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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here .
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so could the quadratic formula replace factoring overall ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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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 . and now notice , if this is plus and we use this minus sign , the plus will become negative and the negative will become positive .
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area of a rectangle is 225m^2 and perimeter is 60 m way is the length ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here .
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you know , whenever c is zero and you use the quadratic formula , does n't that automatically make the origin ( 0 , 0 ) a zero for that function ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here .
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while the quadratic formula can be used for quadratic equations , can it be used in quadratic inequalities ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here .
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can you use the quadratic formula to find an output of a number that is n't zero ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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that 's nice . that '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 .
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with the quadratic equation why do n't i find the correct x values when i try to solve it with the perfect square approach ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here .
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when graphing quadratic equations and also using the formula we look for x , would there be a formula for finding the y intercepts ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here .
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what is a real-life situation where someone would need to know the quadratic formula ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things .
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does the +/- indicate an ( and ) relationship or an ( or ) relationship ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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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 solutions , we 're taking the square root of a negative number .
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i 'm pretty rust on my math , in the second example with no real solutions , is the imaginary solution -1 +/- 2i sqrt ( 21 ) ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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let 's rewrite the formula again , just in case we have n'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 . i 'll supply this to another problem .
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could n't you ( or why could n't you ) simplify part of the formula from 'plus or minus the square root of b squared minus 4ac ' to just 'b minus 4ac ' ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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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 .
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does n't the square root cancel out the extra b ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here .
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why do we need the quadratic formula ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 .
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is there a difference if i apply the ruffini rule ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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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 .
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i.e -4.9 ( x-4 ) +50 = 0 then how do we expand and plug in quadratic ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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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 .
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in specfically where did -3 come from ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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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 is just 2 .
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and what happened to the 2 in the next step from 2 times the sqaure root of 39 ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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we get x , this tells us that x is going to be equal to negative b . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ?
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when putting an already negative number into a negative variable in the formula , do you make the number positive or keep it negative ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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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 .
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for example the quadratic 3x^2-3x-5 , when putting -3 in to the negative -b variable in the quadratic formula , do you make it 3 or -3 ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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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 the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ?
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for example , if i have the equation f ( x ) =x^2-18x+86 , when i plug it into the quadratic formula with a negative b , do i change my b to a postive because it is already negative , or does it have to be a negative either way ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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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 , you know , maybe not so obvious to factor .
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what is the meaning of having two or more answers for one equation or one expression ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here .
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when should you use the quadratic formula ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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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 is just 2 .
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@ 13.37 what happened to + - 2 before square root 39 ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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b squared is 16 , right ? 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 with that -- since this is the first time we 're doing it , let me not skip too many steps .
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i am really confused on what is happening ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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that '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 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 .
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what do i do in order to answer the equations 4b^2+8b+7=4 and 5r^2=80 ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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and the reason why it 's not giving you an answer , 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 .
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how many trees should be planted to produced a maximum number of mangoes per year ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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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 .
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i mean , the roots are the values of x for when y is zero , right ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me graph it .
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so why is it that in the quadratic , the solutions are only for when y = 0 ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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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 the negative sign in front of that -- negative b plus or minus the square root of b squared .
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when you do b ^ how come you do not add in the x ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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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 the negative sign in front of that -- negative b plus or minus the square root of b squared .
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b being 12 x , why is it not 144 x , but only 144 ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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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 is just 2 .
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why is it that when you plugged them into the formula , at the square root , could you put those numbers into parenthesis instead of using the dot ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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we get x , this tells us that x is going to be equal to negative b . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 4 squared is 16 , minus 4 times a , which is 1 , times c , which is negative 21 .
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if b is negative in the starting equation would you make it positive in when it says -b in the quadratic formula ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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well , the first thing we want to do is get it in the form where all of our terms or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here .
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how would i solve a problem like 3x^2-12=0 using the quadratic formula ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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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 .
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what happend to the +-2 behind the ( 39 ) ^1/2 when the top half was being devided by -6 ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , maybe not so obvious to factor . so let 's say we get negative 3x squared plus 12x plus 1 is equal to 0 .
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how does i know if the graphic needs to go up between the points and go down between the points ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify .
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1is 2 plus or minus the square root of 26 an ok answer ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here .
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how can you find the nature of the roots using the quadratic formula ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here .
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how can you find the nature of the roots using the quadratic formula ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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we get x , this tells us that x is going to be equal to negative b . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ?
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what if b is already negative ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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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 .
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what if when you graph the equation and very bottom part goes below the x-axis ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here .
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hi can some one tell me what previous math i need to know in oder to understand the quadratic formula ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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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 is just 2 .
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how does it come out to the square root of 2 times 2 multiplied by the square root of the 39 ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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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 .
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when sal takes out the 2 , why is the -12 also affected and reduced to -6 ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here .
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why does the quadratic formula have to be so long is n't there a simpler way ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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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 .
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for example how do you know that 6x represents the b ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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that is a , this is b and this right here 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 .
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how do you find the explicit/recursive equation for a quadratic function ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here .
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in `` quadratic formula '' practice problems , why does it seem that the coefficient on radicals is ignored ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things .
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what is the dimension of the largest total area martin can enclose ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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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 is just 2 . 2 square roots of 39 , if i did that properly , let 's see , 4 times 39 .
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in the last few minutes , how was the 2 before the square root removed ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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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 .
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considering the way quadratic equations are written , when adding 10 to the left side if i were to replace 3x^2 with 10 , make the b term 3x^2 and make the c term 6x , would that affect the solution of the answer ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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it 's worthless . it just gives me a square root of a negative number . it 's not giving me an answer .
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what happens when u have a negetive number under the sq root ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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we get x , this tells us that x is going to be equal to negative b . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ?
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how does the b work if it is already a negative ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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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 .
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when you have x=-b ... .if the b is already a negative then is it x=b ... , or is it always x=-b ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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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 , right ? that 's what the plus or minus means , it could be this or that or both of them , really .
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is our final answer 2 plus/minus the square root of 39 over 3 ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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so negative 21 , just so you can see how it fit in , and then all of that over 2a . a is 1 , so all of that over 2 . so what does this simplify , or hopefully it simplifies ?
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this also begs another question ... does a second order polynomial really descend by 1/2 or does it de-increment by 1 ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify .
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what is the positive root in an equation ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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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 . i 'll supply this to another problem . let 's say we have the equation 3x squared plus 6x is equal to negative 10 .
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when doing a quadratic formula problem with a negative in the radical , how do you solve such problem ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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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 .
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3x^2=39 what formula is used to solve the equation ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 4 squared is 16 , minus 4 times a , which is 1 , times c , which is negative 21 .
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is n't there a small mistake at 1?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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we get x , this tells us that x is going to be equal to negative b . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ?
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wait ... if the formula says b^2 minus 4ac , why did mr sal put an addition sign ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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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 .
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does the sign change for x ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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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 .
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how come we are solving for x ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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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 ?
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could you have just did negative 12 divided by negative 6 and got 2 plus or minus square root of 39 ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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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 is just 2 .
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why not just find the square root of 156 to make things much easier ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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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 .
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would n't using the slope intercept form `` y=mx+b '' and putting that in your graphing calculator be a better way to solve ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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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 is just 2 .
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what is the square root of 11,390,881 ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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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 , you know , maybe not so obvious to factor .
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what exactly is the concept of `` one repeated solution '' ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here .
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what is the use of the quadratric formula in the sat ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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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 .
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what would you get for 2x-5=-x squared ?
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in this video , i 'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics . and if you 've seen many of my videos , you know that i 'm not a big fan of memorizing things . but i will recommend you memorize it with the caveat that you also remember how to prove it , because i do n'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 zeroes of quadratic equations . so let 's speak in very general terms and 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 and then c , is , you could imagine , the coefficient on the x to the zero 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 . and 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 and that is the quadratic formula , right there . so let 's apply it to some problems . 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 -- let me do that in a different color -- a is equal to 1 , right ? 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 . negative b is negative 4 -- i put the negative sign in front of that -- negative b plus or minus the square root of b squared . b squared is 16 , right ? 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 with that -- since 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 2a . 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 -- let 's see we have a negative times a negative , that 's going to give us a positive . and we had 16 plus , let 's see this is 6 , 4 times 1 is 4 times 21 is 84 . 16 plus 84 is 100 . that 's nice . that '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 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 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 . 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 . and let 's do a couple of those , let 's do some hard-to-factor problems right now . so let 's scroll down to get some fresh real estate . let 's rewrite the formula again , just in case we have n'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 . i 'll supply 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 or on the left-hand side , so let 's add 10 to both sides of this equation . we get 3x squared plus the 6x plus 10 is equal to 0 . and now we can use a quadratic formula . so let 's apply it here . so a is equal to 3 . that is a , this is b and this right here 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 . let 's stretch out the radical 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 -- this is interesting -- minus 4 times 3 times 10 . so this is 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 ? 36 minus 120 is what ? that 's 84 . we make this into a 10 , 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 . we have 36 minus 120 . it 's going to be negative 84 all of that 6 . so you might say , gee , this is crazy . what a this silly quadratic formula you 're introducing me to , sal ? it 's worthless . it just gives 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 , 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 solutions , 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 graphic calculator out and let 's graph this equation right here . so , let 's get the graphs that y is equal to -- that 's what i had there before -- 3x squared plus 6x plus 10 . so that 's the equation and we 're going to see where it intersects the x-axis . where does it equal 0 ? so let me 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 . let 's do one more example , you can never see enough examples here . and i want to do ones that are , you know , 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 , 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 ? it 's a negative times a negative so they cancel out . so i have 144 plus 12 , so that is 156 , right ? 144 plus 12 , all of that over negative 6 . now , i suspect we can simplify this 156 . we could 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 . 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 is 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 . 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 maybe 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 ? 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 does n't matter , right ? we could say minus or plus , that 's the same thing as plus or minus the square root of 39 nine over 3 . i think that 's about as simple as we can get this answered . i want to make a very clear point of what i did that last step . i did not forget about this negative sign . i just said it does n'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 , right ? 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 , right ? 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 , right ? a negative times a negative is a positive . so once again , you have 2 plus or minus the square of 39 over 3 . 2 plus or minus the square root of 39 over 3 are solutions to this equation right there . let verify . i 'm just curious what the graph looks like . so let 's just look at it . let me clear this . where is the clear button ? so we have negative 3 three 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 , you see -- 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 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 2 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 -- it looks close to 0 but maybe a little bit less than that . so anyway , hopefully you found this application of the quadratic formula helpful .
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b squared is 16 , right ? 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 with that -- since this is the first time we 're doing it , let me not skip too many steps .
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why are all the comments from 4 years ago ?
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here 's a simplified diagram for the electrolysis of molten sodium chloride . so if you melt solid sodium chloride , you get molten sodium chloride . so you get sodium ions , liquid sodium ions , and you get liquid chloride anions . so that 's what we have here , we have sodium ions and chloride anions . remember , in an electrolytic cell , the negative terminal of the battery delivers electrons . in this case , delivers electrons to the electrode on the right . and when those liquid sodium ions come in contact with those electrons , we get a reduction half-reaction , so let me write , this is a reduction half-reaction . sodium ions gain electrons , and are reduced to form liquid sodium metal . so liquid sodium metal is one of our products . and this would form at the cathode . remember , reduction occurs at the cathode . let me draw in some liquid sodium metal here forming on the electrode on the right , forming on the cathodes . the other half-reaction , at our other electrode , we know that our battery draws electrons away from this electrode . so oxidation is occurring at this electrode . this would be the anode . so chloride anions , liquid chloride anions are oxidized to chlorine gas , so we lose two electrons . so chloride anions are oxidized to chlorine gas at the anode , and so we 'd have bubbles of chlorine gas forming at this electrode . this is an important reaction in industry . so to form sodium metal , this is a good way to do it . how long does it take to produce 1.00 times 10 to the third kilograms of sodium , so that 's a lot of sodium , with a constant current of 3.00 times 10 to the fourth amps . so we have another quantitative electrolysis problem here . this one 's a little bit harder than the one we did before , but we 're going to start the same way . we 're going to start with the definition of current . so , current is equal to charge over time , so q over t , where charge is in coulombs and time is in seconds . so right now we know the current is three times 10 to the fourth amps . so 3.00 times 10 to the fourth , is equal to the charge over the time . we want to know the time . we want to know how long it takes to make that much sodium . so we 're trying to find the time . if we could find the charge , then we could find the time using this equation . so let 's do that , let 's find the charge , and we know we can start with the mass of sodium here . so we 're given 1.00 times 10 to the third kilograms , we need to get that into grams . so what is 1.00 times 10 to the third kilograms in grams ? so of course , 1.00 times 10 to the sixth grams . next , we can find the moles of sodium that we 're trying to produce . so if we have grams of sodium , this is grams of sodium , how do we figure out moles of sodium ? we could divide by the molar mass , so we divide by the molar mass of sodium , which is 22.99 , and that would be grams per mole . if we divide by the molar mass , the grams cancel out , and we would get moles of sodium . so let 's get out the calculator . so we have 1.00 times 10 to the sixth , that 's how many grams of sodium we have . if we divide that by 22.99 , the molar mass of sodium , we get , this would be , let 's see , 4.35 times 10 to the one , two , three , four . so 4.35 times 10 to the fourth , let 's write that down , 4.35 times 10 to the fourth moles of sodium . so how does that help us ? if we have moles of sodium , how does that help us find our charge ? well , we can relate the moles of sodium to the moles of electrons . so let 's go back up here to our half-reaction where we have our moles of sodium . so right here , so we 're trying to make 4.35 times 10 to the fourth moles of sodium . so how many electrons do we need to do that ? well , the mole ratio is two to two , so you need the same number of moles of electrons , you need 4.35 times 10 to the fourth moles of electrons , and that helps us out . so let me go back down here , we have some more room , and let me write down 4.35 . 4.35 times 10 to the fourth , so because of our mole ratios we know this is the same number of moles of electrons that we need . we 're trying to find charge , we 're trying to find this charge here , and we know we can go from moles of electrons to the total charge using faraday 's constant . faraday 's constant is the charge carried by one mole of electrons . and we know that 's 96,500 . so there 's a charge of 96,500 coulombs for every one mole of electrons . so if we multiply these together , we 'll get charge , because the moles of electrons will cancel , and we will get the charge , so let 's do that . we have 4.35 times 10 to the fourth , and we 're going to multiply that by faraday 's constant , which is ... so , times , 96,500 , and so we get let 's see , that 'd be 4.2 times 10 to the , let 's see , one , two , three , four , five , six , seven , eight , nine , so 4.20 times 10 to the ninth , let 's write that down . 4.20 times 10 to the ninth , that would be coulombs , right ? that would be coulombs , that 's how much charge we need to make this many moles of sodium . so we have the charge , and so now we can plug that back into here , and solve for the time . so let 's do that . we have 3.00 times 10 to the fourth , that was our current , and current is equal to charge over time . so our total charge is 4.20 times 10 to the ninth , and that 's all over the time . so to solve for the time , we just need to divide this number , so let 's do that , so that 's 4.20 times 10 to the ninth , we 're going to divide that number by 3.00 times 10 to the fourth , and that should give us our time in seconds , so our time in seconds this would be 1.4 times 10 to the one , two , three , four , five , so 1.4 times 10 to the fifth . so the time in seconds is 1.40 times 10 to the fifth . so we have our time , i guess we could leave the answer like that . but let 's convert that to something that 's a little bit easier to comprehend . first , let 's convert that into minutes . so if you have 1.40 times 10 to the fifth seconds , how many minutes is that ? we could just divide that by 60 . so if we divide that by 60 , how many minutes ? that 's 2,333 minutes . how many hours is that ? we could just divide that by 60 , so we could figure out how many hours , so 38.9 hours , approximately , so it would take us approximately 38.9 hours so it would take us 38.9 hours to make , let 's go back up here to remind ourselves how much sodium to make 1.00 times 10 to the third kilograms of sodium . we can also figure out how many liters of chlorine gas are produced when we make our sodium . so let 's say we 're at stp here , so standard temperature and pressure . remember , standard temperature is zero degrees c , and standard pressure is one atmosphere . so we 've already found that we 're trying to make 4.35 times 10 to the fourth moles of sodium . so how many moles of chlorine gas will we make ? well , look at our mole ratios here . so this would be a two to one mole ratio . so we can set up a proportion , so we could do sodium to chlorine , so this would be a two to one mole ratio , so for every two moles of sodium that are produced , one mole of chlorine gas is produced . so 2/1 is equal to 4.35 times 10 to the fourth , that 's how many moles of sodium we were making , over x , where x represents the moles of chlorine that we would make . so two x is equal to 4.35 times 10 to the fourth , so x is equal to 4.35 times 10 to the fourth divided by two . so if we find 4.35 times 10 to the fourth , we need to divide that by two , and we get 21,750 moles of chlorine , so this would be 21,750 moles of chlorine gas that are produced . we 're trying to find liters of chlorine gas that are produced , so one thing we could do , just to make our lives easy , would be to say , we know that one mole of an ideal gas occupies 22.4 liters at stp . so if we have this many moles , and we assume this is an ideal gas , we could find out how many liters that is by multiplying that number by 22.4 . so 21,750 moles ... if we multiply that by 22.4 , then we should get the volume in liters . so let 's do it , let 's take that number and multiply it by 22.4 . and we get 4.87 , times 10 to the one , two , three four , five , so we get 4.87 times 10 to the fifth liters of chlorine . so there 's another way to do it , if you do n't want to take the shortcut , you can actually plug everything into pv is equal to nrt , so you could use piv-nert , so pv is equal to nrt , we 're at stp , so standard pressure is one atmosphere , and temperature is zero degrees c , or 273.15 kelvin . so we can plug those in , so our pressure is one atmosphere , one atmosphere , our temperature would be 273.15 , so 273.15 kelvin . we 're trying to find the volume , we 're trying to find liters , so we 're trying to find v , n would be equal to the number of moles , so that would be 21,750 , so we have 21,750 moles . r , r is equal to .0821 ... and i did n't leave room for the units in there . but the units would cancel out to give us liters for the volume , and if you solve this , if you use your calculator to solve this , you will get the same answer . you will get 4.87 times 10 to the fifth liters , so you plug it all into piv-nert or you could use the shortcut way . either way , you 're going to get the same volume of chlorine produced .
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and when those liquid sodium ions come in contact with those electrons , we get a reduction half-reaction , so let me write , this is a reduction half-reaction . sodium ions gain electrons , and are reduced to form liquid sodium metal . so liquid sodium metal is one of our products .
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why do sodium ions gain two electrons instead of one ?
