 minus 2 roots of 15 and 4 plus 2 roots of 15. These are both values of x that'll satisfy this equation. And if this confused you, what I did, turning a plus or minus into minus plus, let me just take a little bit of a side there. I could write this expression up here as two expressions."}, {"video_title": "Example 2 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "Or another way to view it is that the two solutions here are 4 minus 2 roots of 15 and 4 plus 2 roots of 15. These are both values of x that'll satisfy this equation. And if this confused you, what I did, turning a plus or minus into minus plus, let me just take a little bit of a side there. I could write this expression up here as two expressions. That's what the plus or minus really is. There's a negative 8 plus 2 roots of 15 over negative 2. And then there's a negative 8 minus 2 roots 15 over negative 2."}, {"video_title": "Example 2 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "I could write this expression up here as two expressions. That's what the plus or minus really is. There's a negative 8 plus 2 roots of 15 over negative 2. And then there's a negative 8 minus 2 roots 15 over negative 2. This one simplifies to negative 8 divided by negative 2 is 4. 2 divided by negative 2 is negative 1 times 4 minus the square root of 15. And then over here, you have negative 8 divided by negative 2, which is 4, and then negative 2 divided by negative 2, which is plus the square root of 15."}, {"video_title": "Example 2 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "And then there's a negative 8 minus 2 roots 15 over negative 2. This one simplifies to negative 8 divided by negative 2 is 4. 2 divided by negative 2 is negative 1 times 4 minus the square root of 15. And then over here, you have negative 8 divided by negative 2, which is 4, and then negative 2 divided by negative 2, which is plus the square root of 15. And I just realized I made a mistake up here. When we're dividing it 2 divided by negative 2, we don't have this 2 over here. This is just a plus or minus the root of 15."}, {"video_title": "Example 2 Using the quadratic formula Quadratic equations Algebra I Khan Academy.mp3", "Sentence": "And then over here, you have negative 8 divided by negative 2, which is 4, and then negative 2 divided by negative 2, which is plus the square root of 15. And I just realized I made a mistake up here. When we're dividing it 2 divided by negative 2, we don't have this 2 over here. This is just a plus or minus the root of 15. We just saw that when I did it out here. So this is minus the square root of 15, and this is plus the square root of 15. So the two solutions for this equation, it's good that I took that little hiatus there, that little aside there."}, {"video_title": "Piecewise function formula from graph Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "By now we're used to seeing functions defined like h of y is equal to y squared, or f of x is equal to the square root of x. But what we're now going to explore is functions that are defined piece by piece over different intervals. And functions like this are, you'll sometimes view them as a piecewise, or these types of function definitions, they might be called a piecewise function definition. And so let's take a look at this graph right over here. This graph, you can see that the function is constant over this interval for x, and then it jumps up to this interval for x, and then it jumps back down for this interval for x. So let's think about how we would write this using our function notation. So if we say that this right over here is the x-axis, and if this is the y is equal to f of x axis, then let's see, our function f of x is going to be equal to, let's see, there's three different intervals."}, {"video_title": "Piecewise function formula from graph Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And so let's take a look at this graph right over here. This graph, you can see that the function is constant over this interval for x, and then it jumps up to this interval for x, and then it jumps back down for this interval for x. So let's think about how we would write this using our function notation. So if we say that this right over here is the x-axis, and if this is the y is equal to f of x axis, then let's see, our function f of x is going to be equal to, let's see, there's three different intervals. So let me give myself some space for the three different intervals. Now this first interval is from, not including negative nine, I have this open circle here, not a closed in circle, so not including negative nine, but x being greater than negative nine, and all the way up to and including negative five. So I could write that as negative nine is less than x, less than or equal to negative five."}, {"video_title": "Piecewise function formula from graph Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So if we say that this right over here is the x-axis, and if this is the y is equal to f of x axis, then let's see, our function f of x is going to be equal to, let's see, there's three different intervals. So let me give myself some space for the three different intervals. Now this first interval is from, not including negative nine, I have this open circle here, not a closed in circle, so not including negative nine, but x being greater than negative nine, and all the way up to and including negative five. So I could write that as negative nine is less than x, less than or equal to negative five. That's this interval. And what is the value of the function over this interval? Well we see the value of the function is negative nine."