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Polynomial remainder theorem Polynomial and rational functions Algebra II Khan Academy.mp3 | It is going to be, it is going to be f of a. f of a, I know this might seem a little bit abstract right now, I'm talking about f of x's and x minus a's, let's make it a little bit more concrete. So let's say that f of x, f of x is equal to, I'm just gonna make up a, let's say a second degree polynomial. This would be t... |
Polynomial remainder theorem Polynomial and rational functions Algebra II Khan Academy.mp3 | So three x squared minus four x plus seven, and let's say that a is, I don't know, a is one. So we're gonna divide by, we're gonna divide that, we're gonna divide by x minus, x minus one. So a, in this case, is equal to one. So let's just do the polynomial long division, and I encourage you to pause the video, and if y... |
Polynomial remainder theorem Polynomial and rational functions Algebra II Khan Academy.mp3 | So let's just do the polynomial long division, and I encourage you to pause the video, and if you're unfamiliar with polynomial long division, I encourage you to watch that before watching this video because I will assume you know how to do a polynomial long division. So divide three x squared minus four x plus seven, ... |
Polynomial remainder theorem Polynomial and rational functions Algebra II Khan Academy.mp3 | So let's divide x minus one, x minus one into three x squared, into three x squared minus four x plus seven. All right, a little bit of polynomial long division is never a bad way to start your morning. It's morning for me, I don't know what it is for you. All right, so, how many, I'll look at, I always look at the x t... |
Polynomial remainder theorem Polynomial and rational functions Algebra II Khan Academy.mp3 | All right, so, how many, I'll look at, I always look at the x term here, the highest degree term, and then I'll start with the highest degree term here. So how many times is x going to three x squared? Well, it goes three x times. Three x times x is three x squared, so I'll write three x over here. I'm writing it in th... |
Polynomial remainder theorem Polynomial and rational functions Algebra II Khan Academy.mp3 | Three x times x is three x squared, so I'll write three x over here. I'm writing it in the, I guess you could say the first degree place. Three x times x is three x squared. Three x times negative one is negative three x. And now we want to subtract, we want to subtract this thing. And this is just the way that you do ... |
Polynomial remainder theorem Polynomial and rational functions Algebra II Khan Academy.mp3 | Three x times negative one is negative three x. And now we want to subtract, we want to subtract this thing. And this is just the way that you do traditional long division. And so, what do we get? Well, three x squared minus three x squared, that's just going to be zero. So these just add up to zero. And this negative ... |
Polynomial remainder theorem Polynomial and rational functions Algebra II Khan Academy.mp3 | And so, what do we get? Well, three x squared minus three x squared, that's just going to be zero. So these just add up to zero. And this negative four x, this is going to be plus three x, right? A negative of a negative. Negative four x plus three x is going to be negative x. Let me do this in a new color. |
Polynomial remainder theorem Polynomial and rational functions Algebra II Khan Academy.mp3 | And this negative four x, this is going to be plus three x, right? A negative of a negative. Negative four x plus three x is going to be negative x. Let me do this in a new color. So it's going to be negative negative x. And then we can bring down the seven. Complete analogy to how you first learn long division in mayb... |
Polynomial remainder theorem Polynomial and rational functions Algebra II Khan Academy.mp3 | Let me do this in a new color. So it's going to be negative negative x. And then we can bring down the seven. Complete analogy to how you first learn long division in maybe, I don't know, third or fourth grade. So all I did is I multiplied three x times this, you get three x squared minus three x. And then I subtracted... |
Polynomial remainder theorem Polynomial and rational functions Algebra II Khan Academy.mp3 | Complete analogy to how you first learn long division in maybe, I don't know, third or fourth grade. So all I did is I multiplied three x times this, you get three x squared minus three x. And then I subtracted that from three x squared minus four x to get this right over here. Or you could say I subtracted it from thi... |
Polynomial remainder theorem Polynomial and rational functions Algebra II Khan Academy.mp3 | Or you could say I subtracted it from this whole, from the whole polynomial, and then I got negative x plus seven. So now, how many times does x minus one go into negative x plus seven? Well, x goes into negative x negative one times. Negative one times. Negative one times x is negative x. Negative one times negative o... |