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here 's a simplified diagram for the electrolysis of molten sodium chloride . so if you melt solid sodium chloride , you get molten sodium chloride . so you get sodium ions , liquid sodium ions , and you get liquid chloride anions . so that 's what we have here , we have sodium ions and chloride anions . remember , in an electrolytic cell , the negative terminal of the battery delivers electrons . in this case , delivers electrons to the electrode on the right . and when those liquid sodium ions come in contact with those electrons , we get a reduction half-reaction , so let me write , this is a reduction half-reaction . sodium ions gain electrons , and are reduced to form liquid sodium metal . so liquid sodium metal is one of our products . and this would form at the cathode . remember , reduction occurs at the cathode . let me draw in some liquid sodium metal here forming on the electrode on the right , forming on the cathodes . the other half-reaction , at our other electrode , we know that our battery draws electrons away from this electrode . so oxidation is occurring at this electrode . this would be the anode . so chloride anions , liquid chloride anions are oxidized to chlorine gas , so we lose two electrons . so chloride anions are oxidized to chlorine gas at the anode , and so we 'd have bubbles of chlorine gas forming at this electrode . this is an important reaction in industry . so to form sodium metal , this is a good way to do it . how long does it take to produce 1.00 times 10 to the third kilograms of sodium , so that 's a lot of sodium , with a constant current of 3.00 times 10 to the fourth amps . so we have another quantitative electrolysis problem here . this one 's a little bit harder than the one we did before , but we 're going to start the same way . we 're going to start with the definition of current . so , current is equal to charge over time , so q over t , where charge is in coulombs and time is in seconds . so right now we know the current is three times 10 to the fourth amps . so 3.00 times 10 to the fourth , is equal to the charge over the time . we want to know the time . we want to know how long it takes to make that much sodium . so we 're trying to find the time . if we could find the charge , then we could find the time using this equation . so let 's do that , let 's find the charge , and we know we can start with the mass of sodium here . so we 're given 1.00 times 10 to the third kilograms , we need to get that into grams . so what is 1.00 times 10 to the third kilograms in grams ? so of course , 1.00 times 10 to the sixth grams . next , we can find the moles of sodium that we 're trying to produce . so if we have grams of sodium , this is grams of sodium , how do we figure out moles of sodium ? we could divide by the molar mass , so we divide by the molar mass of sodium , which is 22.99 , and that would be grams per mole . if we divide by the molar mass , the grams cancel out , and we would get moles of sodium . so let 's get out the calculator . so we have 1.00 times 10 to the sixth , that 's how many grams of sodium we have . if we divide that by 22.99 , the molar mass of sodium , we get , this would be , let 's see , 4.35 times 10 to the one , two , three , four . so 4.35 times 10 to the fourth , let 's write that down , 4.35 times 10 to the fourth moles of sodium . so how does that help us ? if we have moles of sodium , how does that help us find our charge ? well , we can relate the moles of sodium to the moles of electrons . so let 's go back up here to our half-reaction where we have our moles of sodium . so right here , so we 're trying to make 4.35 times 10 to the fourth moles of sodium . so how many electrons do we need to do that ? well , the mole ratio is two to two , so you need the same number of moles of electrons , you need 4.35 times 10 to the fourth moles of electrons , and that helps us out . so let me go back down here , we have some more room , and let me write down 4.35 . 4.35 times 10 to the fourth , so because of our mole ratios we know this is the same number of moles of electrons that we need . we 're trying to find charge , we 're trying to find this charge here , and we know we can go from moles of electrons to the total charge using faraday 's constant . faraday 's constant is the charge carried by one mole of electrons . and we know that 's 96,500 . so there 's a charge of 96,500 coulombs for every one mole of electrons . so if we multiply these together , we 'll get charge , because the moles of electrons will cancel , and we will get the charge , so let 's do that . we have 4.35 times 10 to the fourth , and we 're going to multiply that by faraday 's constant , which is ... so , times , 96,500 , and so we get let 's see , that 'd be 4.2 times 10 to the , let 's see , one , two , three , four , five , six , seven , eight , nine , so 4.20 times 10 to the ninth , let 's write that down . 4.20 times 10 to the ninth , that would be coulombs , right ? that would be coulombs , that 's how much charge we need to make this many moles of sodium . so we have the charge , and so now we can plug that back into here , and solve for the time . so let 's do that . we have 3.00 times 10 to the fourth , that was our current , and current is equal to charge over time . so our total charge is 4.20 times 10 to the ninth , and that 's all over the time . so to solve for the time , we just need to divide this number , so let 's do that , so that 's 4.20 times 10 to the ninth , we 're going to divide that number by 3.00 times 10 to the fourth , and that should give us our time in seconds , so our time in seconds this would be 1.4 times 10 to the one , two , three , four , five , so 1.4 times 10 to the fifth . so the time in seconds is 1.40 times 10 to the fifth . so we have our time , i guess we could leave the answer like that . but let 's convert that to something that 's a little bit easier to comprehend . first , let 's convert that into minutes . so if you have 1.40 times 10 to the fifth seconds , how many minutes is that ? we could just divide that by 60 . so if we divide that by 60 , how many minutes ? that 's 2,333 minutes . how many hours is that ? we could just divide that by 60 , so we could figure out how many hours , so 38.9 hours , approximately , so it would take us approximately 38.9 hours so it would take us 38.9 hours to make , let 's go back up here to remind ourselves how much sodium to make 1.00 times 10 to the third kilograms of sodium . we can also figure out how many liters of chlorine gas are produced when we make our sodium . so let 's say we 're at stp here , so standard temperature and pressure . remember , standard temperature is zero degrees c , and standard pressure is one atmosphere . so we 've already found that we 're trying to make 4.35 times 10 to the fourth moles of sodium . so how many moles of chlorine gas will we make ? well , look at our mole ratios here . so this would be a two to one mole ratio . so we can set up a proportion , so we could do sodium to chlorine , so this would be a two to one mole ratio , so for every two moles of sodium that are produced , one mole of chlorine gas is produced . so 2/1 is equal to 4.35 times 10 to the fourth , that 's how many moles of sodium we were making , over x , where x represents the moles of chlorine that we would make . so two x is equal to 4.35 times 10 to the fourth , so x is equal to 4.35 times 10 to the fourth divided by two . so if we find 4.35 times 10 to the fourth , we need to divide that by two , and we get 21,750 moles of chlorine , so this would be 21,750 moles of chlorine gas that are produced . we 're trying to find liters of chlorine gas that are produced , so one thing we could do , just to make our lives easy , would be to say , we know that one mole of an ideal gas occupies 22.4 liters at stp . so if we have this many moles , and we assume this is an ideal gas , we could find out how many liters that is by multiplying that number by 22.4 . so 21,750 moles ... if we multiply that by 22.4 , then we should get the volume in liters . so let 's do it , let 's take that number and multiply it by 22.4 . and we get 4.87 , times 10 to the one , two , three four , five , so we get 4.87 times 10 to the fifth liters of chlorine . so there 's another way to do it , if you do n't want to take the shortcut , you can actually plug everything into pv is equal to nrt , so you could use piv-nert , so pv is equal to nrt , we 're at stp , so standard pressure is one atmosphere , and temperature is zero degrees c , or 273.15 kelvin . so we can plug those in , so our pressure is one atmosphere , one atmosphere , our temperature would be 273.15 , so 273.15 kelvin . we 're trying to find the volume , we 're trying to find liters , so we 're trying to find v , n would be equal to the number of moles , so that would be 21,750 , so we have 21,750 moles . r , r is equal to .0821 ... and i did n't leave room for the units in there . but the units would cancel out to give us liters for the volume , and if you solve this , if you use your calculator to solve this , you will get the same answer . you will get 4.87 times 10 to the fifth liters , so you plug it all into piv-nert or you could use the shortcut way . either way , you 're going to get the same volume of chlorine produced .
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if we have moles of sodium , how does that help us find our charge ? well , we can relate the moles of sodium to the moles of electrons . so let 's go back up here to our half-reaction where we have our moles of sodium .
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can you just use the equation moles = current * time / faraday 's constant * moles electrons for the kind of problem presented around 3 ?
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here 's a simplified diagram for the electrolysis of molten sodium chloride . so if you melt solid sodium chloride , you get molten sodium chloride . so you get sodium ions , liquid sodium ions , and you get liquid chloride anions . so that 's what we have here , we have sodium ions and chloride anions . remember , in an electrolytic cell , the negative terminal of the battery delivers electrons . in this case , delivers electrons to the electrode on the right . and when those liquid sodium ions come in contact with those electrons , we get a reduction half-reaction , so let me write , this is a reduction half-reaction . sodium ions gain electrons , and are reduced to form liquid sodium metal . so liquid sodium metal is one of our products . and this would form at the cathode . remember , reduction occurs at the cathode . let me draw in some liquid sodium metal here forming on the electrode on the right , forming on the cathodes . the other half-reaction , at our other electrode , we know that our battery draws electrons away from this electrode . so oxidation is occurring at this electrode . this would be the anode . so chloride anions , liquid chloride anions are oxidized to chlorine gas , so we lose two electrons . so chloride anions are oxidized to chlorine gas at the anode , and so we 'd have bubbles of chlorine gas forming at this electrode . this is an important reaction in industry . so to form sodium metal , this is a good way to do it . how long does it take to produce 1.00 times 10 to the third kilograms of sodium , so that 's a lot of sodium , with a constant current of 3.00 times 10 to the fourth amps . so we have another quantitative electrolysis problem here . this one 's a little bit harder than the one we did before , but we 're going to start the same way . we 're going to start with the definition of current . so , current is equal to charge over time , so q over t , where charge is in coulombs and time is in seconds . so right now we know the current is three times 10 to the fourth amps . so 3.00 times 10 to the fourth , is equal to the charge over the time . we want to know the time . we want to know how long it takes to make that much sodium . so we 're trying to find the time . if we could find the charge , then we could find the time using this equation . so let 's do that , let 's find the charge , and we know we can start with the mass of sodium here . so we 're given 1.00 times 10 to the third kilograms , we need to get that into grams . so what is 1.00 times 10 to the third kilograms in grams ? so of course , 1.00 times 10 to the sixth grams . next , we can find the moles of sodium that we 're trying to produce . so if we have grams of sodium , this is grams of sodium , how do we figure out moles of sodium ? we could divide by the molar mass , so we divide by the molar mass of sodium , which is 22.99 , and that would be grams per mole . if we divide by the molar mass , the grams cancel out , and we would get moles of sodium . so let 's get out the calculator . so we have 1.00 times 10 to the sixth , that 's how many grams of sodium we have . if we divide that by 22.99 , the molar mass of sodium , we get , this would be , let 's see , 4.35 times 10 to the one , two , three , four . so 4.35 times 10 to the fourth , let 's write that down , 4.35 times 10 to the fourth moles of sodium . so how does that help us ? if we have moles of sodium , how does that help us find our charge ? well , we can relate the moles of sodium to the moles of electrons . so let 's go back up here to our half-reaction where we have our moles of sodium . so right here , so we 're trying to make 4.35 times 10 to the fourth moles of sodium . so how many electrons do we need to do that ? well , the mole ratio is two to two , so you need the same number of moles of electrons , you need 4.35 times 10 to the fourth moles of electrons , and that helps us out . so let me go back down here , we have some more room , and let me write down 4.35 . 4.35 times 10 to the fourth , so because of our mole ratios we know this is the same number of moles of electrons that we need . we 're trying to find charge , we 're trying to find this charge here , and we know we can go from moles of electrons to the total charge using faraday 's constant . faraday 's constant is the charge carried by one mole of electrons . and we know that 's 96,500 . so there 's a charge of 96,500 coulombs for every one mole of electrons . so if we multiply these together , we 'll get charge , because the moles of electrons will cancel , and we will get the charge , so let 's do that . we have 4.35 times 10 to the fourth , and we 're going to multiply that by faraday 's constant , which is ... so , times , 96,500 , and so we get let 's see , that 'd be 4.2 times 10 to the , let 's see , one , two , three , four , five , six , seven , eight , nine , so 4.20 times 10 to the ninth , let 's write that down . 4.20 times 10 to the ninth , that would be coulombs , right ? that would be coulombs , that 's how much charge we need to make this many moles of sodium . so we have the charge , and so now we can plug that back into here , and solve for the time . so let 's do that . we have 3.00 times 10 to the fourth , that was our current , and current is equal to charge over time . so our total charge is 4.20 times 10 to the ninth , and that 's all over the time . so to solve for the time , we just need to divide this number , so let 's do that , so that 's 4.20 times 10 to the ninth , we 're going to divide that number by 3.00 times 10 to the fourth , and that should give us our time in seconds , so our time in seconds this would be 1.4 times 10 to the one , two , three , four , five , so 1.4 times 10 to the fifth . so the time in seconds is 1.40 times 10 to the fifth . so we have our time , i guess we could leave the answer like that . but let 's convert that to something that 's a little bit easier to comprehend . first , let 's convert that into minutes . so if you have 1.40 times 10 to the fifth seconds , how many minutes is that ? we could just divide that by 60 . so if we divide that by 60 , how many minutes ? that 's 2,333 minutes . how many hours is that ? we could just divide that by 60 , so we could figure out how many hours , so 38.9 hours , approximately , so it would take us approximately 38.9 hours so it would take us 38.9 hours to make , let 's go back up here to remind ourselves how much sodium to make 1.00 times 10 to the third kilograms of sodium . we can also figure out how many liters of chlorine gas are produced when we make our sodium . so let 's say we 're at stp here , so standard temperature and pressure . remember , standard temperature is zero degrees c , and standard pressure is one atmosphere . so we 've already found that we 're trying to make 4.35 times 10 to the fourth moles of sodium . so how many moles of chlorine gas will we make ? well , look at our mole ratios here . so this would be a two to one mole ratio . so we can set up a proportion , so we could do sodium to chlorine , so this would be a two to one mole ratio , so for every two moles of sodium that are produced , one mole of chlorine gas is produced . so 2/1 is equal to 4.35 times 10 to the fourth , that 's how many moles of sodium we were making , over x , where x represents the moles of chlorine that we would make . so two x is equal to 4.35 times 10 to the fourth , so x is equal to 4.35 times 10 to the fourth divided by two . so if we find 4.35 times 10 to the fourth , we need to divide that by two , and we get 21,750 moles of chlorine , so this would be 21,750 moles of chlorine gas that are produced . we 're trying to find liters of chlorine gas that are produced , so one thing we could do , just to make our lives easy , would be to say , we know that one mole of an ideal gas occupies 22.4 liters at stp . so if we have this many moles , and we assume this is an ideal gas , we could find out how many liters that is by multiplying that number by 22.4 . so 21,750 moles ... if we multiply that by 22.4 , then we should get the volume in liters . so let 's do it , let 's take that number and multiply it by 22.4 . and we get 4.87 , times 10 to the one , two , three four , five , so we get 4.87 times 10 to the fifth liters of chlorine . so there 's another way to do it , if you do n't want to take the shortcut , you can actually plug everything into pv is equal to nrt , so you could use piv-nert , so pv is equal to nrt , we 're at stp , so standard pressure is one atmosphere , and temperature is zero degrees c , or 273.15 kelvin . so we can plug those in , so our pressure is one atmosphere , one atmosphere , our temperature would be 273.15 , so 273.15 kelvin . we 're trying to find the volume , we 're trying to find liters , so we 're trying to find v , n would be equal to the number of moles , so that would be 21,750 , so we have 21,750 moles . r , r is equal to .0821 ... and i did n't leave room for the units in there . but the units would cancel out to give us liters for the volume , and if you solve this , if you use your calculator to solve this , you will get the same answer . you will get 4.87 times 10 to the fifth liters , so you plug it all into piv-nert or you could use the shortcut way . either way , you 're going to get the same volume of chlorine produced .
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we can also figure out how many liters of chlorine gas are produced when we make our sodium . so let 's say we 're at stp here , so standard temperature and pressure . remember , standard temperature is zero degrees c , and standard pressure is one atmosphere .
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is it realistic to ask how much cl2 is produced at stp given molten nacl would be at a far higher temperature and/or pressure than 0oc and 1.0atm ?
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here 's a simplified diagram for the electrolysis of molten sodium chloride . so if you melt solid sodium chloride , you get molten sodium chloride . so you get sodium ions , liquid sodium ions , and you get liquid chloride anions . so that 's what we have here , we have sodium ions and chloride anions . remember , in an electrolytic cell , the negative terminal of the battery delivers electrons . in this case , delivers electrons to the electrode on the right . and when those liquid sodium ions come in contact with those electrons , we get a reduction half-reaction , so let me write , this is a reduction half-reaction . sodium ions gain electrons , and are reduced to form liquid sodium metal . so liquid sodium metal is one of our products . and this would form at the cathode . remember , reduction occurs at the cathode . let me draw in some liquid sodium metal here forming on the electrode on the right , forming on the cathodes . the other half-reaction , at our other electrode , we know that our battery draws electrons away from this electrode . so oxidation is occurring at this electrode . this would be the anode . so chloride anions , liquid chloride anions are oxidized to chlorine gas , so we lose two electrons . so chloride anions are oxidized to chlorine gas at the anode , and so we 'd have bubbles of chlorine gas forming at this electrode . this is an important reaction in industry . so to form sodium metal , this is a good way to do it . how long does it take to produce 1.00 times 10 to the third kilograms of sodium , so that 's a lot of sodium , with a constant current of 3.00 times 10 to the fourth amps . so we have another quantitative electrolysis problem here . this one 's a little bit harder than the one we did before , but we 're going to start the same way . we 're going to start with the definition of current . so , current is equal to charge over time , so q over t , where charge is in coulombs and time is in seconds . so right now we know the current is three times 10 to the fourth amps . so 3.00 times 10 to the fourth , is equal to the charge over the time . we want to know the time . we want to know how long it takes to make that much sodium . so we 're trying to find the time . if we could find the charge , then we could find the time using this equation . so let 's do that , let 's find the charge , and we know we can start with the mass of sodium here . so we 're given 1.00 times 10 to the third kilograms , we need to get that into grams . so what is 1.00 times 10 to the third kilograms in grams ? so of course , 1.00 times 10 to the sixth grams . next , we can find the moles of sodium that we 're trying to produce . so if we have grams of sodium , this is grams of sodium , how do we figure out moles of sodium ? we could divide by the molar mass , so we divide by the molar mass of sodium , which is 22.99 , and that would be grams per mole . if we divide by the molar mass , the grams cancel out , and we would get moles of sodium . so let 's get out the calculator . so we have 1.00 times 10 to the sixth , that 's how many grams of sodium we have . if we divide that by 22.99 , the molar mass of sodium , we get , this would be , let 's see , 4.35 times 10 to the one , two , three , four . so 4.35 times 10 to the fourth , let 's write that down , 4.35 times 10 to the fourth moles of sodium . so how does that help us ? if we have moles of sodium , how does that help us find our charge ? well , we can relate the moles of sodium to the moles of electrons . so let 's go back up here to our half-reaction where we have our moles of sodium . so right here , so we 're trying to make 4.35 times 10 to the fourth moles of sodium . so how many electrons do we need to do that ? well , the mole ratio is two to two , so you need the same number of moles of electrons , you need 4.35 times 10 to the fourth moles of electrons , and that helps us out . so let me go back down here , we have some more room , and let me write down 4.35 . 4.35 times 10 to the fourth , so because of our mole ratios we know this is the same number of moles of electrons that we need . we 're trying to find charge , we 're trying to find this charge here , and we know we can go from moles of electrons to the total charge using faraday 's constant . faraday 's constant is the charge carried by one mole of electrons . and we know that 's 96,500 . so there 's a charge of 96,500 coulombs for every one mole of electrons . so if we multiply these together , we 'll get charge , because the moles of electrons will cancel , and we will get the charge , so let 's do that . we have 4.35 times 10 to the fourth , and we 're going to multiply that by faraday 's constant , which is ... so , times , 96,500 , and so we get let 's see , that 'd be 4.2 times 10 to the , let 's see , one , two , three , four , five , six , seven , eight , nine , so 4.20 times 10 to the ninth , let 's write that down . 4.20 times 10 to the ninth , that would be coulombs , right ? that would be coulombs , that 's how much charge we need to make this many moles of sodium . so we have the charge , and so now we can plug that back into here , and solve for the time . so let 's do that . we have 3.00 times 10 to the fourth , that was our current , and current is equal to charge over time . so our total charge is 4.20 times 10 to the ninth , and that 's all over the time . so to solve for the time , we just need to divide this number , so let 's do that , so that 's 4.20 times 10 to the ninth , we 're going to divide that number by 3.00 times 10 to the fourth , and that should give us our time in seconds , so our time in seconds this would be 1.4 times 10 to the one , two , three , four , five , so 1.4 times 10 to the fifth . so the time in seconds is 1.40 times 10 to the fifth . so we have our time , i guess we could leave the answer like that . but let 's convert that to something that 's a little bit easier to comprehend . first , let 's convert that into minutes . so if you have 1.40 times 10 to the fifth seconds , how many minutes is that ? we could just divide that by 60 . so if we divide that by 60 , how many minutes ? that 's 2,333 minutes . how many hours is that ? we could just divide that by 60 , so we could figure out how many hours , so 38.9 hours , approximately , so it would take us approximately 38.9 hours so it would take us 38.9 hours to make , let 's go back up here to remind ourselves how much sodium to make 1.00 times 10 to the third kilograms of sodium . we can also figure out how many liters of chlorine gas are produced when we make our sodium . so let 's say we 're at stp here , so standard temperature and pressure . remember , standard temperature is zero degrees c , and standard pressure is one atmosphere . so we 've already found that we 're trying to make 4.35 times 10 to the fourth moles of sodium . so how many moles of chlorine gas will we make ? well , look at our mole ratios here . so this would be a two to one mole ratio . so we can set up a proportion , so we could do sodium to chlorine , so this would be a two to one mole ratio , so for every two moles of sodium that are produced , one mole of chlorine gas is produced . so 2/1 is equal to 4.35 times 10 to the fourth , that 's how many moles of sodium we were making , over x , where x represents the moles of chlorine that we would make . so two x is equal to 4.35 times 10 to the fourth , so x is equal to 4.35 times 10 to the fourth divided by two . so if we find 4.35 times 10 to the fourth , we need to divide that by two , and we get 21,750 moles of chlorine , so this would be 21,750 moles of chlorine gas that are produced . we 're trying to find liters of chlorine gas that are produced , so one thing we could do , just to make our lives easy , would be to say , we know that one mole of an ideal gas occupies 22.4 liters at stp . so if we have this many moles , and we assume this is an ideal gas , we could find out how many liters that is by multiplying that number by 22.4 . so 21,750 moles ... if we multiply that by 22.4 , then we should get the volume in liters . so let 's do it , let 's take that number and multiply it by 22.4 . and we get 4.87 , times 10 to the one , two , three four , five , so we get 4.87 times 10 to the fifth liters of chlorine . so there 's another way to do it , if you do n't want to take the shortcut , you can actually plug everything into pv is equal to nrt , so you could use piv-nert , so pv is equal to nrt , we 're at stp , so standard pressure is one atmosphere , and temperature is zero degrees c , or 273.15 kelvin . so we can plug those in , so our pressure is one atmosphere , one atmosphere , our temperature would be 273.15 , so 273.15 kelvin . we 're trying to find the volume , we 're trying to find liters , so we 're trying to find v , n would be equal to the number of moles , so that would be 21,750 , so we have 21,750 moles . r , r is equal to .0821 ... and i did n't leave room for the units in there . but the units would cancel out to give us liters for the volume , and if you solve this , if you use your calculator to solve this , you will get the same answer . you will get 4.87 times 10 to the fifth liters , so you plug it all into piv-nert or you could use the shortcut way . either way , you 're going to get the same volume of chlorine produced .