}, {"video_title": "Piecewise function formula from graph Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So I could write that as negative nine is less than x, less than or equal to negative five. That's this interval. And what is the value of the function over this interval? Well we see the value of the function is negative nine. It's a constant negative nine over that interval. It's a little confusing because the value of the function is actually also the value of the lower bound on this interval right over here. And it's very important to look at, this says negative nine is less than x, not less than or equal."}, {"video_title": "Piecewise function formula from graph Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Well we see the value of the function is negative nine. It's a constant negative nine over that interval. It's a little confusing because the value of the function is actually also the value of the lower bound on this interval right over here. And it's very important to look at, this says negative nine is less than x, not less than or equal. If it was less than or equal, then the function would have been defined at x equals negative nine, but it's not. We have an open circle right over there. But now let's look at the next interval."}, {"video_title": "Piecewise function formula from graph Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And it's very important to look at, this says negative nine is less than x, not less than or equal. If it was less than or equal, then the function would have been defined at x equals negative nine, but it's not. We have an open circle right over there. But now let's look at the next interval. The next interval is from x is greater, or negative five is less than x, which is less than or equal to negative one. And over that interval, the function is equal to, the function is a constant six. It jumps up here."}, {"video_title": "Piecewise function formula from graph Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "But now let's look at the next interval. The next interval is from x is greater, or negative five is less than x, which is less than or equal to negative one. And over that interval, the function is equal to, the function is a constant six. It jumps up here. Sometimes people call this a step function. It steps up. It looks like stairs to some degree."}, {"video_title": "Piecewise function formula from graph Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "It jumps up here. Sometimes people call this a step function. It steps up. It looks like stairs to some degree. Now it's very important here that at x equals negative five, for it to be defined only one place. Here it's defined by this part. It's only defined over here."}, {"video_title": "Piecewise function formula from graph Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "It looks like stairs to some degree. Now it's very important here that at x equals negative five, for it to be defined only one place. Here it's defined by this part. It's only defined over here. And so that's why it's important that this isn't a negative five is less than or equal to, because then if you put negative five into the function, this thing would be filled in, and then the function would be defined to both places, and that's not cool for a function. It wouldn't be a function anymore. So it's very important that this, that when you input negative five in here, you know which of these intervals you are in."}, {"video_title": "Piecewise function formula from graph Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "It's only defined over here. And so that's why it's important that this isn't a negative five is less than or equal to, because then if you put negative five into the function, this thing would be filled in, and then the function would be defined to both places, and that's not cool for a function. It wouldn't be a function anymore. So it's very important that this, that when you input negative five in here, you know which of these intervals you are in. You can't be in two of these intervals. If you are in two of these intervals, the interval should give you the same value, so that the function maps from one input to the same output. Now let's keep going."}, {"video_title": "Piecewise function formula from graph Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So it's very important that this, that when you input negative five in here, you know which of these intervals you are in. You can't be in two of these intervals. If you are in two of these intervals, the interval should give you the same value, so that the function maps from one input to the same output. Now let's keep going. We have this last interval where we're going from negative one, we're going from negative one to nine, from negative one to positive nine, and x, it starts off with negative one less than x, because you have an open circle right over here, and that's good because x equals negative one is defined up here, all the way to x is less than or equal to nine. And over that interval, what is the value of our function? Well, you see the value of our function is a constant negative seven."}, {"video_title": "Piecewise function formula from graph Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Now let's keep going. We have this last interval where we're going from negative one, we're going from negative one to nine, from negative one to positive nine, and x, it starts off with negative one less than x, because you have an open circle right over here, and that's good because x equals negative one is defined up here, all the way to x is less than or equal to nine. And over that interval, what is the value of our function? Well, you see the value of our function is a constant negative seven. A constant negative seven. And we're done. We have just constructed a piece-by-piece definition of this function."