Polynomial remainder theorem Polynomial and rational functions Algebra II Khan Academy.mp3 | Negative one times. Negative one times x is negative x. Negative one times negative one is positive one. But then we're going to want to subtract this thing. We're going to want to subtract this thing, and this is going to give us our remainder. So negative x minus negative x, that's the same thing as negative x plus x... |
Polynomial remainder theorem Polynomial and rational functions Algebra II Khan Academy.mp3 | But then we're going to want to subtract this thing. We're going to want to subtract this thing, and this is going to give us our remainder. So negative x minus negative x, that's the same thing as negative x plus x. So these are just going to add up to zero. And then you have seven. This isn't going to be seven plus o... |
Polynomial remainder theorem Polynomial and rational functions Algebra II Khan Academy.mp3 | So these are just going to add up to zero. And then you have seven. This isn't going to be seven plus one. Remember, we have this negative out. So if you distribute the negative, this is going to be a negative one. Seven minus one is six. So your remainder here is six. |
Polynomial remainder theorem Polynomial and rational functions Algebra II Khan Academy.mp3 | Remember, we have this negative out. So if you distribute the negative, this is going to be a negative one. Seven minus one is six. So your remainder here is six. One way to think about it, you could say, you could say that, you could say, well, actually, I'll save that for a future video. This right over here is the r... |
Polynomial remainder theorem Polynomial and rational functions Algebra II Khan Academy.mp3 | So your remainder here is six. One way to think about it, you could say, you could say that, you could say, well, actually, I'll save that for a future video. This right over here is the remainder. Remainder. And you know when you got to the remainder, and this is just all a review of polynomial long division, is when ... |
Polynomial remainder theorem Polynomial and rational functions Algebra II Khan Academy.mp3 | Remainder. And you know when you got to the remainder, and this is just all a review of polynomial long division, is when you get something that has a lower degree. This is a, I guess you could call this a zero degree polynomial. This has a lower degree than what you are actually dividing into, or then the x minus one,... |
Polynomial remainder theorem Polynomial and rational functions Algebra II Khan Academy.mp3 | This has a lower degree than what you are actually dividing into, or then the x minus one, than your divisor. So this is a lower degree, so this is the remainder. You can't take this into this any more times. Now, by the polynomial remainder theorem, if it's true, and I just picked a random example here, this is by no ... |
Polynomial remainder theorem Polynomial and rational functions Algebra II Khan Academy.mp3 | Now, by the polynomial remainder theorem, if it's true, and I just picked a random example here, this is by no means a proof, but just a kind of a way to make it tangible of what the polynomial remainder theorem is telling us. If the polynomial remainder theorem is true, it's telling us that f of a, in this case one, f... |
Factoring higher degree polynomials Algebra 2 Khan Academy.mp3 | There are many videos on Khan Academy where we talk about factoring polynomials, but what we're going to do in this video is do a few more examples of factoring higher degree polynomials. 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 fou... |
Factoring higher degree polynomials Algebra 2 Khan Academy.mp3 | 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... |
Factoring higher degree polynomials Algebra 2 Khan Academy.mp3 | 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 t... |
Factoring higher degree polynomials Algebra 2 Khan Academy.mp3 | 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? Well, three x times three is nine x. And you can verify that if we were to distribute this three x you wou... |
Factoring higher degree polynomials Algebra 2 Khan Academy.mp3 | 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? Can we factor this? Wel... |
Factoring higher degree polynomials Algebra 2 Khan Academy.mp3 | 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... |
Factoring higher degree polynomials Algebra 2 Khan Academy.mp3 | 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 wr... |
Factoring higher degree polynomials Algebra 2 Khan Academy.mp3 | 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 f... |
Factoring higher degree polynomials Algebra 2 Khan Academy.mp3 | 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. And I'll give you... |
Factoring higher degree polynomials Algebra 2 Khan Academy.mp3 | 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 p... |
Factoring higher degree polynomials Algebra 2 Khan Academy.mp3 | 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... |
Factoring higher degree polynomials Algebra 2 Khan Academy.mp3 | 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 th... |
Factoring higher degree polynomials Algebra 2 Khan Academy.mp3 | 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 squa... |