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and when those liquid sodium ions come in contact with those electrons , we get a reduction half-reaction , so let me write , this is a reduction half-reaction . sodium ions gain electrons , and are reduced to form liquid sodium metal . so liquid sodium metal is one of our products .
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how does electrons from chlorine reaches sodium ?
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here 's a simplified diagram for the electrolysis of molten sodium chloride . so if you melt solid sodium chloride , you get molten sodium chloride . so you get sodium ions , liquid sodium ions , and you get liquid chloride anions . so that 's what we have here , we have sodium ions and chloride anions . remember , in an electrolytic cell , the negative terminal of the battery delivers electrons . in this case , delivers electrons to the electrode on the right . and when those liquid sodium ions come in contact with those electrons , we get a reduction half-reaction , so let me write , this is a reduction half-reaction . sodium ions gain electrons , and are reduced to form liquid sodium metal . so liquid sodium metal is one of our products . and this would form at the cathode . remember , reduction occurs at the cathode . let me draw in some liquid sodium metal here forming on the electrode on the right , forming on the cathodes . the other half-reaction , at our other electrode , we know that our battery draws electrons away from this electrode . so oxidation is occurring at this electrode . this would be the anode . so chloride anions , liquid chloride anions are oxidized to chlorine gas , so we lose two electrons . so chloride anions are oxidized to chlorine gas at the anode , and so we 'd have bubbles of chlorine gas forming at this electrode . this is an important reaction in industry . so to form sodium metal , this is a good way to do it . how long does it take to produce 1.00 times 10 to the third kilograms of sodium , so that 's a lot of sodium , with a constant current of 3.00 times 10 to the fourth amps . so we have another quantitative electrolysis problem here . this one 's a little bit harder than the one we did before , but we 're going to start the same way . we 're going to start with the definition of current . so , current is equal to charge over time , so q over t , where charge is in coulombs and time is in seconds . so right now we know the current is three times 10 to the fourth amps . so 3.00 times 10 to the fourth , is equal to the charge over the time . we want to know the time . we want to know how long it takes to make that much sodium . so we 're trying to find the time . if we could find the charge , then we could find the time using this equation . so let 's do that , let 's find the charge , and we know we can start with the mass of sodium here . so we 're given 1.00 times 10 to the third kilograms , we need to get that into grams . so what is 1.00 times 10 to the third kilograms in grams ? so of course , 1.00 times 10 to the sixth grams . next , we can find the moles of sodium that we 're trying to produce . so if we have grams of sodium , this is grams of sodium , how do we figure out moles of sodium ? we could divide by the molar mass , so we divide by the molar mass of sodium , which is 22.99 , and that would be grams per mole . if we divide by the molar mass , the grams cancel out , and we would get moles of sodium . so let 's get out the calculator . so we have 1.00 times 10 to the sixth , that 's how many grams of sodium we have . if we divide that by 22.99 , the molar mass of sodium , we get , this would be , let 's see , 4.35 times 10 to the one , two , three , four . so 4.35 times 10 to the fourth , let 's write that down , 4.35 times 10 to the fourth moles of sodium . so how does that help us ? if we have moles of sodium , how does that help us find our charge ? well , we can relate the moles of sodium to the moles of electrons . so let 's go back up here to our half-reaction where we have our moles of sodium . so right here , so we 're trying to make 4.35 times 10 to the fourth moles of sodium . so how many electrons do we need to do that ? well , the mole ratio is two to two , so you need the same number of moles of electrons , you need 4.35 times 10 to the fourth moles of electrons , and that helps us out . so let me go back down here , we have some more room , and let me write down 4.35 . 4.35 times 10 to the fourth , so because of our mole ratios we know this is the same number of moles of electrons that we need . we 're trying to find charge , we 're trying to find this charge here , and we know we can go from moles of electrons to the total charge using faraday 's constant . faraday 's constant is the charge carried by one mole of electrons . and we know that 's 96,500 . so there 's a charge of 96,500 coulombs for every one mole of electrons . so if we multiply these together , we 'll get charge , because the moles of electrons will cancel , and we will get the charge , so let 's do that . we have 4.35 times 10 to the fourth , and we 're going to multiply that by faraday 's constant , which is ... so , times , 96,500 , and so we get let 's see , that 'd be 4.2 times 10 to the , let 's see , one , two , three , four , five , six , seven , eight , nine , so 4.20 times 10 to the ninth , let 's write that down . 4.20 times 10 to the ninth , that would be coulombs , right ? that would be coulombs , that 's how much charge we need to make this many moles of sodium . so we have the charge , and so now we can plug that back into here , and solve for the time . so let 's do that . we have 3.00 times 10 to the fourth , that was our current , and current is equal to charge over time . so our total charge is 4.20 times 10 to the ninth , and that 's all over the time . so to solve for the time , we just need to divide this number , so let 's do that , so that 's 4.20 times 10 to the ninth , we 're going to divide that number by 3.00 times 10 to the fourth , and that should give us our time in seconds , so our time in seconds this would be 1.4 times 10 to the one , two , three , four , five , so 1.4 times 10 to the fifth . so the time in seconds is 1.40 times 10 to the fifth . so we have our time , i guess we could leave the answer like that . but let 's convert that to something that 's a little bit easier to comprehend . first , let 's convert that into minutes . so if you have 1.40 times 10 to the fifth seconds , how many minutes is that ? we could just divide that by 60 . so if we divide that by 60 , how many minutes ? that 's 2,333 minutes . how many hours is that ? we could just divide that by 60 , so we could figure out how many hours , so 38.9 hours , approximately , so it would take us approximately 38.9 hours so it would take us 38.9 hours to make , let 's go back up here to remind ourselves how much sodium to make 1.00 times 10 to the third kilograms of sodium . we can also figure out how many liters of chlorine gas are produced when we make our sodium . so let 's say we 're at stp here , so standard temperature and pressure . remember , standard temperature is zero degrees c , and standard pressure is one atmosphere . so we 've already found that we 're trying to make 4.35 times 10 to the fourth moles of sodium . so how many moles of chlorine gas will we make ? well , look at our mole ratios here . so this would be a two to one mole ratio . so we can set up a proportion , so we could do sodium to chlorine , so this would be a two to one mole ratio , so for every two moles of sodium that are produced , one mole of chlorine gas is produced . so 2/1 is equal to 4.35 times 10 to the fourth , that 's how many moles of sodium we were making , over x , where x represents the moles of chlorine that we would make . so two x is equal to 4.35 times 10 to the fourth , so x is equal to 4.35 times 10 to the fourth divided by two . so if we find 4.35 times 10 to the fourth , we need to divide that by two , and we get 21,750 moles of chlorine , so this would be 21,750 moles of chlorine gas that are produced . we 're trying to find liters of chlorine gas that are produced , so one thing we could do , just to make our lives easy , would be to say , we know that one mole of an ideal gas occupies 22.4 liters at stp . so if we have this many moles , and we assume this is an ideal gas , we could find out how many liters that is by multiplying that number by 22.4 . so 21,750 moles ... if we multiply that by 22.4 , then we should get the volume in liters . so let 's do it , let 's take that number and multiply it by 22.4 . and we get 4.87 , times 10 to the one , two , three four , five , so we get 4.87 times 10 to the fifth liters of chlorine . so there 's another way to do it , if you do n't want to take the shortcut , you can actually plug everything into pv is equal to nrt , so you could use piv-nert , so pv is equal to nrt , we 're at stp , so standard pressure is one atmosphere , and temperature is zero degrees c , or 273.15 kelvin . so we can plug those in , so our pressure is one atmosphere , one atmosphere , our temperature would be 273.15 , so 273.15 kelvin . we 're trying to find the volume , we 're trying to find liters , so we 're trying to find v , n would be equal to the number of moles , so that would be 21,750 , so we have 21,750 moles . r , r is equal to .0821 ... and i did n't leave room for the units in there . but the units would cancel out to give us liters for the volume , and if you solve this , if you use your calculator to solve this , you will get the same answer . you will get 4.87 times 10 to the fifth liters , so you plug it all into piv-nert or you could use the shortcut way . either way , you 're going to get the same volume of chlorine produced .
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so if you melt solid sodium chloride , you get molten sodium chloride . so you get sodium ions , liquid sodium ions , and you get liquid chloride anions . so that 's what we have here , we have sodium ions and chloride anions .
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are n't positive ions called cations ?
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here 's a simplified diagram for the electrolysis of molten sodium chloride . so if you melt solid sodium chloride , you get molten sodium chloride . so you get sodium ions , liquid sodium ions , and you get liquid chloride anions . so that 's what we have here , we have sodium ions and chloride anions . remember , in an electrolytic cell , the negative terminal of the battery delivers electrons . in this case , delivers electrons to the electrode on the right . and when those liquid sodium ions come in contact with those electrons , we get a reduction half-reaction , so let me write , this is a reduction half-reaction . sodium ions gain electrons , and are reduced to form liquid sodium metal . so liquid sodium metal is one of our products . and this would form at the cathode . remember , reduction occurs at the cathode . let me draw in some liquid sodium metal here forming on the electrode on the right , forming on the cathodes . the other half-reaction , at our other electrode , we know that our battery draws electrons away from this electrode . so oxidation is occurring at this electrode . this would be the anode . so chloride anions , liquid chloride anions are oxidized to chlorine gas , so we lose two electrons . so chloride anions are oxidized to chlorine gas at the anode , and so we 'd have bubbles of chlorine gas forming at this electrode . this is an important reaction in industry . so to form sodium metal , this is a good way to do it . how long does it take to produce 1.00 times 10 to the third kilograms of sodium , so that 's a lot of sodium , with a constant current of 3.00 times 10 to the fourth amps . so we have another quantitative electrolysis problem here . this one 's a little bit harder than the one we did before , but we 're going to start the same way . we 're going to start with the definition of current . so , current is equal to charge over time , so q over t , where charge is in coulombs and time is in seconds . so right now we know the current is three times 10 to the fourth amps . so 3.00 times 10 to the fourth , is equal to the charge over the time . we want to know the time . we want to know how long it takes to make that much sodium . so we 're trying to find the time . if we could find the charge , then we could find the time using this equation . so let 's do that , let 's find the charge , and we know we can start with the mass of sodium here . so we 're given 1.00 times 10 to the third kilograms , we need to get that into grams . so what is 1.00 times 10 to the third kilograms in grams ? so of course , 1.00 times 10 to the sixth grams . next , we can find the moles of sodium that we 're trying to produce . so if we have grams of sodium , this is grams of sodium , how do we figure out moles of sodium ? we could divide by the molar mass , so we divide by the molar mass of sodium , which is 22.99 , and that would be grams per mole . if we divide by the molar mass , the grams cancel out , and we would get moles of sodium . so let 's get out the calculator . so we have 1.00 times 10 to the sixth , that 's how many grams of sodium we have . if we divide that by 22.99 , the molar mass of sodium , we get , this would be , let 's see , 4.35 times 10 to the one , two , three , four . so 4.35 times 10 to the fourth , let 's write that down , 4.35 times 10 to the fourth moles of sodium . so how does that help us ? if we have moles of sodium , how does that help us find our charge ? well , we can relate the moles of sodium to the moles of electrons . so let 's go back up here to our half-reaction where we have our moles of sodium . so right here , so we 're trying to make 4.35 times 10 to the fourth moles of sodium . so how many electrons do we need to do that ? well , the mole ratio is two to two , so you need the same number of moles of electrons , you need 4.35 times 10 to the fourth moles of electrons , and that helps us out . so let me go back down here , we have some more room , and let me write down 4.35 . 4.35 times 10 to the fourth , so because of our mole ratios we know this is the same number of moles of electrons that we need . we 're trying to find charge , we 're trying to find this charge here , and we know we can go from moles of electrons to the total charge using faraday 's constant . faraday 's constant is the charge carried by one mole of electrons . and we know that 's 96,500 . so there 's a charge of 96,500 coulombs for every one mole of electrons . so if we multiply these together , we 'll get charge , because the moles of electrons will cancel , and we will get the charge , so let 's do that . we have 4.35 times 10 to the fourth , and we 're going to multiply that by faraday 's constant , which is ... so , times , 96,500 , and so we get let 's see , that 'd be 4.2 times 10 to the , let 's see , one , two , three , four , five , six , seven , eight , nine , so 4.20 times 10 to the ninth , let 's write that down . 4.20 times 10 to the ninth , that would be coulombs , right ? that would be coulombs , that 's how much charge we need to make this many moles of sodium . so we have the charge , and so now we can plug that back into here , and solve for the time . so let 's do that . we have 3.00 times 10 to the fourth , that was our current , and current is equal to charge over time . so our total charge is 4.20 times 10 to the ninth , and that 's all over the time . so to solve for the time , we just need to divide this number , so let 's do that , so that 's 4.20 times 10 to the ninth , we 're going to divide that number by 3.00 times 10 to the fourth , and that should give us our time in seconds , so our time in seconds this would be 1.4 times 10 to the one , two , three , four , five , so 1.4 times 10 to the fifth . so the time in seconds is 1.40 times 10 to the fifth . so we have our time , i guess we could leave the answer like that . but let 's convert that to something that 's a little bit easier to comprehend . first , let 's convert that into minutes . so if you have 1.40 times 10 to the fifth seconds , how many minutes is that ? we could just divide that by 60 . so if we divide that by 60 , how many minutes ? that 's 2,333 minutes . how many hours is that ? we could just divide that by 60 , so we could figure out how many hours , so 38.9 hours , approximately , so it would take us approximately 38.9 hours so it would take us 38.9 hours to make , let 's go back up here to remind ourselves how much sodium to make 1.00 times 10 to the third kilograms of sodium . we can also figure out how many liters of chlorine gas are produced when we make our sodium . so let 's say we 're at stp here , so standard temperature and pressure . remember , standard temperature is zero degrees c , and standard pressure is one atmosphere . so we 've already found that we 're trying to make 4.35 times 10 to the fourth moles of sodium . so how many moles of chlorine gas will we make ? well , look at our mole ratios here . so this would be a two to one mole ratio . so we can set up a proportion , so we could do sodium to chlorine , so this would be a two to one mole ratio , so for every two moles of sodium that are produced , one mole of chlorine gas is produced . so 2/1 is equal to 4.35 times 10 to the fourth , that 's how many moles of sodium we were making , over x , where x represents the moles of chlorine that we would make . so two x is equal to 4.35 times 10 to the fourth , so x is equal to 4.35 times 10 to the fourth divided by two . so if we find 4.35 times 10 to the fourth , we need to divide that by two , and we get 21,750 moles of chlorine , so this would be 21,750 moles of chlorine gas that are produced . we 're trying to find liters of chlorine gas that are produced , so one thing we could do , just to make our lives easy , would be to say , we know that one mole of an ideal gas occupies 22.4 liters at stp . so if we have this many moles , and we assume this is an ideal gas , we could find out how many liters that is by multiplying that number by 22.4 . so 21,750 moles ... if we multiply that by 22.4 , then we should get the volume in liters . so let 's do it , let 's take that number and multiply it by 22.4 . and we get 4.87 , times 10 to the one , two , three four , five , so we get 4.87 times 10 to the fifth liters of chlorine . so there 's another way to do it , if you do n't want to take the shortcut , you can actually plug everything into pv is equal to nrt , so you could use piv-nert , so pv is equal to nrt , we 're at stp , so standard pressure is one atmosphere , and temperature is zero degrees c , or 273.15 kelvin . so we can plug those in , so our pressure is one atmosphere , one atmosphere , our temperature would be 273.15 , so 273.15 kelvin . we 're trying to find the volume , we 're trying to find liters , so we 're trying to find v , n would be equal to the number of moles , so that would be 21,750 , so we have 21,750 moles . r , r is equal to .0821 ... and i did n't leave room for the units in there . but the units would cancel out to give us liters for the volume , and if you solve this , if you use your calculator to solve this , you will get the same answer . you will get 4.87 times 10 to the fifth liters , so you plug it all into piv-nert or you could use the shortcut way . either way , you 're going to get the same volume of chlorine produced .
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so chloride anions , liquid chloride anions are oxidized to chlorine gas , so we lose two electrons . so chloride anions are oxidized to chlorine gas at the anode , and so we 'd have bubbles of chlorine gas forming at this electrode . this is an important reaction in industry .
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why does n't sodium react with chlorine gas again ?