}, {"video_title": "Piecewise function formula from graph Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Well, you see the value of our function is a constant negative seven. A constant negative seven. And we're done. We have just constructed a piece-by-piece definition of this function. And actually, when you see this type of function notation, it becomes a lot clearer why function notation is useful, even. And hopefully, well anyway, hopefully you enjoyed that. I always find these piecewise functions a lot of fun."}, {"video_title": "Parallel lines from equation (example 2) Mathematics I High School Math Khan Academy.mp3", "Sentence": "So the equation for line A is y is equal to 3 4ths x minus 4. Line B is 4y minus 20 is equal to negative 3x. And then line C is negative 3x plus 4y is equal to 40. So to figure out if any of these lines are parallel to any of the other lines, we just have to compare their slopes. If any two of these lines have the same slope and they're different lines, they have different y-intercepts, then they're going to be parallel. Now line A, it's very easy to figure out its slope. It's already in slope-intercept form."}, {"video_title": "Parallel lines from equation (example 2) Mathematics I High School Math Khan Academy.mp3", "Sentence": "So to figure out if any of these lines are parallel to any of the other lines, we just have to compare their slopes. If any two of these lines have the same slope and they're different lines, they have different y-intercepts, then they're going to be parallel. Now line A, it's very easy to figure out its slope. It's already in slope-intercept form. This is mx plus b. The slope is 3 4ths and the y-intercept, which isn't as relevant when you're figuring out parallel lines, is negative 4. So let's see what the other character's slopes are."}, {"video_title": "Parallel lines from equation (example 2) Mathematics I High School Math Khan Academy.mp3", "Sentence": "It's already in slope-intercept form. This is mx plus b. The slope is 3 4ths and the y-intercept, which isn't as relevant when you're figuring out parallel lines, is negative 4. So let's see what the other character's slopes are. This isn't in any kind of standard form. It's not in standard form, slope-intercept, or point-slope form. But let's see what the slope of this line is."}, {"video_title": "Parallel lines from equation (example 2) Mathematics I High School Math Khan Academy.mp3", "Sentence": "So let's see what the other character's slopes are. This isn't in any kind of standard form. It's not in standard form, slope-intercept, or point-slope form. But let's see what the slope of this line is. So to get it into slope-intercept form, which is really the easiest one to pick out the slope from, let's add 20 to both sides of this equation. So let's add 20 to both sides. The left-hand side, those cancel out, that was the whole point."}, {"video_title": "Parallel lines from equation (example 2) Mathematics I High School Math Khan Academy.mp3", "Sentence": "But let's see what the slope of this line is. So to get it into slope-intercept form, which is really the easiest one to pick out the slope from, let's add 20 to both sides of this equation. So let's add 20 to both sides. The left-hand side, those cancel out, that was the whole point. You get 4y is equal to negative 3x plus 20. And now we can divide everything by 4. Just dividing both sides of this equation by 4."}, {"video_title": "Parallel lines from equation (example 2) Mathematics I High School Math Khan Academy.mp3", "Sentence": "The left-hand side, those cancel out, that was the whole point. You get 4y is equal to negative 3x plus 20. And now we can divide everything by 4. Just dividing both sides of this equation by 4. We are left with y is equal to negative 3 4ths x plus 5. So in this case, y-intercept is 5, but most importantly, the slope is negative 3 4ths. So it's different than this guy."}, {"video_title": "Parallel lines from equation (example 2) Mathematics I High School Math Khan Academy.mp3", "Sentence": "Just dividing both sides of this equation by 4. We are left with y is equal to negative 3 4ths x plus 5. So in this case, y-intercept is 5, but most importantly, the slope is negative 3 4ths. So it's different than this guy. This is negative 3 4ths, this is positive 3 4ths. So these two guys definitely aren't parallel. Let's move on to this guy."}, {"video_title": "Parallel lines from equation (example 2) Mathematics I High School Math Khan Academy.mp3", "Sentence": "So it's different than this guy. This is negative 3 4ths, this is positive 3 4ths. So these two guys definitely aren't parallel. Let's move on to this guy. This guy written in standard form. Let's get the x term on the other side. So let's add 3x to both sides of this equation."}, {"video_title": "Parallel lines from equation (example 2) Mathematics I High School Math Khan Academy.mp3", "Sentence": "Let's move on to this guy. This guy written in standard form. Let's get the x term on the other side. So let's add 3x to both sides of this equation. Left-hand side, these cancel out. We're just left with 4y is equal to 3x plus 40, or 40 plus 3x, either way. Now we can divide both sides by 4."