Factoring higher degree polynomials Algebra 2 Khan Academy.mp3 | 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. So let's factor out a six here. So plus six times x minus four. |
Zeros of polynomials matching equation to graph Polynomial graphs Algebra 2 Khan Academy.mp3 | We are asked, what could be the equation of p? And we have the graph of our polynomial p right over here. You could view this as the graph of y is equal to p of x. So pause this video and see if you can figure that out. All right, now let's work on this together. And you can see that all the choices have p of x in fact... |
Zeros of polynomials matching equation to graph Polynomial graphs Algebra 2 Khan Academy.mp3 | So pause this video and see if you can figure that out. All right, now let's work on this together. And you can see that all the choices have p of x in factored form, where it's very easy to identify the zeros, or the x values that would make our polynomial equal to zero. And we could also look at this graph, and we ca... |
Zeros of polynomials matching equation to graph Polynomial graphs Algebra 2 Khan Academy.mp3 | And we could also look at this graph, and we can see what the zeros are. This is where we're going to intersect the x-axis, also known as the x-intercepts. So you can see when x is equal to negative four, we have a zero, because our polynomial is zero there. So we know p of negative four is equal to zero. We also know ... |
Zeros of polynomials matching equation to graph Polynomial graphs Algebra 2 Khan Academy.mp3 | So we know p of negative four is equal to zero. We also know that p of, it looks like 1 1⁄2, or I could say 3⁄2, p of 3⁄2 is equal to zero. And we also know that p of three is equal to zero. So let's look for an expression where that is true. And because it's in factored form, each of the parts of the product will prob... |
Zeros of polynomials matching equation to graph Polynomial graphs Algebra 2 Khan Academy.mp3 | So let's look for an expression where that is true. And because it's in factored form, each of the parts of the product will probably make our polynomial zero for one of these zeros. So let's see, if in order for our polynomial to be equal to zero when x is equal to negative four, we probably want to have a term that h... |
Zeros of polynomials matching equation to graph Polynomial graphs Algebra 2 Khan Academy.mp3 | Or we wanna have, I should say, a product that has an x plus four in it, because x plus four is equal to zero when x is equal to negative four. Well, we have an x plus four there, and we have an x plus four there. So I'm liking choices B and D so far. Now, for this second root, we have p of 3⁄2 is equal to zero. So I w... |
Zeros of polynomials matching equation to graph Polynomial graphs Algebra 2 Khan Academy.mp3 | Now, for this second root, we have p of 3⁄2 is equal to zero. So I would look for something like x minus 3⁄2 in our product. I don't see an x minus 3⁄2 here. But as we've mentioned in other videos, you can also multiply these times constants. So if I were to multiply, let's see, to get rid of this fraction here, if I m... |
Zeros of polynomials matching equation to graph Polynomial graphs Algebra 2 Khan Academy.mp3 | But as we've mentioned in other videos, you can also multiply these times constants. So if I were to multiply, let's see, to get rid of this fraction here, if I multiply by two, this would be the same thing as, let me scroll down a little bit, same thing as 2x minus three. And you could test that out. 2x minus three is... |
Zeros of polynomials matching equation to graph Polynomial graphs Algebra 2 Khan Academy.mp3 | 2x minus three is equal to zero when x is equal to 3⁄2. And let's see, we have a 2x minus three right over there. So choice D is looking awfully good, but let's just verify it with this last one. For p of three to be equal to zero, we could have an expression like x minus three in the product because this is equal to z... |
Zeros of polynomials matching equation to graph Polynomial graphs Algebra 2 Khan Academy.mp3 | For p of three to be equal to zero, we could have an expression like x minus three in the product because this is equal to zero when x is equal to three. And we indeed have that right over there. So choice D is looking very good. When x is equal to negative four, this part of our product is equal to zero, which makes t... |
Dividing polynomials by x (no remainders) Algebra 2 Khan Academy.mp3 | All right, so if we're saying what is this top expression divided by this bottom expression, another way to think about it is what do I have to multiply? So I'm going to multiply something, I'll put that in parentheses, if I multiply that something times x, I should get x to the fourth minus two x to the third plus fiv... |
Dividing polynomials by x (no remainders) Algebra 2 Khan Academy.mp3 | Well, there's two ways that I could tackle it. One way is I could just rewrite this expression as being, and I will just make this x in yellow so I can keep track of it. I could just rewrite this as one over x times, times x to the fourth minus two x to the third plus five x, and then I can distribute the one over x. A... |