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here 's a simplified diagram for the electrolysis of molten sodium chloride . so if you melt solid sodium chloride , you get molten sodium chloride . so you get sodium ions , liquid sodium ions , and you get liquid chloride anions . so that 's what we have here , we have sodium ions and chloride anions . remember , in an electrolytic cell , the negative terminal of the battery delivers electrons . in this case , delivers electrons to the electrode on the right . and when those liquid sodium ions come in contact with those electrons , we get a reduction half-reaction , so let me write , this is a reduction half-reaction . sodium ions gain electrons , and are reduced to form liquid sodium metal . so liquid sodium metal is one of our products . and this would form at the cathode . remember , reduction occurs at the cathode . let me draw in some liquid sodium metal here forming on the electrode on the right , forming on the cathodes . the other half-reaction , at our other electrode , we know that our battery draws electrons away from this electrode . so oxidation is occurring at this electrode . this would be the anode . so chloride anions , liquid chloride anions are oxidized to chlorine gas , so we lose two electrons . so chloride anions are oxidized to chlorine gas at the anode , and so we 'd have bubbles of chlorine gas forming at this electrode . this is an important reaction in industry . so to form sodium metal , this is a good way to do it . how long does it take to produce 1.00 times 10 to the third kilograms of sodium , so that 's a lot of sodium , with a constant current of 3.00 times 10 to the fourth amps . so we have another quantitative electrolysis problem here . this one 's a little bit harder than the one we did before , but we 're going to start the same way . we 're going to start with the definition of current . so , current is equal to charge over time , so q over t , where charge is in coulombs and time is in seconds . so right now we know the current is three times 10 to the fourth amps . so 3.00 times 10 to the fourth , is equal to the charge over the time . we want to know the time . we want to know how long it takes to make that much sodium . so we 're trying to find the time . if we could find the charge , then we could find the time using this equation . so let 's do that , let 's find the charge , and we know we can start with the mass of sodium here . so we 're given 1.00 times 10 to the third kilograms , we need to get that into grams . so what is 1.00 times 10 to the third kilograms in grams ? so of course , 1.00 times 10 to the sixth grams . next , we can find the moles of sodium that we 're trying to produce . so if we have grams of sodium , this is grams of sodium , how do we figure out moles of sodium ? we could divide by the molar mass , so we divide by the molar mass of sodium , which is 22.99 , and that would be grams per mole . if we divide by the molar mass , the grams cancel out , and we would get moles of sodium . so let 's get out the calculator . so we have 1.00 times 10 to the sixth , that 's how many grams of sodium we have . if we divide that by 22.99 , the molar mass of sodium , we get , this would be , let 's see , 4.35 times 10 to the one , two , three , four . so 4.35 times 10 to the fourth , let 's write that down , 4.35 times 10 to the fourth moles of sodium . so how does that help us ? if we have moles of sodium , how does that help us find our charge ? well , we can relate the moles of sodium to the moles of electrons . so let 's go back up here to our half-reaction where we have our moles of sodium . so right here , so we 're trying to make 4.35 times 10 to the fourth moles of sodium . so how many electrons do we need to do that ? well , the mole ratio is two to two , so you need the same number of moles of electrons , you need 4.35 times 10 to the fourth moles of electrons , and that helps us out . so let me go back down here , we have some more room , and let me write down 4.35 . 4.35 times 10 to the fourth , so because of our mole ratios we know this is the same number of moles of electrons that we need . we 're trying to find charge , we 're trying to find this charge here , and we know we can go from moles of electrons to the total charge using faraday 's constant . faraday 's constant is the charge carried by one mole of electrons . and we know that 's 96,500 . so there 's a charge of 96,500 coulombs for every one mole of electrons . so if we multiply these together , we 'll get charge , because the moles of electrons will cancel , and we will get the charge , so let 's do that . we have 4.35 times 10 to the fourth , and we 're going to multiply that by faraday 's constant , which is ... so , times , 96,500 , and so we get let 's see , that 'd be 4.2 times 10 to the , let 's see , one , two , three , four , five , six , seven , eight , nine , so 4.20 times 10 to the ninth , let 's write that down . 4.20 times 10 to the ninth , that would be coulombs , right ? that would be coulombs , that 's how much charge we need to make this many moles of sodium . so we have the charge , and so now we can plug that back into here , and solve for the time . so let 's do that . we have 3.00 times 10 to the fourth , that was our current , and current is equal to charge over time . so our total charge is 4.20 times 10 to the ninth , and that 's all over the time . so to solve for the time , we just need to divide this number , so let 's do that , so that 's 4.20 times 10 to the ninth , we 're going to divide that number by 3.00 times 10 to the fourth , and that should give us our time in seconds , so our time in seconds this would be 1.4 times 10 to the one , two , three , four , five , so 1.4 times 10 to the fifth . so the time in seconds is 1.40 times 10 to the fifth . so we have our time , i guess we could leave the answer like that . but let 's convert that to something that 's a little bit easier to comprehend . first , let 's convert that into minutes . so if you have 1.40 times 10 to the fifth seconds , how many minutes is that ? we could just divide that by 60 . so if we divide that by 60 , how many minutes ? that 's 2,333 minutes . how many hours is that ? we could just divide that by 60 , so we could figure out how many hours , so 38.9 hours , approximately , so it would take us approximately 38.9 hours so it would take us 38.9 hours to make , let 's go back up here to remind ourselves how much sodium to make 1.00 times 10 to the third kilograms of sodium . we can also figure out how many liters of chlorine gas are produced when we make our sodium . so let 's say we 're at stp here , so standard temperature and pressure . remember , standard temperature is zero degrees c , and standard pressure is one atmosphere . so we 've already found that we 're trying to make 4.35 times 10 to the fourth moles of sodium . so how many moles of chlorine gas will we make ? well , look at our mole ratios here . so this would be a two to one mole ratio . so we can set up a proportion , so we could do sodium to chlorine , so this would be a two to one mole ratio , so for every two moles of sodium that are produced , one mole of chlorine gas is produced . so 2/1 is equal to 4.35 times 10 to the fourth , that 's how many moles of sodium we were making , over x , where x represents the moles of chlorine that we would make . so two x is equal to 4.35 times 10 to the fourth , so x is equal to 4.35 times 10 to the fourth divided by two . so if we find 4.35 times 10 to the fourth , we need to divide that by two , and we get 21,750 moles of chlorine , so this would be 21,750 moles of chlorine gas that are produced . we 're trying to find liters of chlorine gas that are produced , so one thing we could do , just to make our lives easy , would be to say , we know that one mole of an ideal gas occupies 22.4 liters at stp . so if we have this many moles , and we assume this is an ideal gas , we could find out how many liters that is by multiplying that number by 22.4 . so 21,750 moles ... if we multiply that by 22.4 , then we should get the volume in liters . so let 's do it , let 's take that number and multiply it by 22.4 . and we get 4.87 , times 10 to the one , two , three four , five , so we get 4.87 times 10 to the fifth liters of chlorine . so there 's another way to do it , if you do n't want to take the shortcut , you can actually plug everything into pv is equal to nrt , so you could use piv-nert , so pv is equal to nrt , we 're at stp , so standard pressure is one atmosphere , and temperature is zero degrees c , or 273.15 kelvin . so we can plug those in , so our pressure is one atmosphere , one atmosphere , our temperature would be 273.15 , so 273.15 kelvin . we 're trying to find the volume , we 're trying to find liters , so we 're trying to find v , n would be equal to the number of moles , so that would be 21,750 , so we have 21,750 moles . r , r is equal to .0821 ... and i did n't leave room for the units in there . but the units would cancel out to give us liters for the volume , and if you solve this , if you use your calculator to solve this , you will get the same answer . you will get 4.87 times 10 to the fifth liters , so you plug it all into piv-nert or you could use the shortcut way . either way , you 're going to get the same volume of chlorine produced .
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how long does it take to produce 1.00 times 10 to the third kilograms of sodium , so that 's a lot of sodium , with a constant current of 3.00 times 10 to the fourth amps . so we have another quantitative electrolysis problem here . this one 's a little bit harder than the one we did before , but we 're going to start the same way .
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is electrolysis of water endothermic or not ?
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here 's a simplified diagram for the electrolysis of molten sodium chloride . so if you melt solid sodium chloride , you get molten sodium chloride . so you get sodium ions , liquid sodium ions , and you get liquid chloride anions . so that 's what we have here , we have sodium ions and chloride anions . remember , in an electrolytic cell , the negative terminal of the battery delivers electrons . in this case , delivers electrons to the electrode on the right . and when those liquid sodium ions come in contact with those electrons , we get a reduction half-reaction , so let me write , this is a reduction half-reaction . sodium ions gain electrons , and are reduced to form liquid sodium metal . so liquid sodium metal is one of our products . and this would form at the cathode . remember , reduction occurs at the cathode . let me draw in some liquid sodium metal here forming on the electrode on the right , forming on the cathodes . the other half-reaction , at our other electrode , we know that our battery draws electrons away from this electrode . so oxidation is occurring at this electrode . this would be the anode . so chloride anions , liquid chloride anions are oxidized to chlorine gas , so we lose two electrons . so chloride anions are oxidized to chlorine gas at the anode , and so we 'd have bubbles of chlorine gas forming at this electrode . this is an important reaction in industry . so to form sodium metal , this is a good way to do it . how long does it take to produce 1.00 times 10 to the third kilograms of sodium , so that 's a lot of sodium , with a constant current of 3.00 times 10 to the fourth amps . so we have another quantitative electrolysis problem here . this one 's a little bit harder than the one we did before , but we 're going to start the same way . we 're going to start with the definition of current . so , current is equal to charge over time , so q over t , where charge is in coulombs and time is in seconds . so right now we know the current is three times 10 to the fourth amps . so 3.00 times 10 to the fourth , is equal to the charge over the time . we want to know the time . we want to know how long it takes to make that much sodium . so we 're trying to find the time . if we could find the charge , then we could find the time using this equation . so let 's do that , let 's find the charge , and we know we can start with the mass of sodium here . so we 're given 1.00 times 10 to the third kilograms , we need to get that into grams . so what is 1.00 times 10 to the third kilograms in grams ? so of course , 1.00 times 10 to the sixth grams . next , we can find the moles of sodium that we 're trying to produce . so if we have grams of sodium , this is grams of sodium , how do we figure out moles of sodium ? we could divide by the molar mass , so we divide by the molar mass of sodium , which is 22.99 , and that would be grams per mole . if we divide by the molar mass , the grams cancel out , and we would get moles of sodium . so let 's get out the calculator . so we have 1.00 times 10 to the sixth , that 's how many grams of sodium we have . if we divide that by 22.99 , the molar mass of sodium , we get , this would be , let 's see , 4.35 times 10 to the one , two , three , four . so 4.35 times 10 to the fourth , let 's write that down , 4.35 times 10 to the fourth moles of sodium . so how does that help us ? if we have moles of sodium , how does that help us find our charge ? well , we can relate the moles of sodium to the moles of electrons . so let 's go back up here to our half-reaction where we have our moles of sodium . so right here , so we 're trying to make 4.35 times 10 to the fourth moles of sodium . so how many electrons do we need to do that ? well , the mole ratio is two to two , so you need the same number of moles of electrons , you need 4.35 times 10 to the fourth moles of electrons , and that helps us out . so let me go back down here , we have some more room , and let me write down 4.35 . 4.35 times 10 to the fourth , so because of our mole ratios we know this is the same number of moles of electrons that we need . we 're trying to find charge , we 're trying to find this charge here , and we know we can go from moles of electrons to the total charge using faraday 's constant . faraday 's constant is the charge carried by one mole of electrons . and we know that 's 96,500 . so there 's a charge of 96,500 coulombs for every one mole of electrons . so if we multiply these together , we 'll get charge , because the moles of electrons will cancel , and we will get the charge , so let 's do that . we have 4.35 times 10 to the fourth , and we 're going to multiply that by faraday 's constant , which is ... so , times , 96,500 , and so we get let 's see , that 'd be 4.2 times 10 to the , let 's see , one , two , three , four , five , six , seven , eight , nine , so 4.20 times 10 to the ninth , let 's write that down . 4.20 times 10 to the ninth , that would be coulombs , right ? that would be coulombs , that 's how much charge we need to make this many moles of sodium . so we have the charge , and so now we can plug that back into here , and solve for the time . so let 's do that . we have 3.00 times 10 to the fourth , that was our current , and current is equal to charge over time . so our total charge is 4.20 times 10 to the ninth , and that 's all over the time . so to solve for the time , we just need to divide this number , so let 's do that , so that 's 4.20 times 10 to the ninth , we 're going to divide that number by 3.00 times 10 to the fourth , and that should give us our time in seconds , so our time in seconds this would be 1.4 times 10 to the one , two , three , four , five , so 1.4 times 10 to the fifth . so the time in seconds is 1.40 times 10 to the fifth . so we have our time , i guess we could leave the answer like that . but let 's convert that to something that 's a little bit easier to comprehend . first , let 's convert that into minutes . so if you have 1.40 times 10 to the fifth seconds , how many minutes is that ? we could just divide that by 60 . so if we divide that by 60 , how many minutes ? that 's 2,333 minutes . how many hours is that ? we could just divide that by 60 , so we could figure out how many hours , so 38.9 hours , approximately , so it would take us approximately 38.9 hours so it would take us 38.9 hours to make , let 's go back up here to remind ourselves how much sodium to make 1.00 times 10 to the third kilograms of sodium . we can also figure out how many liters of chlorine gas are produced when we make our sodium . so let 's say we 're at stp here , so standard temperature and pressure . remember , standard temperature is zero degrees c , and standard pressure is one atmosphere . so we 've already found that we 're trying to make 4.35 times 10 to the fourth moles of sodium . so how many moles of chlorine gas will we make ? well , look at our mole ratios here . so this would be a two to one mole ratio . so we can set up a proportion , so we could do sodium to chlorine , so this would be a two to one mole ratio , so for every two moles of sodium that are produced , one mole of chlorine gas is produced . so 2/1 is equal to 4.35 times 10 to the fourth , that 's how many moles of sodium we were making , over x , where x represents the moles of chlorine that we would make . so two x is equal to 4.35 times 10 to the fourth , so x is equal to 4.35 times 10 to the fourth divided by two . so if we find 4.35 times 10 to the fourth , we need to divide that by two , and we get 21,750 moles of chlorine , so this would be 21,750 moles of chlorine gas that are produced . we 're trying to find liters of chlorine gas that are produced , so one thing we could do , just to make our lives easy , would be to say , we know that one mole of an ideal gas occupies 22.4 liters at stp . so if we have this many moles , and we assume this is an ideal gas , we could find out how many liters that is by multiplying that number by 22.4 . so 21,750 moles ... if we multiply that by 22.4 , then we should get the volume in liters . so let 's do it , let 's take that number and multiply it by 22.4 . and we get 4.87 , times 10 to the one , two , three four , five , so we get 4.87 times 10 to the fifth liters of chlorine . so there 's another way to do it , if you do n't want to take the shortcut , you can actually plug everything into pv is equal to nrt , so you could use piv-nert , so pv is equal to nrt , we 're at stp , so standard pressure is one atmosphere , and temperature is zero degrees c , or 273.15 kelvin . so we can plug those in , so our pressure is one atmosphere , one atmosphere , our temperature would be 273.15 , so 273.15 kelvin . we 're trying to find the volume , we 're trying to find liters , so we 're trying to find v , n would be equal to the number of moles , so that would be 21,750 , so we have 21,750 moles . r , r is equal to .0821 ... and i did n't leave room for the units in there . but the units would cancel out to give us liters for the volume , and if you solve this , if you use your calculator to solve this , you will get the same answer . you will get 4.87 times 10 to the fifth liters , so you plug it all into piv-nert or you could use the shortcut way . either way , you 're going to get the same volume of chlorine produced .
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we 're trying to find the volume , we 're trying to find liters , so we 're trying to find v , n would be equal to the number of moles , so that would be 21,750 , so we have 21,750 moles . r , r is equal to .0821 ... and i did n't leave room for the units in there . but the units would cancel out to give us liters for the volume , and if you solve this , if you use your calculator to solve this , you will get the same answer .
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same ques for formation of n2 from n atom ?
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here 's a simplified diagram for the electrolysis of molten sodium chloride . so if you melt solid sodium chloride , you get molten sodium chloride . so you get sodium ions , liquid sodium ions , and you get liquid chloride anions . so that 's what we have here , we have sodium ions and chloride anions . remember , in an electrolytic cell , the negative terminal of the battery delivers electrons . in this case , delivers electrons to the electrode on the right . and when those liquid sodium ions come in contact with those electrons , we get a reduction half-reaction , so let me write , this is a reduction half-reaction . sodium ions gain electrons , and are reduced to form liquid sodium metal . so liquid sodium metal is one of our products . and this would form at the cathode . remember , reduction occurs at the cathode . let me draw in some liquid sodium metal here forming on the electrode on the right , forming on the cathodes . the other half-reaction , at our other electrode , we know that our battery draws electrons away from this electrode . so oxidation is occurring at this electrode . this would be the anode . so chloride anions , liquid chloride anions are oxidized to chlorine gas , so we lose two electrons . so chloride anions are oxidized to chlorine gas at the anode , and so we 'd have bubbles of chlorine gas forming at this electrode . this is an important reaction in industry . so to form sodium metal , this is a good way to do it . how long does it take to produce 1.00 times 10 to the third kilograms of sodium , so that 's a lot of sodium , with a constant current of 3.00 times 10 to the fourth amps . so we have another quantitative electrolysis problem here . this one 's a little bit harder than the one we did before , but we 're going to start the same way . we 're going to start with the definition of current . so , current is equal to charge over time , so q over t , where charge is in coulombs and time is in seconds . so right now we know the current is three times 10 to the fourth amps . so 3.00 times 10 to the fourth , is equal to the charge over the time . we want to know the time . we want to know how long it takes to make that much sodium . so we 're trying to find the time . if we could find the charge , then we could find the time using this equation . so let 's do that , let 's find the charge , and we know we can start with the mass of sodium here . so we 're given 1.00 times 10 to the third kilograms , we need to get that into grams . so what is 1.00 times 10 to the third kilograms in grams ? so of course , 1.00 times 10 to the sixth grams . next , we can find the moles of sodium that we 're trying to produce . so if we have grams of sodium , this is grams of sodium , how do we figure out moles of sodium ? we could divide by the molar mass , so we divide by the molar mass of sodium , which is 22.99 , and that would be grams per mole . if we divide by the molar mass , the grams cancel out , and we would get moles of sodium . so let 's get out the calculator . so we have 1.00 times 10 to the sixth , that 's how many grams of sodium we have . if we divide that by 22.99 , the molar mass of sodium , we get , this would be , let 's see , 4.35 times 10 to the one , two , three , four . so 4.35 times 10 to the fourth , let 's write that down , 4.35 times 10 to the fourth moles of sodium . so how does that help us ? if we have moles of sodium , how does that help us find our charge ? well , we can relate the moles of sodium to the moles of electrons . so let 's go back up here to our half-reaction where we have our moles of sodium . so right here , so we 're trying to make 4.35 times 10 to the fourth moles of sodium . so how many electrons do we need to do that ? well , the mole ratio is two to two , so you need the same number of moles of electrons , you need 4.35 times 10 to the fourth moles of electrons , and that helps us out . so let me go back down here , we have some more room , and let me write down 4.35 . 4.35 times 10 to the fourth , so because of our mole ratios we know this is the same number of moles of electrons that we need . we 're trying to find charge , we 're trying to find this charge here , and we know we can go from moles of electrons to the total charge using faraday 's constant . faraday 's constant is the charge carried by one mole of electrons . and we know that 's 96,500 . so there 's a charge of 96,500 coulombs for every one mole of electrons . so if we multiply these together , we 'll get charge , because the moles of electrons will cancel , and we will get the charge , so let 's do that . we have 4.35 times 10 to the fourth , and we 're going to multiply that by faraday 's constant , which is ... so , times , 96,500 , and so we get let 's see , that 'd be 4.2 times 10 to the , let 's see , one , two , three , four , five , six , seven , eight , nine , so 4.20 times 10 to the ninth , let 's write that down . 4.20 times 10 to the ninth , that would be coulombs , right ? that would be coulombs , that 's how much charge we need to make this many moles of sodium . so we have the charge , and so now we can plug that back into here , and solve for the time . so let 's do that . we have 3.00 times 10 to the fourth , that was our current , and current is equal to charge over time . so our total charge is 4.20 times 10 to the ninth , and that 's all over the time . so to solve for the time , we just need to divide this number , so let 's do that , so that 's 4.20 times 10 to the ninth , we 're going to divide that number by 3.00 times 10 to the fourth , and that should give us our time in seconds , so our time in seconds this would be 1.4 times 10 to the one , two , three , four , five , so 1.4 times 10 to the fifth . so the time in seconds is 1.40 times 10 to the fifth . so we have our time , i guess we could leave the answer like that . but let 's convert that to something that 's a little bit easier to comprehend . first , let 's convert that into minutes . so if you have 1.40 times 10 to the fifth seconds , how many minutes is that ? we could just divide that by 60 . so if we divide that by 60 , how many minutes ? that 's 2,333 minutes . how many hours is that ? we could just divide that by 60 , so we could figure out how many hours , so 38.9 hours , approximately , so it would take us approximately 38.9 hours so it would take us 38.9 hours to make , let 's go back up here to remind ourselves how much sodium to make 1.00 times 10 to the third kilograms of sodium . we can also figure out how many liters of chlorine gas are produced when we make our sodium . so let 's say we 're at stp here , so standard temperature and pressure . remember , standard temperature is zero degrees c , and standard pressure is one atmosphere . so we 've already found that we 're trying to make 4.35 times 10 to the fourth moles of sodium . so how many moles of chlorine gas will we make ? well , look at our mole ratios here . so this would be a two to one mole ratio . so we can set up a proportion , so we could do sodium to chlorine , so this would be a two to one mole ratio , so for every two moles of sodium that are produced , one mole of chlorine gas is produced . so 2/1 is equal to 4.35 times 10 to the fourth , that 's how many moles of sodium we were making , over x , where x represents the moles of chlorine that we would make . so two x is equal to 4.35 times 10 to the fourth , so x is equal to 4.35 times 10 to the fourth divided by two . so if we find 4.35 times 10 to the fourth , we need to divide that by two , and we get 21,750 moles of chlorine , so this would be 21,750 moles of chlorine gas that are produced . we 're trying to find liters of chlorine gas that are produced , so one thing we could do , just to make our lives easy , would be to say , we know that one mole of an ideal gas occupies 22.4 liters at stp . so if we have this many moles , and we assume this is an ideal gas , we could find out how many liters that is by multiplying that number by 22.4 . so 21,750 moles ... if we multiply that by 22.4 , then we should get the volume in liters . so let 's do it , let 's take that number and multiply it by 22.4 . and we get 4.87 , times 10 to the one , two , three four , five , so we get 4.87 times 10 to the fifth liters of chlorine . so there 's another way to do it , if you do n't want to take the shortcut , you can actually plug everything into pv is equal to nrt , so you could use piv-nert , so pv is equal to nrt , we 're at stp , so standard pressure is one atmosphere , and temperature is zero degrees c , or 273.15 kelvin . so we can plug those in , so our pressure is one atmosphere , one atmosphere , our temperature would be 273.15 , so 273.15 kelvin . we 're trying to find the volume , we 're trying to find liters , so we 're trying to find v , n would be equal to the number of moles , so that would be 21,750 , so we have 21,750 moles . r , r is equal to .0821 ... and i did n't leave room for the units in there . but the units would cancel out to give us liters for the volume , and if you solve this , if you use your calculator to solve this , you will get the same answer . you will get 4.87 times 10 to the fifth liters , so you plug it all into piv-nert or you could use the shortcut way . either way , you 're going to get the same volume of chlorine produced .