}, {"video_title": "Parallel lines from equation (example 2) Mathematics I High School Math Khan Academy.mp3", "Sentence": "So let's add 3x to both sides of this equation. Left-hand side, these cancel out. We're just left with 4y is equal to 3x plus 40, or 40 plus 3x, either way. Now we can divide both sides by 4. We have to divide every term by 4. The left-hand side, you're left with y. The right-hand side, you have 3 4ths x plus 10."}, {"video_title": "Parallel lines from equation (example 2) Mathematics I High School Math Khan Academy.mp3", "Sentence": "Now we can divide both sides by 4. We have to divide every term by 4. The left-hand side, you're left with y. The right-hand side, you have 3 4ths x plus 10. So here, our slope is 3 4ths, and our y-intercept, if we care about it, is 10. So this line and this line have the exact same slope, 3 4ths, and they're different lines because their y-intercept is different. So we know that A and C are parallel lines, and B is not parallel to neither one of the other two."}, {"video_title": "Writing in logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "So if we want to write the same information, really, in logarithmic form, we could say that the power, the power that I need to raise 10 to, to get to 100 is equal to two, or log base 10, log base 10 of 100 is equal to two. Notice, these are equivalent statements. This is just an exponential form. This is in logarithmic form. This is saying the power that I need to raise 10 to, to get to 100, is equal to two, which is the same thing as saying that 10 to the second power is 100. 10 to the second power is 100. And the way that I specified the base is by doing this underscore right over here."}, {"video_title": "Writing in logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "This is in logarithmic form. This is saying the power that I need to raise 10 to, to get to 100, is equal to two, which is the same thing as saying that 10 to the second power is 100. 10 to the second power is 100. And the way that I specified the base is by doing this underscore right over here. So underscore 10, log base 10 of 100 is equal to two. Now they, here they ask us to rewrite the following equation in exponential form. So this is log base five of one over 125 is equal to negative three."}, {"video_title": "Writing in logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "And the way that I specified the base is by doing this underscore right over here. So underscore 10, log base 10 of 100 is equal to two. Now they, here they ask us to rewrite the following equation in exponential form. So this is log base five of one over 125 is equal to negative three. This is, one way to think about it is saying the power that I need to raise five to, to get to one over 125 is equal to negative three, or that five to the negative three power, five to the negative three power is equal to one over 125. And we can verify that this has formatted it the right way. Five to the negative three power is one over 125."}, {"video_title": "Writing in logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "So this is log base five of one over 125 is equal to negative three. This is, one way to think about it is saying the power that I need to raise five to, to get to one over 125 is equal to negative three, or that five to the negative three power, five to the negative three power is equal to one over 125. And we can verify that this has formatted it the right way. Five to the negative three power is one over 125. The exact same truth about the universe, just in different forms, logarithmic form and exponential form. So let me check my answer, make sure I got it right. And I did."}, {"video_title": "Creativity break Why is creativity important in algebra Algebra 1 Khan Academy.mp3", "Sentence": "It's not like memorizing how to solve problems. It's learning the tools of how to solve problems and then using them and building them up in creative ways. So it's kind of like, it really does remind me of art because if you are doing like a painting or something, you have like a specific tools about like maybe paint brushes or different techniques you can use to create something bigger. So I feel like algebra is really all about combining things that you already know how to do into something bigger. So the word algebra comes from the Arabic word al-jabr that literally translates to reunion of broken parts. This breaking apart of mathematical representations and putting it back together in the form of balanced equations is the essence of algebra. And the beautiful thing is that this balancing and solving of equations can be done in many different ways, which is where creativity comes in."}, {"video_title": "Shifting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And what we're going to start off doing is just graph a plain vanilla function, f of x is equal to x squared. That looks as we would expect it to look. But now let's think about how we could shift it up or down. Well, one thought is, well, to shift it up, we just have to make the value of f of x higher so we could add a value. And that does look like it shifted it up by one. Whatever f of x was before, we're now adding one to it, so it shifts the graph up by one. That's pretty intuitive."}, {"video_title": "Shifting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Well, one thought is, well, to shift it up, we just have to make the value of f of x higher so we could add a value. And that does look like it shifted it up by one. Whatever f of x was before, we're now adding one to it, so it shifts the graph up by one. That's pretty intuitive. If we subtract one, or actually let's subtract three, notice it shifted it down. The vertex was right over here at zero, zero. Now it is at zero, negative three, so it shifted it down."