Dividing polynomials by x (no remainders) Algebra 2 Khan Academy.mp3 | And so what is that going to be equal to? Well, it's going to be equal to x to the fourth. Let me do this. X to the fourth over x minus two x to the third over x plus five x, plus five x over x. And so what are each of these going to be equal to? X to the fourth divided by x, if I have four x's that I'm multiplying tog... |
Dividing polynomials by x (no remainders) Algebra 2 Khan Academy.mp3 | X to the fourth over x minus two x to the third over x plus five x, plus five x over x. And so what are each of these going to be equal to? X to the fourth divided by x, if I have four x's that I'm multiplying together and then I divide by x, that's going to be equivalent to x to the third power. So this right over her... |
Dividing polynomials by x (no remainders) Algebra 2 Khan Academy.mp3 | So this right over here is equal to x to the third. You could also get there from your exponent properties. In the denominator, you have an x to the first power, and so you would subtract the exponents. You have the same base here, so that's x to the third. And then this part right over here, what would that equal to? ... |
Dividing polynomials by x (no remainders) Algebra 2 Khan Academy.mp3 | You have the same base here, so that's x to the third. And then this part right over here, what would that equal to? Well, it's going to be minus two x to the third divided by x to the first. Well, by the same property, that's going to be x squared. And then last but not least, if you take five x's and then you divide ... |
Dividing polynomials by x (no remainders) Algebra 2 Khan Academy.mp3 | Well, by the same property, that's going to be x squared. And then last but not least, if you take five x's and then you divide by x, you are just going to be left with five. And you can verify that this indeed, if I were to multiply it by x, I'm going to get x to the fourth minus two x to the third plus five x. Let me... |
Dividing polynomials by x (no remainders) Algebra 2 Khan Academy.mp3 | Let me do that. If I put x to the third minus two x squared plus five times x, what I can do is distribute the x. X times x to the third is x to the fourth. X times negative two x squared is negative two x to the third. X times five is five x. Now I mentioned there's two ways that I could do it. Another way that I coul... |
Dividing polynomials by x (no remainders) Algebra 2 Khan Academy.mp3 | X times five is five x. Now I mentioned there's two ways that I could do it. Another way that I could try to tackle it is I could look at this numerator and try to factor an x out. I would try to factor out whatever I see in the denominator. So if I do that, actually let me just rewrite the numerator. So I can rewrite ... |
Dividing polynomials by x (no remainders) Algebra 2 Khan Academy.mp3 | I would try to factor out whatever I see in the denominator. So if I do that, actually let me just rewrite the numerator. So I can rewrite x to the fourth as x times x to the third. And then I can rewrite the minus two x to the third as, let me write it this way, as plus x times negative two x squared. And then I could... |
Dividing polynomials by x (no remainders) Algebra 2 Khan Academy.mp3 | And then I can rewrite the minus two x to the third as, let me write it this way, as plus x times negative two x squared. And then I could write this five x as being equal to plus x times five. And then I'm gonna divide everything by x. Divide everything by x. I just rewrote the numerator here, but for each of those te... |
Dividing polynomials by x (no remainders) Algebra 2 Khan Academy.mp3 | And now I can factor out x out of the whole thing. So I sometimes think of factoring out an x out of the whole thing is reverse distributive property. So if I factor out this x out of every term, what am I left with? I'm left with an x times x to the third minus two x squared plus five. I ended up doing that in the sam... |
Dividing polynomials by x (no remainders) Algebra 2 Khan Academy.mp3 | I'm left with an x times x to the third minus two x squared plus five. I ended up doing that in the same color, but hopefully you're following. Plus five. And then all of that is divided by x. And as long as x does not equal zero, x divided by x is going to be equal to one, and we're left with what we had to begin with... |
Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3 | 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... |
Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3 | 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 is... |
Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3 | 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. Otherwis... |
Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3 | 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 sq... |
Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3 | 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 t... |
Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3 | 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... |
Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3 | 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 ... |
Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3 | 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... |
Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3 | 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. |
Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3 | 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. |
Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3 | 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... |
Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3 | 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. S... |
Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3 | 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. |
Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3 | 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. |
Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3 | 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. |
Extraneous solutions Equations Algebra 2 Khan Academy.mp3 | But just as a bit of a refresher, it's the idea that you do a bunch of legitimate algebraic operations, you get a solution or some solutions at the end, but then when you test it in the original equation, it doesn't satisfy the original equation. And so the key of this video is why do extraneous solutions even occur? A... |
Extraneous solutions Equations Algebra 2 Khan Academy.mp3 | There's certain operations in algebra that you can do in one direction, but you can't, and it'll always be true in one direction, but it isn't always true in the other direction. And I'll show you those two operations. One is squaring, and the other is multiplying both sides by a variable expression. So let's see the e... |
Extraneous solutions Equations Algebra 2 Khan Academy.mp3 | So let's see the example of squaring, and then we're going to see it in an actual scenario where you're dealing with an extraneous solution. So we know, for example, that if a is equal to b, I could square both sides, and then a squared is going to be equal to b squared. But the other way is not true. For example, if a... |
Extraneous solutions Equations Algebra 2 Khan Academy.mp3 | For example, if a squared is equal to b squared, it is not always the case that a is equal to b. What's an example that shows that this is not always the case, but actually pause the video, try to think about it. Well, negative two squared is indeed equal to two squared, but negative two is not equal to two. So this sh... |
Extraneous solutions Equations Algebra 2 Khan Academy.mp3 | So this shows that you can square both sides of an equation and deduce something that is true, but the other way around is not necessarily going to be true. Another non-reversible operation sometimes is multiplying both sides by a variable expression. So multiplying both sides, actually, that got us confused, it looks ... |
Extraneous solutions Equations Algebra 2 Khan Academy.mp3 | Multiply both sides by variable. I'll just write variable, but it could be a variable expression as well. For example, we know that if a is equal to b, that if we multiply both sides by a variable, that's still going to be true. Xa is going to be equal to xb. But the other, the reverse, isn't always the case. If xa is ... |
Extraneous solutions Equations Algebra 2 Khan Academy.mp3 | Xa is going to be equal to xb. But the other, the reverse, isn't always the case. If xa is equal to xb, is it always the case that a is equal to b? Well, the simple answer is no, and I always encourage you, pause this video and see if you can find an example where this doesn't work. Well, what if a was two and b is thr... |
Extraneous solutions Equations Algebra 2 Khan Academy.mp3 | Well, the simple answer is no, and I always encourage you, pause this video and see if you can find an example where this doesn't work. Well, what if a was two and b is three and the variable x just happened to take on the value zero? So we know that zero times two is indeed equal to zero times three, but two is not eq... |
Extraneous solutions Equations Algebra 2 Khan Academy.mp3 | Now, how does all of this connect to the extraneous solutions you've seen when you're solving radical equations or when you're solving some rational or equations with rational expressions on both sides? Well, let's look at an example. Let's solve a radical equation. If I wanted to solve the equation, the square root of... |
Extraneous solutions Equations Algebra 2 Khan Academy.mp3 | If I wanted to solve the equation, the square root of five x minus four is equal to x minus two. A typical first step is, hey, let's get rid of this radical by squaring both sides. So I am going to square both sides, and then I'm going to get five x minus four is equal to x squared minus four x plus four. Once again, i... |
Extraneous solutions Equations Algebra 2 Khan Academy.mp3 | Once again, if this looks completely unfamiliar to you, we go into much more depth in other videos where we introduce the idea of radical equations. And let's see, we can subtract five x from both sides. We can add four to both sides. I'm just trying to get a zero on the left-hand side. And so I'm going to be left with... |