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here 's a simplified diagram for the electrolysis of molten sodium chloride . so if you melt solid sodium chloride , you get molten sodium chloride .
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what kind of material must have the electrodes for electrolysis molten salt ?
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here 's a simplified diagram for the electrolysis of molten sodium chloride . so if you melt solid sodium chloride , you get molten sodium chloride . so you get sodium ions , liquid sodium ions , and you get liquid chloride anions . so that 's what we have here , we have sodium ions and chloride anions . remember , in an electrolytic cell , the negative terminal of the battery delivers electrons . in this case , delivers electrons to the electrode on the right . and when those liquid sodium ions come in contact with those electrons , we get a reduction half-reaction , so let me write , this is a reduction half-reaction . sodium ions gain electrons , and are reduced to form liquid sodium metal . so liquid sodium metal is one of our products . and this would form at the cathode . remember , reduction occurs at the cathode . let me draw in some liquid sodium metal here forming on the electrode on the right , forming on the cathodes . the other half-reaction , at our other electrode , we know that our battery draws electrons away from this electrode . so oxidation is occurring at this electrode . this would be the anode . so chloride anions , liquid chloride anions are oxidized to chlorine gas , so we lose two electrons . so chloride anions are oxidized to chlorine gas at the anode , and so we 'd have bubbles of chlorine gas forming at this electrode . this is an important reaction in industry . so to form sodium metal , this is a good way to do it . how long does it take to produce 1.00 times 10 to the third kilograms of sodium , so that 's a lot of sodium , with a constant current of 3.00 times 10 to the fourth amps . so we have another quantitative electrolysis problem here . this one 's a little bit harder than the one we did before , but we 're going to start the same way . we 're going to start with the definition of current . so , current is equal to charge over time , so q over t , where charge is in coulombs and time is in seconds . so right now we know the current is three times 10 to the fourth amps . so 3.00 times 10 to the fourth , is equal to the charge over the time . we want to know the time . we want to know how long it takes to make that much sodium . so we 're trying to find the time . if we could find the charge , then we could find the time using this equation . so let 's do that , let 's find the charge , and we know we can start with the mass of sodium here . so we 're given 1.00 times 10 to the third kilograms , we need to get that into grams . so what is 1.00 times 10 to the third kilograms in grams ? so of course , 1.00 times 10 to the sixth grams . next , we can find the moles of sodium that we 're trying to produce . so if we have grams of sodium , this is grams of sodium , how do we figure out moles of sodium ? we could divide by the molar mass , so we divide by the molar mass of sodium , which is 22.99 , and that would be grams per mole . if we divide by the molar mass , the grams cancel out , and we would get moles of sodium . so let 's get out the calculator . so we have 1.00 times 10 to the sixth , that 's how many grams of sodium we have . if we divide that by 22.99 , the molar mass of sodium , we get , this would be , let 's see , 4.35 times 10 to the one , two , three , four . so 4.35 times 10 to the fourth , let 's write that down , 4.35 times 10 to the fourth moles of sodium . so how does that help us ? if we have moles of sodium , how does that help us find our charge ? well , we can relate the moles of sodium to the moles of electrons . so let 's go back up here to our half-reaction where we have our moles of sodium . so right here , so we 're trying to make 4.35 times 10 to the fourth moles of sodium . so how many electrons do we need to do that ? well , the mole ratio is two to two , so you need the same number of moles of electrons , you need 4.35 times 10 to the fourth moles of electrons , and that helps us out . so let me go back down here , we have some more room , and let me write down 4.35 . 4.35 times 10 to the fourth , so because of our mole ratios we know this is the same number of moles of electrons that we need . we 're trying to find charge , we 're trying to find this charge here , and we know we can go from moles of electrons to the total charge using faraday 's constant . faraday 's constant is the charge carried by one mole of electrons . and we know that 's 96,500 . so there 's a charge of 96,500 coulombs for every one mole of electrons . so if we multiply these together , we 'll get charge , because the moles of electrons will cancel , and we will get the charge , so let 's do that . we have 4.35 times 10 to the fourth , and we 're going to multiply that by faraday 's constant , which is ... so , times , 96,500 , and so we get let 's see , that 'd be 4.2 times 10 to the , let 's see , one , two , three , four , five , six , seven , eight , nine , so 4.20 times 10 to the ninth , let 's write that down . 4.20 times 10 to the ninth , that would be coulombs , right ? that would be coulombs , that 's how much charge we need to make this many moles of sodium . so we have the charge , and so now we can plug that back into here , and solve for the time . so let 's do that . we have 3.00 times 10 to the fourth , that was our current , and current is equal to charge over time . so our total charge is 4.20 times 10 to the ninth , and that 's all over the time . so to solve for the time , we just need to divide this number , so let 's do that , so that 's 4.20 times 10 to the ninth , we 're going to divide that number by 3.00 times 10 to the fourth , and that should give us our time in seconds , so our time in seconds this would be 1.4 times 10 to the one , two , three , four , five , so 1.4 times 10 to the fifth . so the time in seconds is 1.40 times 10 to the fifth . so we have our time , i guess we could leave the answer like that . but let 's convert that to something that 's a little bit easier to comprehend . first , let 's convert that into minutes . so if you have 1.40 times 10 to the fifth seconds , how many minutes is that ? we could just divide that by 60 . so if we divide that by 60 , how many minutes ? that 's 2,333 minutes . how many hours is that ? we could just divide that by 60 , so we could figure out how many hours , so 38.9 hours , approximately , so it would take us approximately 38.9 hours so it would take us 38.9 hours to make , let 's go back up here to remind ourselves how much sodium to make 1.00 times 10 to the third kilograms of sodium . we can also figure out how many liters of chlorine gas are produced when we make our sodium . so let 's say we 're at stp here , so standard temperature and pressure . remember , standard temperature is zero degrees c , and standard pressure is one atmosphere . so we 've already found that we 're trying to make 4.35 times 10 to the fourth moles of sodium . so how many moles of chlorine gas will we make ? well , look at our mole ratios here . so this would be a two to one mole ratio . so we can set up a proportion , so we could do sodium to chlorine , so this would be a two to one mole ratio , so for every two moles of sodium that are produced , one mole of chlorine gas is produced . so 2/1 is equal to 4.35 times 10 to the fourth , that 's how many moles of sodium we were making , over x , where x represents the moles of chlorine that we would make . so two x is equal to 4.35 times 10 to the fourth , so x is equal to 4.35 times 10 to the fourth divided by two . so if we find 4.35 times 10 to the fourth , we need to divide that by two , and we get 21,750 moles of chlorine , so this would be 21,750 moles of chlorine gas that are produced . we 're trying to find liters of chlorine gas that are produced , so one thing we could do , just to make our lives easy , would be to say , we know that one mole of an ideal gas occupies 22.4 liters at stp . so if we have this many moles , and we assume this is an ideal gas , we could find out how many liters that is by multiplying that number by 22.4 . so 21,750 moles ... if we multiply that by 22.4 , then we should get the volume in liters . so let 's do it , let 's take that number and multiply it by 22.4 . and we get 4.87 , times 10 to the one , two , three four , five , so we get 4.87 times 10 to the fifth liters of chlorine . so there 's another way to do it , if you do n't want to take the shortcut , you can actually plug everything into pv is equal to nrt , so you could use piv-nert , so pv is equal to nrt , we 're at stp , so standard pressure is one atmosphere , and temperature is zero degrees c , or 273.15 kelvin . so we can plug those in , so our pressure is one atmosphere , one atmosphere , our temperature would be 273.15 , so 273.15 kelvin . we 're trying to find the volume , we 're trying to find liters , so we 're trying to find v , n would be equal to the number of moles , so that would be 21,750 , so we have 21,750 moles . r , r is equal to .0821 ... and i did n't leave room for the units in there . but the units would cancel out to give us liters for the volume , and if you solve this , if you use your calculator to solve this , you will get the same answer . you will get 4.87 times 10 to the fifth liters , so you plug it all into piv-nert or you could use the shortcut way . either way , you 're going to get the same volume of chlorine produced .
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the other half-reaction , at our other electrode , we know that our battery draws electrons away from this electrode . so oxidation is occurring at this electrode . this would be the anode . so chloride anions , liquid chloride anions are oxidized to chlorine gas , so we lose two electrons .
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how would you go about figuring which species undergoes reduction or oxidation if the solution is aqueous instead of molten ?
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here 's a simplified diagram for the electrolysis of molten sodium chloride . so if you melt solid sodium chloride , you get molten sodium chloride . so you get sodium ions , liquid sodium ions , and you get liquid chloride anions . so that 's what we have here , we have sodium ions and chloride anions . remember , in an electrolytic cell , the negative terminal of the battery delivers electrons . in this case , delivers electrons to the electrode on the right . and when those liquid sodium ions come in contact with those electrons , we get a reduction half-reaction , so let me write , this is a reduction half-reaction . sodium ions gain electrons , and are reduced to form liquid sodium metal . so liquid sodium metal is one of our products . and this would form at the cathode . remember , reduction occurs at the cathode . let me draw in some liquid sodium metal here forming on the electrode on the right , forming on the cathodes . the other half-reaction , at our other electrode , we know that our battery draws electrons away from this electrode . so oxidation is occurring at this electrode . this would be the anode . so chloride anions , liquid chloride anions are oxidized to chlorine gas , so we lose two electrons . so chloride anions are oxidized to chlorine gas at the anode , and so we 'd have bubbles of chlorine gas forming at this electrode . this is an important reaction in industry . so to form sodium metal , this is a good way to do it . how long does it take to produce 1.00 times 10 to the third kilograms of sodium , so that 's a lot of sodium , with a constant current of 3.00 times 10 to the fourth amps . so we have another quantitative electrolysis problem here . this one 's a little bit harder than the one we did before , but we 're going to start the same way . we 're going to start with the definition of current . so , current is equal to charge over time , so q over t , where charge is in coulombs and time is in seconds . so right now we know the current is three times 10 to the fourth amps . so 3.00 times 10 to the fourth , is equal to the charge over the time . we want to know the time . we want to know how long it takes to make that much sodium . so we 're trying to find the time . if we could find the charge , then we could find the time using this equation . so let 's do that , let 's find the charge , and we know we can start with the mass of sodium here . so we 're given 1.00 times 10 to the third kilograms , we need to get that into grams . so what is 1.00 times 10 to the third kilograms in grams ? so of course , 1.00 times 10 to the sixth grams . next , we can find the moles of sodium that we 're trying to produce . so if we have grams of sodium , this is grams of sodium , how do we figure out moles of sodium ? we could divide by the molar mass , so we divide by the molar mass of sodium , which is 22.99 , and that would be grams per mole . if we divide by the molar mass , the grams cancel out , and we would get moles of sodium . so let 's get out the calculator . so we have 1.00 times 10 to the sixth , that 's how many grams of sodium we have . if we divide that by 22.99 , the molar mass of sodium , we get , this would be , let 's see , 4.35 times 10 to the one , two , three , four . so 4.35 times 10 to the fourth , let 's write that down , 4.35 times 10 to the fourth moles of sodium . so how does that help us ? if we have moles of sodium , how does that help us find our charge ? well , we can relate the moles of sodium to the moles of electrons . so let 's go back up here to our half-reaction where we have our moles of sodium . so right here , so we 're trying to make 4.35 times 10 to the fourth moles of sodium . so how many electrons do we need to do that ? well , the mole ratio is two to two , so you need the same number of moles of electrons , you need 4.35 times 10 to the fourth moles of electrons , and that helps us out . so let me go back down here , we have some more room , and let me write down 4.35 . 4.35 times 10 to the fourth , so because of our mole ratios we know this is the same number of moles of electrons that we need . we 're trying to find charge , we 're trying to find this charge here , and we know we can go from moles of electrons to the total charge using faraday 's constant . faraday 's constant is the charge carried by one mole of electrons . and we know that 's 96,500 . so there 's a charge of 96,500 coulombs for every one mole of electrons . so if we multiply these together , we 'll get charge , because the moles of electrons will cancel , and we will get the charge , so let 's do that . we have 4.35 times 10 to the fourth , and we 're going to multiply that by faraday 's constant , which is ... so , times , 96,500 , and so we get let 's see , that 'd be 4.2 times 10 to the , let 's see , one , two , three , four , five , six , seven , eight , nine , so 4.20 times 10 to the ninth , let 's write that down . 4.20 times 10 to the ninth , that would be coulombs , right ? that would be coulombs , that 's how much charge we need to make this many moles of sodium . so we have the charge , and so now we can plug that back into here , and solve for the time . so let 's do that . we have 3.00 times 10 to the fourth , that was our current , and current is equal to charge over time . so our total charge is 4.20 times 10 to the ninth , and that 's all over the time . so to solve for the time , we just need to divide this number , so let 's do that , so that 's 4.20 times 10 to the ninth , we 're going to divide that number by 3.00 times 10 to the fourth , and that should give us our time in seconds , so our time in seconds this would be 1.4 times 10 to the one , two , three , four , five , so 1.4 times 10 to the fifth . so the time in seconds is 1.40 times 10 to the fifth . so we have our time , i guess we could leave the answer like that . but let 's convert that to something that 's a little bit easier to comprehend . first , let 's convert that into minutes . so if you have 1.40 times 10 to the fifth seconds , how many minutes is that ? we could just divide that by 60 . so if we divide that by 60 , how many minutes ? that 's 2,333 minutes . how many hours is that ? we could just divide that by 60 , so we could figure out how many hours , so 38.9 hours , approximately , so it would take us approximately 38.9 hours so it would take us 38.9 hours to make , let 's go back up here to remind ourselves how much sodium to make 1.00 times 10 to the third kilograms of sodium . we can also figure out how many liters of chlorine gas are produced when we make our sodium . so let 's say we 're at stp here , so standard temperature and pressure . remember , standard temperature is zero degrees c , and standard pressure is one atmosphere . so we 've already found that we 're trying to make 4.35 times 10 to the fourth moles of sodium . so how many moles of chlorine gas will we make ? well , look at our mole ratios here . so this would be a two to one mole ratio . so we can set up a proportion , so we could do sodium to chlorine , so this would be a two to one mole ratio , so for every two moles of sodium that are produced , one mole of chlorine gas is produced . so 2/1 is equal to 4.35 times 10 to the fourth , that 's how many moles of sodium we were making , over x , where x represents the moles of chlorine that we would make . so two x is equal to 4.35 times 10 to the fourth , so x is equal to 4.35 times 10 to the fourth divided by two . so if we find 4.35 times 10 to the fourth , we need to divide that by two , and we get 21,750 moles of chlorine , so this would be 21,750 moles of chlorine gas that are produced . we 're trying to find liters of chlorine gas that are produced , so one thing we could do , just to make our lives easy , would be to say , we know that one mole of an ideal gas occupies 22.4 liters at stp . so if we have this many moles , and we assume this is an ideal gas , we could find out how many liters that is by multiplying that number by 22.4 . so 21,750 moles ... if we multiply that by 22.4 , then we should get the volume in liters . so let 's do it , let 's take that number and multiply it by 22.4 . and we get 4.87 , times 10 to the one , two , three four , five , so we get 4.87 times 10 to the fifth liters of chlorine . so there 's another way to do it , if you do n't want to take the shortcut , you can actually plug everything into pv is equal to nrt , so you could use piv-nert , so pv is equal to nrt , we 're at stp , so standard pressure is one atmosphere , and temperature is zero degrees c , or 273.15 kelvin . so we can plug those in , so our pressure is one atmosphere , one atmosphere , our temperature would be 273.15 , so 273.15 kelvin . we 're trying to find the volume , we 're trying to find liters , so we 're trying to find v , n would be equal to the number of moles , so that would be 21,750 , so we have 21,750 moles . r , r is equal to .0821 ... and i did n't leave room for the units in there . but the units would cancel out to give us liters for the volume , and if you solve this , if you use your calculator to solve this , you will get the same answer . you will get 4.87 times 10 to the fifth liters , so you plug it all into piv-nert or you could use the shortcut way . either way , you 're going to get the same volume of chlorine produced .
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and when those liquid sodium ions come in contact with those electrons , we get a reduction half-reaction , so let me write , this is a reduction half-reaction . sodium ions gain electrons , and are reduced to form liquid sodium metal . so liquid sodium metal is one of our products .
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why does sodium gains an electron , sodium has an extra electron already ?