}, {"video_title": "Shifting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "That's pretty intuitive. If we subtract one, or actually let's subtract three, notice it shifted it down. The vertex was right over here at zero, zero. Now it is at zero, negative three, so it shifted it down. And we can set up a slider here to make that a little bit clearer. So if I just replace this with, if I just replace this with the variable k, then, let me delete this little thing here, that little subscript thing that happened. Then we can add a slider k here."}, {"video_title": "Shifting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Now it is at zero, negative three, so it shifted it down. And we can set up a slider here to make that a little bit clearer. So if I just replace this with, if I just replace this with the variable k, then, let me delete this little thing here, that little subscript thing that happened. Then we can add a slider k here. And this is just allowing us to set what k is equal to. So here k is equal to one. So this is x squared plus one."}, {"video_title": "Shifting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Then we can add a slider k here. And this is just allowing us to set what k is equal to. So here k is equal to one. So this is x squared plus one. And notice, we have shifted up. And if we increase the value of k, notice how it shifts the graph up. And as we decrease the value of k, if k is zero, we're back where our vertex is right at the origin."}, {"video_title": "Shifting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So this is x squared plus one. And notice, we have shifted up. And if we increase the value of k, notice how it shifts the graph up. And as we decrease the value of k, if k is zero, we're back where our vertex is right at the origin. And then as we decrease the value of k, it shifts our graph down. And that's pretty intuitive, because we're adding or subtracting that amount to x squared so it changes, we could say, the y value. It shifts it up or down."}, {"video_title": "Shifting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And as we decrease the value of k, if k is zero, we're back where our vertex is right at the origin. And then as we decrease the value of k, it shifts our graph down. And that's pretty intuitive, because we're adding or subtracting that amount to x squared so it changes, we could say, the y value. It shifts it up or down. But how do we shift to the left or to the right? So what's interesting here is, to shift to the left or to the right, we can replace our x with an x minus something. So let's see how that might work."}, {"video_title": "Shifting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "It shifts it up or down. But how do we shift to the left or to the right? So what's interesting here is, to shift to the left or to the right, we can replace our x with an x minus something. So let's see how that might work. So I'm going to replace our x with an x minus, let's replace it with an x minus one. What do you think's going to happen? Do you think that's going to shift it one to the right or one to the left?"}, {"video_title": "Shifting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So let's see how that might work. So I'm going to replace our x with an x minus, let's replace it with an x minus one. What do you think's going to happen? Do you think that's going to shift it one to the right or one to the left? So let's just put the one in. Well, that's interesting. Before, our vertex was at zero, zero."}, {"video_title": "Shifting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Do you think that's going to shift it one to the right or one to the left? So let's just put the one in. Well, that's interesting. Before, our vertex was at zero, zero. Now, our vertex is at one, zero. So by replacing our x with an x minus one, we actually shifted one to the right. Now, why does that make sense?"}, {"video_title": "Shifting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Before, our vertex was at zero, zero. Now, our vertex is at one, zero. So by replacing our x with an x minus one, we actually shifted one to the right. Now, why does that make sense? Well, one way to think about it, before we put this x, before we replaced our x with an x minus one, the vertex was when we were squaring zero. Now, in order to square zero, squaring zero happens when x is equal to one. When x is equal to one, you do one minus one, you get zero, and then that's when you are squaring zero."}, {"video_title": "Shifting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Now, why does that make sense? Well, one way to think about it, before we put this x, before we replaced our x with an x minus one, the vertex was when we were squaring zero. Now, in order to square zero, squaring zero happens when x is equal to one. When x is equal to one, you do one minus one, you get zero, and then that's when you are squaring zero. So it makes sense that you have a similar behavior of the graph at the vertex now when x equals one as before you had when x equals zero. And to see how this can be generalized, let's put another variable here and let's add a slider for h, and then we can see that when h is zero and k is zero, our function is really then just x squared. And then if h increases, we're replacing our x with x minus a larger value, that's shifting to the right, and then as h decreases, as it becomes negative, that shifts to the left."