Extraneous solutions Equations Algebra 2 Khan Academy.mp3 | I'm just trying to get a zero on the left-hand side. And so I'm going to be left with zero is equal to x squared minus nine x plus eight, or zero is equal to x minus eight times x minus one, or we could say that x minus one is equal to zero, or x minus eight is equal to zero. We get x equals one or x equals eight. So l... |
Extraneous solutions Equations Algebra 2 Khan Academy.mp3 | So let's test these solutions. If x equals eight, we would get, and I'll color-code it a little bit. For x equals eight, if I test it in the original equation, I get the square root of 36 is equal to six, which is absolutely true, so that one works. But what about x equals one? I get the square root of five times one m... |
Extraneous solutions Equations Algebra 2 Khan Academy.mp3 | But what about x equals one? I get the square root of five times one minus four is one is equal to one minus two, which is equal to negative one. That did not work. This right over here is an extraneous solution. If someone said, what are all the x values that satisfy this equation, you would not say x equals one, even... |
Extraneous solutions Equations Algebra 2 Khan Academy.mp3 | This right over here is an extraneous solution. If someone said, what are all the x values that satisfy this equation, you would not say x equals one, even though you got there with legitimate algebraic steps. And the reason that is true is, actually, pause this video. Look back. For which of these steps does x equal o... |
Extraneous solutions Equations Algebra 2 Khan Academy.mp3 | Look back. For which of these steps does x equal one still work, and what step does it not work? Well, you'll see that x equals one works for all of these equations below this purple line. It just doesn't work for the square root of five minus four x is equal to x minus two. In fact, you could start with x minus one, a... |
Extraneous solutions Equations Algebra 2 Khan Academy.mp3 | It just doesn't work for the square root of five minus four x is equal to x minus two. In fact, you could start with x minus one, and then you could deduce all the way up to this line here. But the issue here is that squaring is not a reversible operation. This is analogous to saying, hey, we know that a squared is equ... |
Extraneous solutions Equations Algebra 2 Khan Academy.mp3 | This is analogous to saying, hey, we know that a squared is equal to b squared. We know that this is equal to this, but then that doesn't mean that a is necessarily equal to b for x equals one. And we could do the same thing with a rational or an equation that deals with rational expressions. So, for example, we might ... |
Extraneous solutions Equations Algebra 2 Khan Academy.mp3 | So, for example, we might have to deal with, and let me make sure I have some space here. If I had to solve x squared over x minus one is equal to one over x minus one, the first thing I might wanna do is multiply both sides by x minus one. So multiply, multiply by x minus one. Now, notice, I'm multiplying both sides b... |
Extraneous solutions Equations Algebra 2 Khan Academy.mp3 | Now, notice, I'm multiplying both sides by a variable expression, so we have to be a little bit conscientious now, but if I multiply both sides by x minus one, I'm going to get x squared is equal to one, or I could say that x equals one, or x is equal to negative one. But we could test these for x equals one. If I go u... |
Extraneous solutions Equations Algebra 2 Khan Academy.mp3 | The key here is that we multiplied both sides by a variable expression. In this case, we multiplied both sides by x minus one. You can do that. You can multiply both sides by a variable expression, and it is a legitimate algebraic operation. It's completely analogous to what we saw right over here. Just because zero ti... |
Extraneous solutions Equations Algebra 2 Khan Academy.mp3 | You can multiply both sides by a variable expression, and it is a legitimate algebraic operation. It's completely analogous to what we saw right over here. Just because zero times two is equal to zero times three does not mean that two is equal to three. It's completely analogous because we multiplied by a variable exp... |
Simplifying quotient of powers (rational exponents) Algebra I High School Math Khan Academy.mp3 | 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 usi... |
Simplifying quotient of powers (rational exponents) Algebra I High School Math Khan Academy.mp3 | 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 realiza... |
Simplifying quotient of powers (rational exponents) Algebra I High School Math Khan Academy.mp3 | 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 one over x to the b, which is the sam... |
Simplifying quotient of powers (rational exponents) Algebra I High School Math Khan Academy.mp3 | 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 yo... |
Simplifying quotient of powers (rational exponents) Algebra I High School Math Khan Academy.mp3 | 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... |
Simplifying quotient of powers (rational exponents) Algebra I High School Math Khan Academy.mp3 | 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 e... |
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