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here 's a simplified diagram for the electrolysis of molten sodium chloride . so if you melt solid sodium chloride , you get molten sodium chloride . so you get sodium ions , liquid sodium ions , and you get liquid chloride anions . so that 's what we have here , we have sodium ions and chloride anions . remember , in an electrolytic cell , the negative terminal of the battery delivers electrons . in this case , delivers electrons to the electrode on the right . and when those liquid sodium ions come in contact with those electrons , we get a reduction half-reaction , so let me write , this is a reduction half-reaction . sodium ions gain electrons , and are reduced to form liquid sodium metal . so liquid sodium metal is one of our products . and this would form at the cathode . remember , reduction occurs at the cathode . let me draw in some liquid sodium metal here forming on the electrode on the right , forming on the cathodes . the other half-reaction , at our other electrode , we know that our battery draws electrons away from this electrode . so oxidation is occurring at this electrode . this would be the anode . so chloride anions , liquid chloride anions are oxidized to chlorine gas , so we lose two electrons . so chloride anions are oxidized to chlorine gas at the anode , and so we 'd have bubbles of chlorine gas forming at this electrode . this is an important reaction in industry . so to form sodium metal , this is a good way to do it . how long does it take to produce 1.00 times 10 to the third kilograms of sodium , so that 's a lot of sodium , with a constant current of 3.00 times 10 to the fourth amps . so we have another quantitative electrolysis problem here . this one 's a little bit harder than the one we did before , but we 're going to start the same way . we 're going to start with the definition of current . so , current is equal to charge over time , so q over t , where charge is in coulombs and time is in seconds . so right now we know the current is three times 10 to the fourth amps . so 3.00 times 10 to the fourth , is equal to the charge over the time . we want to know the time . we want to know how long it takes to make that much sodium . so we 're trying to find the time . if we could find the charge , then we could find the time using this equation . so let 's do that , let 's find the charge , and we know we can start with the mass of sodium here . so we 're given 1.00 times 10 to the third kilograms , we need to get that into grams . so what is 1.00 times 10 to the third kilograms in grams ? so of course , 1.00 times 10 to the sixth grams . next , we can find the moles of sodium that we 're trying to produce . so if we have grams of sodium , this is grams of sodium , how do we figure out moles of sodium ? we could divide by the molar mass , so we divide by the molar mass of sodium , which is 22.99 , and that would be grams per mole . if we divide by the molar mass , the grams cancel out , and we would get moles of sodium . so let 's get out the calculator . so we have 1.00 times 10 to the sixth , that 's how many grams of sodium we have . if we divide that by 22.99 , the molar mass of sodium , we get , this would be , let 's see , 4.35 times 10 to the one , two , three , four . so 4.35 times 10 to the fourth , let 's write that down , 4.35 times 10 to the fourth moles of sodium . so how does that help us ? if we have moles of sodium , how does that help us find our charge ? well , we can relate the moles of sodium to the moles of electrons . so let 's go back up here to our half-reaction where we have our moles of sodium . so right here , so we 're trying to make 4.35 times 10 to the fourth moles of sodium . so how many electrons do we need to do that ? well , the mole ratio is two to two , so you need the same number of moles of electrons , you need 4.35 times 10 to the fourth moles of electrons , and that helps us out . so let me go back down here , we have some more room , and let me write down 4.35 . 4.35 times 10 to the fourth , so because of our mole ratios we know this is the same number of moles of electrons that we need . we 're trying to find charge , we 're trying to find this charge here , and we know we can go from moles of electrons to the total charge using faraday 's constant . faraday 's constant is the charge carried by one mole of electrons . and we know that 's 96,500 . so there 's a charge of 96,500 coulombs for every one mole of electrons . so if we multiply these together , we 'll get charge , because the moles of electrons will cancel , and we will get the charge , so let 's do that . we have 4.35 times 10 to the fourth , and we 're going to multiply that by faraday 's constant , which is ... so , times , 96,500 , and so we get let 's see , that 'd be 4.2 times 10 to the , let 's see , one , two , three , four , five , six , seven , eight , nine , so 4.20 times 10 to the ninth , let 's write that down . 4.20 times 10 to the ninth , that would be coulombs , right ? that would be coulombs , that 's how much charge we need to make this many moles of sodium . so we have the charge , and so now we can plug that back into here , and solve for the time . so let 's do that . we have 3.00 times 10 to the fourth , that was our current , and current is equal to charge over time . so our total charge is 4.20 times 10 to the ninth , and that 's all over the time . so to solve for the time , we just need to divide this number , so let 's do that , so that 's 4.20 times 10 to the ninth , we 're going to divide that number by 3.00 times 10 to the fourth , and that should give us our time in seconds , so our time in seconds this would be 1.4 times 10 to the one , two , three , four , five , so 1.4 times 10 to the fifth . so the time in seconds is 1.40 times 10 to the fifth . so we have our time , i guess we could leave the answer like that . but let 's convert that to something that 's a little bit easier to comprehend . first , let 's convert that into minutes . so if you have 1.40 times 10 to the fifth seconds , how many minutes is that ? we could just divide that by 60 . so if we divide that by 60 , how many minutes ? that 's 2,333 minutes . how many hours is that ? we could just divide that by 60 , so we could figure out how many hours , so 38.9 hours , approximately , so it would take us approximately 38.9 hours so it would take us 38.9 hours to make , let 's go back up here to remind ourselves how much sodium to make 1.00 times 10 to the third kilograms of sodium . we can also figure out how many liters of chlorine gas are produced when we make our sodium . so let 's say we 're at stp here , so standard temperature and pressure . remember , standard temperature is zero degrees c , and standard pressure is one atmosphere . so we 've already found that we 're trying to make 4.35 times 10 to the fourth moles of sodium . so how many moles of chlorine gas will we make ? well , look at our mole ratios here . so this would be a two to one mole ratio . so we can set up a proportion , so we could do sodium to chlorine , so this would be a two to one mole ratio , so for every two moles of sodium that are produced , one mole of chlorine gas is produced . so 2/1 is equal to 4.35 times 10 to the fourth , that 's how many moles of sodium we were making , over x , where x represents the moles of chlorine that we would make . so two x is equal to 4.35 times 10 to the fourth , so x is equal to 4.35 times 10 to the fourth divided by two . so if we find 4.35 times 10 to the fourth , we need to divide that by two , and we get 21,750 moles of chlorine , so this would be 21,750 moles of chlorine gas that are produced . we 're trying to find liters of chlorine gas that are produced , so one thing we could do , just to make our lives easy , would be to say , we know that one mole of an ideal gas occupies 22.4 liters at stp . so if we have this many moles , and we assume this is an ideal gas , we could find out how many liters that is by multiplying that number by 22.4 . so 21,750 moles ... if we multiply that by 22.4 , then we should get the volume in liters . so let 's do it , let 's take that number and multiply it by 22.4 . and we get 4.87 , times 10 to the one , two , three four , five , so we get 4.87 times 10 to the fifth liters of chlorine . so there 's another way to do it , if you do n't want to take the shortcut , you can actually plug everything into pv is equal to nrt , so you could use piv-nert , so pv is equal to nrt , we 're at stp , so standard pressure is one atmosphere , and temperature is zero degrees c , or 273.15 kelvin . so we can plug those in , so our pressure is one atmosphere , one atmosphere , our temperature would be 273.15 , so 273.15 kelvin . we 're trying to find the volume , we 're trying to find liters , so we 're trying to find v , n would be equal to the number of moles , so that would be 21,750 , so we have 21,750 moles . r , r is equal to .0821 ... and i did n't leave room for the units in there . but the units would cancel out to give us liters for the volume , and if you solve this , if you use your calculator to solve this , you will get the same answer . you will get 4.87 times 10 to the fifth liters , so you plug it all into piv-nert or you could use the shortcut way . either way , you 're going to get the same volume of chlorine produced .
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so chloride anions , liquid chloride anions are oxidized to chlorine gas , so we lose two electrons . so chloride anions are oxidized to chlorine gas at the anode , and so we 'd have bubbles of chlorine gas forming at this electrode . this is an important reaction in industry .
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it makes chlorine gas while making metal right ?
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so the main thing i wanted to do in this video is just show you the timing of a heartbeat . and we 've been drawing the left ventricular pressure . and so far , i 've just sketched it out . but now , i want to try to be a little bit more careful with how i draw it so that you can get a real appreciation for how long everything seems to take . so these numbers i 'm going to write up are just estimates . they 're not exact numbers . and of course , you know that many things change how fast or slow a heartbeat can be . but they give you a real sense for the timing . so let 's get started . the left ventricle -- it begins , we know , with a pretty low pressure . let say this is about 50 millimeters mercury . i 'm just going to estimate this is about 10 . and we know that it 's going to begin contracting . and i have to pick a point somewhere , so i 'm just going to pick this point right here . it begins contracting at a pretty low pressure , about 10 millimeters of mercury , let 's say . and right before it contracts , the last thing that happens , remember , is atrial systole . and it 's a lump in the pressure . the pressure goes up , and then it slightly goes down . and that 's because of the atria contracting . and before that -- i 'm going backwards now , you see that ? before that , the atria and the ventricle are just slowly filling up with blood . and so the pressure 's just slowly creeping up . so that 's the first step in terms of what the left ventricle pressure looks like . it creeps up and then has that little bump in the end . now , it has to go from that point to a very high point . when the ventricle contracts , it 's going to skyrocket in pressure . and let 's say it 's going to get up to around 80 or so . and it does n't take much time . it actually does it all in about 0.05 seconds . it just shoots up like that . so it just skyrockets up . so that 's the next step . it rockets up . and really , let 's talk about these two points real quickly . these points here -- let 's call this b and i 'll call this a . and so between a and b , what 's happening exactly ? well , the big event at a is that the mitral valve closed . so i 'm going to write mitral closed . and then , of course , at b , the big event is that the aortic valve opened . so i 'm going to write aortic opened . and you know when i write aortic , i mean the valve , not the artery , because of course , the aorta is just an artery . it 's always open . but the aortic valve opened at that point . now , between those two -- and this is actually really a cool thing to think about -- between those two , what 's going on ? well here , when the mitral valve closes to the point where the other one opens , you 've got a chamber , the left ventricle , with kind of a room with two doors that are closed . there 's nothing open between these two spots where i 've drawn the red . and so because all of it 's closed , we actually have a special name for this because there 's contraction going on . the left ventricle is contracting . so we call that contraction . but because there 's a room with two doors closed , the blood has nowhere to go . so the volume of blood is not going to change . it 's going to be the same . and the medical word for that is `` iso . '' `` iso '' means same . so isovolumetric , same volume . so contraction . and all that means is that hey , the left ventricle 's contracting . and oh , by the way , the blood volume is not changing , because there 's nowhere for the blood to go . so a fancy word , but that 's what it means . so now the blood is going to start entering the aorta up here . and it 's going to really get a high pressure . and we know that it 's going to take a little bit of time for all that blood to go into the aorta . in fact , it takes about a quarter of a second . and at the end of it , we know that the pressure is going to be somewhere around 100 . it 's going to be around 100 , but my blood pressure is going to peak out somewhere higher than that , somewhere around 120 . in fact , i know that , because whenever i go to the doctor 's , they always check my blood pressure . and they tell me , hey , rishi , your blood pressure 's around 120/80 . so this is helping me draw my graph . so i can say , well , i know it has to get up to about 120 , because that 's what my doctor told me it was . and at some point , it 's going to dip down again to this point . and i 'm drawing it at 100 , because we know that the aortic valve is going to close at some point . and then , of course , you remember that whole dicrotic notch bit . but that 's roughly what it might look like . in fact , let me actually just draw that with a yellow . and the yellow just reminds us that now , at this point , blood is entering the aorta . this whole yellow bit is blood going into the aorta . and of course , you get to another important spot here . and you remember , let 's call this spot c. here , the aortic valve closes . so this is where the valve now says , hey , enough is enough . let 's shut down , because pressure is going the other way . and then , our blood pressure is going to fall . it 's going to fall , and it 's going to fall over a bit of time . it 's actually going to take about 0.15 seconds . and it 's going to go to about , let 's say -- i 'm just sketching it out -- let 's say about here . so about that point , let 's say , and it 's going to fall , fall , fall , fall , fall . and why did i choose this point ? well , that 's the point where we say , well , this is where what happens ? what happens at this point ? the mitral valve opens . the mitral valve opens at that spot . and then , of course , the pressure continues to fall . it gets pretty low , and then eventually it has to get back up to where we started . otherwise , the next heartbeat is not ready to go . so we have to creep our way back up as the blood fills in , and we 're done . so this part right here , this third segment , where i 'm going to use green , again you have the aortic valve is closed , but the mitral has not yet opened . and so does blood have anywhere to go ? nope . again , it 's stuck in a room . so if the blood is stuck in the left ventricle and the left ventricle is relaxing , then you better believe we 're going to have a fancy word for it . we 're going to call that relaxation . i guess not that fancy , but the first part of it is `` iso , '' same , volumetric . so same volume . isovolumetric relaxation , and that 's this part right here . because again , the blood has nowhere to go , and the left ventricle is relaxing . and the last chunk out here , i 'm going to do in a different color -- blue , let 's say -- is where blood is slowly just filling back into the left atrium . and obviously , since the mitral valve is open , also the left ventricle . so these are the four segments . and you might think well , wait a second . what about that first segment ? i did n't color that part in . and what i 'm going to do instead is i 'm going to say , well , let 's say this is 0.2 . so i 'm going to do the same thing over here . i 'm going to say what about the point 1.2 ? that would be equivalent . i 'm going to say blood pressure keeps rising . and let 's say we have our little atrial systole , something like that . so just to make it continuous instead of drawing two separate chunks , i think this will prove to you that it 's basically the same thing . so this is getting ready for the next heartbeat . but in terms of time , you would agree that that 's the same segment . so , then if i was actually to chunk it out , this part right here -- i 'm just going to draw it on the timeline axis -- is about 0.5 seconds . i 'm going to try to draw it big . i said 0.5 , sorry . 0.05 seconds . my mistake , sorry . so 0.05 seconds . the next chunk , this bit right here , is about 0.25 seconds . so you could say about a quarter of a second is right there . this is about a quarter of a second . and then you 've got the next chunk . this is about 0.15 seconds . and again , these numbers are not super important , but i just want you get a rough sense of simply the fact that this is actually not the same as the contraction bit . so it 's a little bit longer to relax . so just get that intuitive feel for that . and then finally , this last bit . this is obviously the longest bit . this is just going on and on and on . this is going to be about -- holy cow , it 's long -- 0.55 seconds . and of course , if you add up these four numbers , if you add them all up , they should add up to 1 second . because the whole point is that this is all happening in about a second . and that 's if we assume , of course , that our heart rate is 60 beats a minute . now , that is not always true , of course . certainly sometimes it 's much faster than that . but if we assume that , just for the sake of getting a sense or a feel for this stuff , then this might be a rough estimate of how it might look . now , one thing that i 've always thought is kind of interesting . when you look at this stuff , you think , ok , well , which parts are systole and which parts are diastole ? and these two , if you chunk them together , if you add them up , this makes up your systole . so that 's a nice way of thinking about it . and if you then add up the rest of it , this whole bit right here , this is your diastole . and you remember , we have talked about how the fact that diastole is about 2/3 of the time and systole is about 1/3 of the time . and you can see how that 's basically true here . and now , the final thing -- this is actually something that always threw me off , confused me a little bit -- is this chunk right here . i 've always wondered why this is n't part of systole . it certainly looks like it 's part of the lump , or the big mountain drawing . but the truth is that we have to remember that the left ventricle is relaxing during this time . and diastole is all about relaxation for the left ventricle . so because it 's part of relaxation , it is technically and truthfully part of diastole , even though it looks like it 's part of the lump . so just keep that in mind .
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so isovolumetric , same volume . so contraction . and all that means is that hey , the left ventricle 's contracting .
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how much do the relative proportions of contraction and relaxation discussed in the video change with rising heart rate ?
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so the main thing i wanted to do in this video is just show you the timing of a heartbeat . and we 've been drawing the left ventricular pressure . and so far , i 've just sketched it out . but now , i want to try to be a little bit more careful with how i draw it so that you can get a real appreciation for how long everything seems to take . so these numbers i 'm going to write up are just estimates . they 're not exact numbers . and of course , you know that many things change how fast or slow a heartbeat can be . but they give you a real sense for the timing . so let 's get started . the left ventricle -- it begins , we know , with a pretty low pressure . let say this is about 50 millimeters mercury . i 'm just going to estimate this is about 10 . and we know that it 's going to begin contracting . and i have to pick a point somewhere , so i 'm just going to pick this point right here . it begins contracting at a pretty low pressure , about 10 millimeters of mercury , let 's say . and right before it contracts , the last thing that happens , remember , is atrial systole . and it 's a lump in the pressure . the pressure goes up , and then it slightly goes down . and that 's because of the atria contracting . and before that -- i 'm going backwards now , you see that ? before that , the atria and the ventricle are just slowly filling up with blood . and so the pressure 's just slowly creeping up . so that 's the first step in terms of what the left ventricle pressure looks like . it creeps up and then has that little bump in the end . now , it has to go from that point to a very high point . when the ventricle contracts , it 's going to skyrocket in pressure . and let 's say it 's going to get up to around 80 or so . and it does n't take much time . it actually does it all in about 0.05 seconds . it just shoots up like that . so it just skyrockets up . so that 's the next step . it rockets up . and really , let 's talk about these two points real quickly . these points here -- let 's call this b and i 'll call this a . and so between a and b , what 's happening exactly ? well , the big event at a is that the mitral valve closed . so i 'm going to write mitral closed . and then , of course , at b , the big event is that the aortic valve opened . so i 'm going to write aortic opened . and you know when i write aortic , i mean the valve , not the artery , because of course , the aorta is just an artery . it 's always open . but the aortic valve opened at that point . now , between those two -- and this is actually really a cool thing to think about -- between those two , what 's going on ? well here , when the mitral valve closes to the point where the other one opens , you 've got a chamber , the left ventricle , with kind of a room with two doors that are closed . there 's nothing open between these two spots where i 've drawn the red . and so because all of it 's closed , we actually have a special name for this because there 's contraction going on . the left ventricle is contracting . so we call that contraction . but because there 's a room with two doors closed , the blood has nowhere to go . so the volume of blood is not going to change . it 's going to be the same . and the medical word for that is `` iso . '' `` iso '' means same . so isovolumetric , same volume . so contraction . and all that means is that hey , the left ventricle 's contracting . and oh , by the way , the blood volume is not changing , because there 's nowhere for the blood to go . so a fancy word , but that 's what it means . so now the blood is going to start entering the aorta up here . and it 's going to really get a high pressure . and we know that it 's going to take a little bit of time for all that blood to go into the aorta . in fact , it takes about a quarter of a second . and at the end of it , we know that the pressure is going to be somewhere around 100 . it 's going to be around 100 , but my blood pressure is going to peak out somewhere higher than that , somewhere around 120 . in fact , i know that , because whenever i go to the doctor 's , they always check my blood pressure . and they tell me , hey , rishi , your blood pressure 's around 120/80 . so this is helping me draw my graph . so i can say , well , i know it has to get up to about 120 , because that 's what my doctor told me it was . and at some point , it 's going to dip down again to this point . and i 'm drawing it at 100 , because we know that the aortic valve is going to close at some point . and then , of course , you remember that whole dicrotic notch bit . but that 's roughly what it might look like . in fact , let me actually just draw that with a yellow . and the yellow just reminds us that now , at this point , blood is entering the aorta . this whole yellow bit is blood going into the aorta . and of course , you get to another important spot here . and you remember , let 's call this spot c. here , the aortic valve closes . so this is where the valve now says , hey , enough is enough . let 's shut down , because pressure is going the other way . and then , our blood pressure is going to fall . it 's going to fall , and it 's going to fall over a bit of time . it 's actually going to take about 0.15 seconds . and it 's going to go to about , let 's say -- i 'm just sketching it out -- let 's say about here . so about that point , let 's say , and it 's going to fall , fall , fall , fall , fall . and why did i choose this point ? well , that 's the point where we say , well , this is where what happens ? what happens at this point ? the mitral valve opens . the mitral valve opens at that spot . and then , of course , the pressure continues to fall . it gets pretty low , and then eventually it has to get back up to where we started . otherwise , the next heartbeat is not ready to go . so we have to creep our way back up as the blood fills in , and we 're done . so this part right here , this third segment , where i 'm going to use green , again you have the aortic valve is closed , but the mitral has not yet opened . and so does blood have anywhere to go ? nope . again , it 's stuck in a room . so if the blood is stuck in the left ventricle and the left ventricle is relaxing , then you better believe we 're going to have a fancy word for it . we 're going to call that relaxation . i guess not that fancy , but the first part of it is `` iso , '' same , volumetric . so same volume . isovolumetric relaxation , and that 's this part right here . because again , the blood has nowhere to go , and the left ventricle is relaxing . and the last chunk out here , i 'm going to do in a different color -- blue , let 's say -- is where blood is slowly just filling back into the left atrium . and obviously , since the mitral valve is open , also the left ventricle . so these are the four segments . and you might think well , wait a second . what about that first segment ? i did n't color that part in . and what i 'm going to do instead is i 'm going to say , well , let 's say this is 0.2 . so i 'm going to do the same thing over here . i 'm going to say what about the point 1.2 ? that would be equivalent . i 'm going to say blood pressure keeps rising . and let 's say we have our little atrial systole , something like that . so just to make it continuous instead of drawing two separate chunks , i think this will prove to you that it 's basically the same thing . so this is getting ready for the next heartbeat . but in terms of time , you would agree that that 's the same segment . so , then if i was actually to chunk it out , this part right here -- i 'm just going to draw it on the timeline axis -- is about 0.5 seconds . i 'm going to try to draw it big . i said 0.5 , sorry . 0.05 seconds . my mistake , sorry . so 0.05 seconds . the next chunk , this bit right here , is about 0.25 seconds . so you could say about a quarter of a second is right there . this is about a quarter of a second . and then you 've got the next chunk . this is about 0.15 seconds . and again , these numbers are not super important , but i just want you get a rough sense of simply the fact that this is actually not the same as the contraction bit . so it 's a little bit longer to relax . so just get that intuitive feel for that . and then finally , this last bit . this is obviously the longest bit . this is just going on and on and on . this is going to be about -- holy cow , it 's long -- 0.55 seconds . and of course , if you add up these four numbers , if you add them all up , they should add up to 1 second . because the whole point is that this is all happening in about a second . and that 's if we assume , of course , that our heart rate is 60 beats a minute . now , that is not always true , of course . certainly sometimes it 's much faster than that . but if we assume that , just for the sake of getting a sense or a feel for this stuff , then this might be a rough estimate of how it might look . now , one thing that i 've always thought is kind of interesting . when you look at this stuff , you think , ok , well , which parts are systole and which parts are diastole ? and these two , if you chunk them together , if you add them up , this makes up your systole . so that 's a nice way of thinking about it . and if you then add up the rest of it , this whole bit right here , this is your diastole . and you remember , we have talked about how the fact that diastole is about 2/3 of the time and systole is about 1/3 of the time . and you can see how that 's basically true here . and now , the final thing -- this is actually something that always threw me off , confused me a little bit -- is this chunk right here . i 've always wondered why this is n't part of systole . it certainly looks like it 's part of the lump , or the big mountain drawing . but the truth is that we have to remember that the left ventricle is relaxing during this time . and diastole is all about relaxation for the left ventricle . so because it 's part of relaxation , it is technically and truthfully part of diastole , even though it looks like it 's part of the lump . so just keep that in mind .