}, {"video_title": "Shifting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "When x is equal to one, you do one minus one, you get zero, and then that's when you are squaring zero. So it makes sense that you have a similar behavior of the graph at the vertex now when x equals one as before you had when x equals zero. And to see how this can be generalized, let's put another variable here and let's add a slider for h, and then we can see that when h is zero and k is zero, our function is really then just x squared. And then if h increases, we're replacing our x with x minus a larger value, that's shifting to the right, and then as h decreases, as it becomes negative, that shifts to the left. Now, right here, h is equal to negative five. You typically won't see x minus negative five. You would see that written as x plus five."}, {"video_title": "Shifting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And then if h increases, we're replacing our x with x minus a larger value, that's shifting to the right, and then as h decreases, as it becomes negative, that shifts to the left. Now, right here, h is equal to negative five. You typically won't see x minus negative five. You would see that written as x plus five. So if you replace your x's with an x plus five, that actually shifts everything five units to the left. And of course, we can shift both of them together like this. So here, we're shifting it up, and then we can get back to our neutral horizontal shift, and then we can shift it to the right like that."}, {"video_title": "Shifting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "You would see that written as x plus five. So if you replace your x's with an x plus five, that actually shifts everything five units to the left. And of course, we can shift both of them together like this. So here, we're shifting it up, and then we can get back to our neutral horizontal shift, and then we can shift it to the right like that. And everything we did just now is with the x squared function as our core function, but you could do it with all sorts of functions. You could do it with an absolute value function. Let's do absolute value."}, {"video_title": "Shifting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So here, we're shifting it up, and then we can get back to our neutral horizontal shift, and then we can shift it to the right like that. And everything we did just now is with the x squared function as our core function, but you could do it with all sorts of functions. You could do it with an absolute value function. Let's do absolute value. That's always a fun one. So instead of squaring all this business, let's have an absolute value here. So I'm gonna put an absolute, whoops, absolute value, and there you have it."}, {"video_title": "Shifting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Let's do absolute value. That's always a fun one. So instead of squaring all this business, let's have an absolute value here. So I'm gonna put an absolute, whoops, absolute value, and there you have it. You can start at, let me make both of these variables equal to zero. So that would just be the graph of f of x is equal to the absolute value of x. But let's say you wanted to shift it so that this point right over here that's at the origin is at the point negative five, negative five, which is right over there."}, {"video_title": "Shifting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So I'm gonna put an absolute, whoops, absolute value, and there you have it. You can start at, let me make both of these variables equal to zero. So that would just be the graph of f of x is equal to the absolute value of x. But let's say you wanted to shift it so that this point right over here that's at the origin is at the point negative five, negative five, which is right over there. So what you would do is you would replace your x with x plus five, or you would make this h variable to negative five right over here, because notice, if you replace your h with a negative five, inside the absolute value, you would have an x plus five, and then if you wanna shift it down, you just reduce the value of k. And if you wanna shift it down by five, you reduce it by five, and you could get something like that. So I encourage you, go to desmos.com, try this out for yourself, and really play around with these functions to give yourself an intuition of how things and why things shift up or down when you add a constant, and why things shift to the left or the right when you replace your x's with an x minus, in this case, an x minus h, but it really could be x minus some type of a constant."}, {"video_title": "Shifting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "But let's say you wanted to shift it so that this point right over here that's at the origin is at the point negative five, negative five, which is right over there. So what you would do is you would replace your x with x plus five, or you would make this h variable to negative five right over here, because notice, if you replace your h with a negative five, inside the absolute value, you would have an x plus five, and then if you wanna shift it down, you just reduce the value of k. And if you wanna shift it down by five, you reduce it by five, and you could get something like that. So I encourage you, go to desmos.com, try this out for yourself, and really play around with these functions to give yourself an intuition of how things and why things shift up or down when you add a constant, and why things shift to the left or the right when you replace your x's with an x minus, in this case, an x minus h, but it really could be x minus some type of a constant."}]