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but the truth is that we have to remember that the left ventricle is relaxing during this time . and diastole is all about relaxation for the left ventricle . so because it 's part of relaxation , it is technically and truthfully part of diastole , even though it looks like it 's part of the lump .
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during the second half part of the yellow line ( at the top ) , there is relaxation of the ventricles therefore the left ventricle pressure drops , then why it is considered under the systole at the end of the video ?
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so the main thing i wanted to do in this video is just show you the timing of a heartbeat . and we 've been drawing the left ventricular pressure . and so far , i 've just sketched it out . but now , i want to try to be a little bit more careful with how i draw it so that you can get a real appreciation for how long everything seems to take . so these numbers i 'm going to write up are just estimates . they 're not exact numbers . and of course , you know that many things change how fast or slow a heartbeat can be . but they give you a real sense for the timing . so let 's get started . the left ventricle -- it begins , we know , with a pretty low pressure . let say this is about 50 millimeters mercury . i 'm just going to estimate this is about 10 . and we know that it 's going to begin contracting . and i have to pick a point somewhere , so i 'm just going to pick this point right here . it begins contracting at a pretty low pressure , about 10 millimeters of mercury , let 's say . and right before it contracts , the last thing that happens , remember , is atrial systole . and it 's a lump in the pressure . the pressure goes up , and then it slightly goes down . and that 's because of the atria contracting . and before that -- i 'm going backwards now , you see that ? before that , the atria and the ventricle are just slowly filling up with blood . and so the pressure 's just slowly creeping up . so that 's the first step in terms of what the left ventricle pressure looks like . it creeps up and then has that little bump in the end . now , it has to go from that point to a very high point . when the ventricle contracts , it 's going to skyrocket in pressure . and let 's say it 's going to get up to around 80 or so . and it does n't take much time . it actually does it all in about 0.05 seconds . it just shoots up like that . so it just skyrockets up . so that 's the next step . it rockets up . and really , let 's talk about these two points real quickly . these points here -- let 's call this b and i 'll call this a . and so between a and b , what 's happening exactly ? well , the big event at a is that the mitral valve closed . so i 'm going to write mitral closed . and then , of course , at b , the big event is that the aortic valve opened . so i 'm going to write aortic opened . and you know when i write aortic , i mean the valve , not the artery , because of course , the aorta is just an artery . it 's always open . but the aortic valve opened at that point . now , between those two -- and this is actually really a cool thing to think about -- between those two , what 's going on ? well here , when the mitral valve closes to the point where the other one opens , you 've got a chamber , the left ventricle , with kind of a room with two doors that are closed . there 's nothing open between these two spots where i 've drawn the red . and so because all of it 's closed , we actually have a special name for this because there 's contraction going on . the left ventricle is contracting . so we call that contraction . but because there 's a room with two doors closed , the blood has nowhere to go . so the volume of blood is not going to change . it 's going to be the same . and the medical word for that is `` iso . '' `` iso '' means same . so isovolumetric , same volume . so contraction . and all that means is that hey , the left ventricle 's contracting . and oh , by the way , the blood volume is not changing , because there 's nowhere for the blood to go . so a fancy word , but that 's what it means . so now the blood is going to start entering the aorta up here . and it 's going to really get a high pressure . and we know that it 's going to take a little bit of time for all that blood to go into the aorta . in fact , it takes about a quarter of a second . and at the end of it , we know that the pressure is going to be somewhere around 100 . it 's going to be around 100 , but my blood pressure is going to peak out somewhere higher than that , somewhere around 120 . in fact , i know that , because whenever i go to the doctor 's , they always check my blood pressure . and they tell me , hey , rishi , your blood pressure 's around 120/80 . so this is helping me draw my graph . so i can say , well , i know it has to get up to about 120 , because that 's what my doctor told me it was . and at some point , it 's going to dip down again to this point . and i 'm drawing it at 100 , because we know that the aortic valve is going to close at some point . and then , of course , you remember that whole dicrotic notch bit . but that 's roughly what it might look like . in fact , let me actually just draw that with a yellow . and the yellow just reminds us that now , at this point , blood is entering the aorta . this whole yellow bit is blood going into the aorta . and of course , you get to another important spot here . and you remember , let 's call this spot c. here , the aortic valve closes . so this is where the valve now says , hey , enough is enough . let 's shut down , because pressure is going the other way . and then , our blood pressure is going to fall . it 's going to fall , and it 's going to fall over a bit of time . it 's actually going to take about 0.15 seconds . and it 's going to go to about , let 's say -- i 'm just sketching it out -- let 's say about here . so about that point , let 's say , and it 's going to fall , fall , fall , fall , fall . and why did i choose this point ? well , that 's the point where we say , well , this is where what happens ? what happens at this point ? the mitral valve opens . the mitral valve opens at that spot . and then , of course , the pressure continues to fall . it gets pretty low , and then eventually it has to get back up to where we started . otherwise , the next heartbeat is not ready to go . so we have to creep our way back up as the blood fills in , and we 're done . so this part right here , this third segment , where i 'm going to use green , again you have the aortic valve is closed , but the mitral has not yet opened . and so does blood have anywhere to go ? nope . again , it 's stuck in a room . so if the blood is stuck in the left ventricle and the left ventricle is relaxing , then you better believe we 're going to have a fancy word for it . we 're going to call that relaxation . i guess not that fancy , but the first part of it is `` iso , '' same , volumetric . so same volume . isovolumetric relaxation , and that 's this part right here . because again , the blood has nowhere to go , and the left ventricle is relaxing . and the last chunk out here , i 'm going to do in a different color -- blue , let 's say -- is where blood is slowly just filling back into the left atrium . and obviously , since the mitral valve is open , also the left ventricle . so these are the four segments . and you might think well , wait a second . what about that first segment ? i did n't color that part in . and what i 'm going to do instead is i 'm going to say , well , let 's say this is 0.2 . so i 'm going to do the same thing over here . i 'm going to say what about the point 1.2 ? that would be equivalent . i 'm going to say blood pressure keeps rising . and let 's say we have our little atrial systole , something like that . so just to make it continuous instead of drawing two separate chunks , i think this will prove to you that it 's basically the same thing . so this is getting ready for the next heartbeat . but in terms of time , you would agree that that 's the same segment . so , then if i was actually to chunk it out , this part right here -- i 'm just going to draw it on the timeline axis -- is about 0.5 seconds . i 'm going to try to draw it big . i said 0.5 , sorry . 0.05 seconds . my mistake , sorry . so 0.05 seconds . the next chunk , this bit right here , is about 0.25 seconds . so you could say about a quarter of a second is right there . this is about a quarter of a second . and then you 've got the next chunk . this is about 0.15 seconds . and again , these numbers are not super important , but i just want you get a rough sense of simply the fact that this is actually not the same as the contraction bit . so it 's a little bit longer to relax . so just get that intuitive feel for that . and then finally , this last bit . this is obviously the longest bit . this is just going on and on and on . this is going to be about -- holy cow , it 's long -- 0.55 seconds . and of course , if you add up these four numbers , if you add them all up , they should add up to 1 second . because the whole point is that this is all happening in about a second . and that 's if we assume , of course , that our heart rate is 60 beats a minute . now , that is not always true , of course . certainly sometimes it 's much faster than that . but if we assume that , just for the sake of getting a sense or a feel for this stuff , then this might be a rough estimate of how it might look . now , one thing that i 've always thought is kind of interesting . when you look at this stuff , you think , ok , well , which parts are systole and which parts are diastole ? and these two , if you chunk them together , if you add them up , this makes up your systole . so that 's a nice way of thinking about it . and if you then add up the rest of it , this whole bit right here , this is your diastole . and you remember , we have talked about how the fact that diastole is about 2/3 of the time and systole is about 1/3 of the time . and you can see how that 's basically true here . and now , the final thing -- this is actually something that always threw me off , confused me a little bit -- is this chunk right here . i 've always wondered why this is n't part of systole . it certainly looks like it 's part of the lump , or the big mountain drawing . but the truth is that we have to remember that the left ventricle is relaxing during this time . and diastole is all about relaxation for the left ventricle . so because it 's part of relaxation , it is technically and truthfully part of diastole , even though it looks like it 's part of the lump . so just keep that in mind .
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because the whole point is that this is all happening in about a second . and that 's if we assume , of course , that our heart rate is 60 beats a minute . now , that is not always true , of course .
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can you explain me about right ventricle of heart and about heart beat ?
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so the main thing i wanted to do in this video is just show you the timing of a heartbeat . and we 've been drawing the left ventricular pressure . and so far , i 've just sketched it out . but now , i want to try to be a little bit more careful with how i draw it so that you can get a real appreciation for how long everything seems to take . so these numbers i 'm going to write up are just estimates . they 're not exact numbers . and of course , you know that many things change how fast or slow a heartbeat can be . but they give you a real sense for the timing . so let 's get started . the left ventricle -- it begins , we know , with a pretty low pressure . let say this is about 50 millimeters mercury . i 'm just going to estimate this is about 10 . and we know that it 's going to begin contracting . and i have to pick a point somewhere , so i 'm just going to pick this point right here . it begins contracting at a pretty low pressure , about 10 millimeters of mercury , let 's say . and right before it contracts , the last thing that happens , remember , is atrial systole . and it 's a lump in the pressure . the pressure goes up , and then it slightly goes down . and that 's because of the atria contracting . and before that -- i 'm going backwards now , you see that ? before that , the atria and the ventricle are just slowly filling up with blood . and so the pressure 's just slowly creeping up . so that 's the first step in terms of what the left ventricle pressure looks like . it creeps up and then has that little bump in the end . now , it has to go from that point to a very high point . when the ventricle contracts , it 's going to skyrocket in pressure . and let 's say it 's going to get up to around 80 or so . and it does n't take much time . it actually does it all in about 0.05 seconds . it just shoots up like that . so it just skyrockets up . so that 's the next step . it rockets up . and really , let 's talk about these two points real quickly . these points here -- let 's call this b and i 'll call this a . and so between a and b , what 's happening exactly ? well , the big event at a is that the mitral valve closed . so i 'm going to write mitral closed . and then , of course , at b , the big event is that the aortic valve opened . so i 'm going to write aortic opened . and you know when i write aortic , i mean the valve , not the artery , because of course , the aorta is just an artery . it 's always open . but the aortic valve opened at that point . now , between those two -- and this is actually really a cool thing to think about -- between those two , what 's going on ? well here , when the mitral valve closes to the point where the other one opens , you 've got a chamber , the left ventricle , with kind of a room with two doors that are closed . there 's nothing open between these two spots where i 've drawn the red . and so because all of it 's closed , we actually have a special name for this because there 's contraction going on . the left ventricle is contracting . so we call that contraction . but because there 's a room with two doors closed , the blood has nowhere to go . so the volume of blood is not going to change . it 's going to be the same . and the medical word for that is `` iso . '' `` iso '' means same . so isovolumetric , same volume . so contraction . and all that means is that hey , the left ventricle 's contracting . and oh , by the way , the blood volume is not changing , because there 's nowhere for the blood to go . so a fancy word , but that 's what it means . so now the blood is going to start entering the aorta up here . and it 's going to really get a high pressure . and we know that it 's going to take a little bit of time for all that blood to go into the aorta . in fact , it takes about a quarter of a second . and at the end of it , we know that the pressure is going to be somewhere around 100 . it 's going to be around 100 , but my blood pressure is going to peak out somewhere higher than that , somewhere around 120 . in fact , i know that , because whenever i go to the doctor 's , they always check my blood pressure . and they tell me , hey , rishi , your blood pressure 's around 120/80 . so this is helping me draw my graph . so i can say , well , i know it has to get up to about 120 , because that 's what my doctor told me it was . and at some point , it 's going to dip down again to this point . and i 'm drawing it at 100 , because we know that the aortic valve is going to close at some point . and then , of course , you remember that whole dicrotic notch bit . but that 's roughly what it might look like . in fact , let me actually just draw that with a yellow . and the yellow just reminds us that now , at this point , blood is entering the aorta . this whole yellow bit is blood going into the aorta . and of course , you get to another important spot here . and you remember , let 's call this spot c. here , the aortic valve closes . so this is where the valve now says , hey , enough is enough . let 's shut down , because pressure is going the other way . and then , our blood pressure is going to fall . it 's going to fall , and it 's going to fall over a bit of time . it 's actually going to take about 0.15 seconds . and it 's going to go to about , let 's say -- i 'm just sketching it out -- let 's say about here . so about that point , let 's say , and it 's going to fall , fall , fall , fall , fall . and why did i choose this point ? well , that 's the point where we say , well , this is where what happens ? what happens at this point ? the mitral valve opens . the mitral valve opens at that spot . and then , of course , the pressure continues to fall . it gets pretty low , and then eventually it has to get back up to where we started . otherwise , the next heartbeat is not ready to go . so we have to creep our way back up as the blood fills in , and we 're done . so this part right here , this third segment , where i 'm going to use green , again you have the aortic valve is closed , but the mitral has not yet opened . and so does blood have anywhere to go ? nope . again , it 's stuck in a room . so if the blood is stuck in the left ventricle and the left ventricle is relaxing , then you better believe we 're going to have a fancy word for it . we 're going to call that relaxation . i guess not that fancy , but the first part of it is `` iso , '' same , volumetric . so same volume . isovolumetric relaxation , and that 's this part right here . because again , the blood has nowhere to go , and the left ventricle is relaxing . and the last chunk out here , i 'm going to do in a different color -- blue , let 's say -- is where blood is slowly just filling back into the left atrium . and obviously , since the mitral valve is open , also the left ventricle . so these are the four segments . and you might think well , wait a second . what about that first segment ? i did n't color that part in . and what i 'm going to do instead is i 'm going to say , well , let 's say this is 0.2 . so i 'm going to do the same thing over here . i 'm going to say what about the point 1.2 ? that would be equivalent . i 'm going to say blood pressure keeps rising . and let 's say we have our little atrial systole , something like that . so just to make it continuous instead of drawing two separate chunks , i think this will prove to you that it 's basically the same thing . so this is getting ready for the next heartbeat . but in terms of time , you would agree that that 's the same segment . so , then if i was actually to chunk it out , this part right here -- i 'm just going to draw it on the timeline axis -- is about 0.5 seconds . i 'm going to try to draw it big . i said 0.5 , sorry . 0.05 seconds . my mistake , sorry . so 0.05 seconds . the next chunk , this bit right here , is about 0.25 seconds . so you could say about a quarter of a second is right there . this is about a quarter of a second . and then you 've got the next chunk . this is about 0.15 seconds . and again , these numbers are not super important , but i just want you get a rough sense of simply the fact that this is actually not the same as the contraction bit . so it 's a little bit longer to relax . so just get that intuitive feel for that . and then finally , this last bit . this is obviously the longest bit . this is just going on and on and on . this is going to be about -- holy cow , it 's long -- 0.55 seconds . and of course , if you add up these four numbers , if you add them all up , they should add up to 1 second . because the whole point is that this is all happening in about a second . and that 's if we assume , of course , that our heart rate is 60 beats a minute . now , that is not always true , of course . certainly sometimes it 's much faster than that . but if we assume that , just for the sake of getting a sense or a feel for this stuff , then this might be a rough estimate of how it might look . now , one thing that i 've always thought is kind of interesting . when you look at this stuff , you think , ok , well , which parts are systole and which parts are diastole ? and these two , if you chunk them together , if you add them up , this makes up your systole . so that 's a nice way of thinking about it . and if you then add up the rest of it , this whole bit right here , this is your diastole . and you remember , we have talked about how the fact that diastole is about 2/3 of the time and systole is about 1/3 of the time . and you can see how that 's basically true here . and now , the final thing -- this is actually something that always threw me off , confused me a little bit -- is this chunk right here . i 've always wondered why this is n't part of systole . it certainly looks like it 's part of the lump , or the big mountain drawing . but the truth is that we have to remember that the left ventricle is relaxing during this time . and diastole is all about relaxation for the left ventricle . so because it 's part of relaxation , it is technically and truthfully part of diastole , even though it looks like it 's part of the lump . so just keep that in mind .
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so the main thing i wanted to do in this video is just show you the timing of a heartbeat . and we 've been drawing the left ventricular pressure . and so far , i 've just sketched it out .
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how to draw right ventricular and atrial pressure curve ?
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so the main thing i wanted to do in this video is just show you the timing of a heartbeat . and we 've been drawing the left ventricular pressure . and so far , i 've just sketched it out . but now , i want to try to be a little bit more careful with how i draw it so that you can get a real appreciation for how long everything seems to take . so these numbers i 'm going to write up are just estimates . they 're not exact numbers . and of course , you know that many things change how fast or slow a heartbeat can be . but they give you a real sense for the timing . so let 's get started . the left ventricle -- it begins , we know , with a pretty low pressure . let say this is about 50 millimeters mercury . i 'm just going to estimate this is about 10 . and we know that it 's going to begin contracting . and i have to pick a point somewhere , so i 'm just going to pick this point right here . it begins contracting at a pretty low pressure , about 10 millimeters of mercury , let 's say . and right before it contracts , the last thing that happens , remember , is atrial systole . and it 's a lump in the pressure . the pressure goes up , and then it slightly goes down . and that 's because of the atria contracting . and before that -- i 'm going backwards now , you see that ? before that , the atria and the ventricle are just slowly filling up with blood . and so the pressure 's just slowly creeping up . so that 's the first step in terms of what the left ventricle pressure looks like . it creeps up and then has that little bump in the end . now , it has to go from that point to a very high point . when the ventricle contracts , it 's going to skyrocket in pressure . and let 's say it 's going to get up to around 80 or so . and it does n't take much time . it actually does it all in about 0.05 seconds . it just shoots up like that . so it just skyrockets up . so that 's the next step . it rockets up . and really , let 's talk about these two points real quickly . these points here -- let 's call this b and i 'll call this a . and so between a and b , what 's happening exactly ? well , the big event at a is that the mitral valve closed . so i 'm going to write mitral closed . and then , of course , at b , the big event is that the aortic valve opened . so i 'm going to write aortic opened . and you know when i write aortic , i mean the valve , not the artery , because of course , the aorta is just an artery . it 's always open . but the aortic valve opened at that point . now , between those two -- and this is actually really a cool thing to think about -- between those two , what 's going on ? well here , when the mitral valve closes to the point where the other one opens , you 've got a chamber , the left ventricle , with kind of a room with two doors that are closed . there 's nothing open between these two spots where i 've drawn the red . and so because all of it 's closed , we actually have a special name for this because there 's contraction going on . the left ventricle is contracting . so we call that contraction . but because there 's a room with two doors closed , the blood has nowhere to go . so the volume of blood is not going to change . it 's going to be the same . and the medical word for that is `` iso . '' `` iso '' means same . so isovolumetric , same volume . so contraction . and all that means is that hey , the left ventricle 's contracting . and oh , by the way , the blood volume is not changing , because there 's nowhere for the blood to go . so a fancy word , but that 's what it means . so now the blood is going to start entering the aorta up here . and it 's going to really get a high pressure . and we know that it 's going to take a little bit of time for all that blood to go into the aorta . in fact , it takes about a quarter of a second . and at the end of it , we know that the pressure is going to be somewhere around 100 . it 's going to be around 100 , but my blood pressure is going to peak out somewhere higher than that , somewhere around 120 . in fact , i know that , because whenever i go to the doctor 's , they always check my blood pressure . and they tell me , hey , rishi , your blood pressure 's around 120/80 . so this is helping me draw my graph . so i can say , well , i know it has to get up to about 120 , because that 's what my doctor told me it was . and at some point , it 's going to dip down again to this point . and i 'm drawing it at 100 , because we know that the aortic valve is going to close at some point . and then , of course , you remember that whole dicrotic notch bit . but that 's roughly what it might look like . in fact , let me actually just draw that with a yellow . and the yellow just reminds us that now , at this point , blood is entering the aorta . this whole yellow bit is blood going into the aorta . and of course , you get to another important spot here . and you remember , let 's call this spot c. here , the aortic valve closes . so this is where the valve now says , hey , enough is enough . let 's shut down , because pressure is going the other way . and then , our blood pressure is going to fall . it 's going to fall , and it 's going to fall over a bit of time . it 's actually going to take about 0.15 seconds . and it 's going to go to about , let 's say -- i 'm just sketching it out -- let 's say about here . so about that point , let 's say , and it 's going to fall , fall , fall , fall , fall . and why did i choose this point ? well , that 's the point where we say , well , this is where what happens ? what happens at this point ? the mitral valve opens . the mitral valve opens at that spot . and then , of course , the pressure continues to fall . it gets pretty low , and then eventually it has to get back up to where we started . otherwise , the next heartbeat is not ready to go . so we have to creep our way back up as the blood fills in , and we 're done . so this part right here , this third segment , where i 'm going to use green , again you have the aortic valve is closed , but the mitral has not yet opened . and so does blood have anywhere to go ? nope . again , it 's stuck in a room . so if the blood is stuck in the left ventricle and the left ventricle is relaxing , then you better believe we 're going to have a fancy word for it . we 're going to call that relaxation . i guess not that fancy , but the first part of it is `` iso , '' same , volumetric . so same volume . isovolumetric relaxation , and that 's this part right here . because again , the blood has nowhere to go , and the left ventricle is relaxing . and the last chunk out here , i 'm going to do in a different color -- blue , let 's say -- is where blood is slowly just filling back into the left atrium . and obviously , since the mitral valve is open , also the left ventricle . so these are the four segments . and you might think well , wait a second . what about that first segment ? i did n't color that part in . and what i 'm going to do instead is i 'm going to say , well , let 's say this is 0.2 . so i 'm going to do the same thing over here . i 'm going to say what about the point 1.2 ? that would be equivalent . i 'm going to say blood pressure keeps rising . and let 's say we have our little atrial systole , something like that . so just to make it continuous instead of drawing two separate chunks , i think this will prove to you that it 's basically the same thing . so this is getting ready for the next heartbeat . but in terms of time , you would agree that that 's the same segment . so , then if i was actually to chunk it out , this part right here -- i 'm just going to draw it on the timeline axis -- is about 0.5 seconds . i 'm going to try to draw it big . i said 0.5 , sorry . 0.05 seconds . my mistake , sorry . so 0.05 seconds . the next chunk , this bit right here , is about 0.25 seconds . so you could say about a quarter of a second is right there . this is about a quarter of a second . and then you 've got the next chunk . this is about 0.15 seconds . and again , these numbers are not super important , but i just want you get a rough sense of simply the fact that this is actually not the same as the contraction bit . so it 's a little bit longer to relax . so just get that intuitive feel for that . and then finally , this last bit . this is obviously the longest bit . this is just going on and on and on . this is going to be about -- holy cow , it 's long -- 0.55 seconds . and of course , if you add up these four numbers , if you add them all up , they should add up to 1 second . because the whole point is that this is all happening in about a second . and that 's if we assume , of course , that our heart rate is 60 beats a minute . now , that is not always true , of course . certainly sometimes it 's much faster than that . but if we assume that , just for the sake of getting a sense or a feel for this stuff , then this might be a rough estimate of how it might look . now , one thing that i 've always thought is kind of interesting . when you look at this stuff , you think , ok , well , which parts are systole and which parts are diastole ? and these two , if you chunk them together , if you add them up , this makes up your systole . so that 's a nice way of thinking about it . and if you then add up the rest of it , this whole bit right here , this is your diastole . and you remember , we have talked about how the fact that diastole is about 2/3 of the time and systole is about 1/3 of the time . and you can see how that 's basically true here . and now , the final thing -- this is actually something that always threw me off , confused me a little bit -- is this chunk right here . i 've always wondered why this is n't part of systole . it certainly looks like it 's part of the lump , or the big mountain drawing . but the truth is that we have to remember that the left ventricle is relaxing during this time . and diastole is all about relaxation for the left ventricle . so because it 's part of relaxation , it is technically and truthfully part of diastole , even though it looks like it 's part of the lump . so just keep that in mind .
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so the main thing i wanted to do in this video is just show you the timing of a heartbeat . and we 've been drawing the left ventricular pressure . and so far , i 've just sketched it out .
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can you explain me why did you put the ( atrial ) systole in the curve left ( ventricular ) curve , please ?
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so the main thing i wanted to do in this video is just show you the timing of a heartbeat . and we 've been drawing the left ventricular pressure . and so far , i 've just sketched it out . but now , i want to try to be a little bit more careful with how i draw it so that you can get a real appreciation for how long everything seems to take . so these numbers i 'm going to write up are just estimates . they 're not exact numbers . and of course , you know that many things change how fast or slow a heartbeat can be . but they give you a real sense for the timing . so let 's get started . the left ventricle -- it begins , we know , with a pretty low pressure . let say this is about 50 millimeters mercury . i 'm just going to estimate this is about 10 . and we know that it 's going to begin contracting . and i have to pick a point somewhere , so i 'm just going to pick this point right here . it begins contracting at a pretty low pressure , about 10 millimeters of mercury , let 's say . and right before it contracts , the last thing that happens , remember , is atrial systole . and it 's a lump in the pressure . the pressure goes up , and then it slightly goes down . and that 's because of the atria contracting . and before that -- i 'm going backwards now , you see that ? before that , the atria and the ventricle are just slowly filling up with blood . and so the pressure 's just slowly creeping up . so that 's the first step in terms of what the left ventricle pressure looks like . it creeps up and then has that little bump in the end . now , it has to go from that point to a very high point . when the ventricle contracts , it 's going to skyrocket in pressure . and let 's say it 's going to get up to around 80 or so . and it does n't take much time . it actually does it all in about 0.05 seconds . it just shoots up like that . so it just skyrockets up . so that 's the next step . it rockets up . and really , let 's talk about these two points real quickly . these points here -- let 's call this b and i 'll call this a . and so between a and b , what 's happening exactly ? well , the big event at a is that the mitral valve closed . so i 'm going to write mitral closed . and then , of course , at b , the big event is that the aortic valve opened . so i 'm going to write aortic opened . and you know when i write aortic , i mean the valve , not the artery , because of course , the aorta is just an artery . it 's always open . but the aortic valve opened at that point . now , between those two -- and this is actually really a cool thing to think about -- between those two , what 's going on ? well here , when the mitral valve closes to the point where the other one opens , you 've got a chamber , the left ventricle , with kind of a room with two doors that are closed . there 's nothing open between these two spots where i 've drawn the red . and so because all of it 's closed , we actually have a special name for this because there 's contraction going on . the left ventricle is contracting . so we call that contraction . but because there 's a room with two doors closed , the blood has nowhere to go . so the volume of blood is not going to change . it 's going to be the same . and the medical word for that is `` iso . '' `` iso '' means same . so isovolumetric , same volume . so contraction . and all that means is that hey , the left ventricle 's contracting . and oh , by the way , the blood volume is not changing , because there 's nowhere for the blood to go . so a fancy word , but that 's what it means . so now the blood is going to start entering the aorta up here . and it 's going to really get a high pressure . and we know that it 's going to take a little bit of time for all that blood to go into the aorta . in fact , it takes about a quarter of a second . and at the end of it , we know that the pressure is going to be somewhere around 100 . it 's going to be around 100 , but my blood pressure is going to peak out somewhere higher than that , somewhere around 120 . in fact , i know that , because whenever i go to the doctor 's , they always check my blood pressure . and they tell me , hey , rishi , your blood pressure 's around 120/80 . so this is helping me draw my graph . so i can say , well , i know it has to get up to about 120 , because that 's what my doctor told me it was . and at some point , it 's going to dip down again to this point . and i 'm drawing it at 100 , because we know that the aortic valve is going to close at some point . and then , of course , you remember that whole dicrotic notch bit . but that 's roughly what it might look like . in fact , let me actually just draw that with a yellow . and the yellow just reminds us that now , at this point , blood is entering the aorta . this whole yellow bit is blood going into the aorta . and of course , you get to another important spot here . and you remember , let 's call this spot c. here , the aortic valve closes . so this is where the valve now says , hey , enough is enough . let 's shut down , because pressure is going the other way . and then , our blood pressure is going to fall . it 's going to fall , and it 's going to fall over a bit of time . it 's actually going to take about 0.15 seconds . and it 's going to go to about , let 's say -- i 'm just sketching it out -- let 's say about here . so about that point , let 's say , and it 's going to fall , fall , fall , fall , fall . and why did i choose this point ? well , that 's the point where we say , well , this is where what happens ? what happens at this point ? the mitral valve opens . the mitral valve opens at that spot . and then , of course , the pressure continues to fall . it gets pretty low , and then eventually it has to get back up to where we started . otherwise , the next heartbeat is not ready to go . so we have to creep our way back up as the blood fills in , and we 're done . so this part right here , this third segment , where i 'm going to use green , again you have the aortic valve is closed , but the mitral has not yet opened . and so does blood have anywhere to go ? nope . again , it 's stuck in a room . so if the blood is stuck in the left ventricle and the left ventricle is relaxing , then you better believe we 're going to have a fancy word for it . we 're going to call that relaxation . i guess not that fancy , but the first part of it is `` iso , '' same , volumetric . so same volume . isovolumetric relaxation , and that 's this part right here . because again , the blood has nowhere to go , and the left ventricle is relaxing . and the last chunk out here , i 'm going to do in a different color -- blue , let 's say -- is where blood is slowly just filling back into the left atrium . and obviously , since the mitral valve is open , also the left ventricle . so these are the four segments . and you might think well , wait a second . what about that first segment ? i did n't color that part in . and what i 'm going to do instead is i 'm going to say , well , let 's say this is 0.2 . so i 'm going to do the same thing over here . i 'm going to say what about the point 1.2 ? that would be equivalent . i 'm going to say blood pressure keeps rising . and let 's say we have our little atrial systole , something like that . so just to make it continuous instead of drawing two separate chunks , i think this will prove to you that it 's basically the same thing . so this is getting ready for the next heartbeat . but in terms of time , you would agree that that 's the same segment . so , then if i was actually to chunk it out , this part right here -- i 'm just going to draw it on the timeline axis -- is about 0.5 seconds . i 'm going to try to draw it big . i said 0.5 , sorry . 0.05 seconds . my mistake , sorry . so 0.05 seconds . the next chunk , this bit right here , is about 0.25 seconds . so you could say about a quarter of a second is right there . this is about a quarter of a second . and then you 've got the next chunk . this is about 0.15 seconds . and again , these numbers are not super important , but i just want you get a rough sense of simply the fact that this is actually not the same as the contraction bit . so it 's a little bit longer to relax . so just get that intuitive feel for that . and then finally , this last bit . this is obviously the longest bit . this is just going on and on and on . this is going to be about -- holy cow , it 's long -- 0.55 seconds . and of course , if you add up these four numbers , if you add them all up , they should add up to 1 second . because the whole point is that this is all happening in about a second . and that 's if we assume , of course , that our heart rate is 60 beats a minute . now , that is not always true , of course . certainly sometimes it 's much faster than that . but if we assume that , just for the sake of getting a sense or a feel for this stuff , then this might be a rough estimate of how it might look . now , one thing that i 've always thought is kind of interesting . when you look at this stuff , you think , ok , well , which parts are systole and which parts are diastole ? and these two , if you chunk them together , if you add them up , this makes up your systole . so that 's a nice way of thinking about it . and if you then add up the rest of it , this whole bit right here , this is your diastole . and you remember , we have talked about how the fact that diastole is about 2/3 of the time and systole is about 1/3 of the time . and you can see how that 's basically true here . and now , the final thing -- this is actually something that always threw me off , confused me a little bit -- is this chunk right here . i 've always wondered why this is n't part of systole . it certainly looks like it 's part of the lump , or the big mountain drawing . but the truth is that we have to remember that the left ventricle is relaxing during this time . and diastole is all about relaxation for the left ventricle . so because it 's part of relaxation , it is technically and truthfully part of diastole , even though it looks like it 's part of the lump . so just keep that in mind .
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so this is where the valve now says , hey , enough is enough . let 's shut down , because pressure is going the other way . and then , our blood pressure is going to fall . it 's going to fall , and it 's going to fall over a bit of time .
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is the temperature of the blood going up during this time with the two valves closed , pressure going up but volume staying the same ?
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so the main thing i wanted to do in this video is just show you the timing of a heartbeat . and we 've been drawing the left ventricular pressure . and so far , i 've just sketched it out . but now , i want to try to be a little bit more careful with how i draw it so that you can get a real appreciation for how long everything seems to take . so these numbers i 'm going to write up are just estimates . they 're not exact numbers . and of course , you know that many things change how fast or slow a heartbeat can be . but they give you a real sense for the timing . so let 's get started . the left ventricle -- it begins , we know , with a pretty low pressure . let say this is about 50 millimeters mercury . i 'm just going to estimate this is about 10 . and we know that it 's going to begin contracting . and i have to pick a point somewhere , so i 'm just going to pick this point right here . it begins contracting at a pretty low pressure , about 10 millimeters of mercury , let 's say . and right before it contracts , the last thing that happens , remember , is atrial systole . and it 's a lump in the pressure . the pressure goes up , and then it slightly goes down . and that 's because of the atria contracting . and before that -- i 'm going backwards now , you see that ? before that , the atria and the ventricle are just slowly filling up with blood . and so the pressure 's just slowly creeping up . so that 's the first step in terms of what the left ventricle pressure looks like . it creeps up and then has that little bump in the end . now , it has to go from that point to a very high point . when the ventricle contracts , it 's going to skyrocket in pressure . and let 's say it 's going to get up to around 80 or so . and it does n't take much time . it actually does it all in about 0.05 seconds . it just shoots up like that . so it just skyrockets up . so that 's the next step . it rockets up . and really , let 's talk about these two points real quickly . these points here -- let 's call this b and i 'll call this a . and so between a and b , what 's happening exactly ? well , the big event at a is that the mitral valve closed . so i 'm going to write mitral closed . and then , of course , at b , the big event is that the aortic valve opened . so i 'm going to write aortic opened . and you know when i write aortic , i mean the valve , not the artery , because of course , the aorta is just an artery . it 's always open . but the aortic valve opened at that point . now , between those two -- and this is actually really a cool thing to think about -- between those two , what 's going on ? well here , when the mitral valve closes to the point where the other one opens , you 've got a chamber , the left ventricle , with kind of a room with two doors that are closed . there 's nothing open between these two spots where i 've drawn the red . and so because all of it 's closed , we actually have a special name for this because there 's contraction going on . the left ventricle is contracting . so we call that contraction . but because there 's a room with two doors closed , the blood has nowhere to go . so the volume of blood is not going to change . it 's going to be the same . and the medical word for that is `` iso . '' `` iso '' means same . so isovolumetric , same volume . so contraction . and all that means is that hey , the left ventricle 's contracting . and oh , by the way , the blood volume is not changing , because there 's nowhere for the blood to go . so a fancy word , but that 's what it means . so now the blood is going to start entering the aorta up here . and it 's going to really get a high pressure . and we know that it 's going to take a little bit of time for all that blood to go into the aorta . in fact , it takes about a quarter of a second . and at the end of it , we know that the pressure is going to be somewhere around 100 . it 's going to be around 100 , but my blood pressure is going to peak out somewhere higher than that , somewhere around 120 . in fact , i know that , because whenever i go to the doctor 's , they always check my blood pressure . and they tell me , hey , rishi , your blood pressure 's around 120/80 . so this is helping me draw my graph . so i can say , well , i know it has to get up to about 120 , because that 's what my doctor told me it was . and at some point , it 's going to dip down again to this point . and i 'm drawing it at 100 , because we know that the aortic valve is going to close at some point . and then , of course , you remember that whole dicrotic notch bit . but that 's roughly what it might look like . in fact , let me actually just draw that with a yellow . and the yellow just reminds us that now , at this point , blood is entering the aorta . this whole yellow bit is blood going into the aorta . and of course , you get to another important spot here . and you remember , let 's call this spot c. here , the aortic valve closes . so this is where the valve now says , hey , enough is enough . let 's shut down , because pressure is going the other way . and then , our blood pressure is going to fall . it 's going to fall , and it 's going to fall over a bit of time . it 's actually going to take about 0.15 seconds . and it 's going to go to about , let 's say -- i 'm just sketching it out -- let 's say about here . so about that point , let 's say , and it 's going to fall , fall , fall , fall , fall . and why did i choose this point ? well , that 's the point where we say , well , this is where what happens ? what happens at this point ? the mitral valve opens . the mitral valve opens at that spot . and then , of course , the pressure continues to fall . it gets pretty low , and then eventually it has to get back up to where we started . otherwise , the next heartbeat is not ready to go . so we have to creep our way back up as the blood fills in , and we 're done . so this part right here , this third segment , where i 'm going to use green , again you have the aortic valve is closed , but the mitral has not yet opened . and so does blood have anywhere to go ? nope . again , it 's stuck in a room . so if the blood is stuck in the left ventricle and the left ventricle is relaxing , then you better believe we 're going to have a fancy word for it . we 're going to call that relaxation . i guess not that fancy , but the first part of it is `` iso , '' same , volumetric . so same volume . isovolumetric relaxation , and that 's this part right here . because again , the blood has nowhere to go , and the left ventricle is relaxing . and the last chunk out here , i 'm going to do in a different color -- blue , let 's say -- is where blood is slowly just filling back into the left atrium . and obviously , since the mitral valve is open , also the left ventricle . so these are the four segments . and you might think well , wait a second . what about that first segment ? i did n't color that part in . and what i 'm going to do instead is i 'm going to say , well , let 's say this is 0.2 . so i 'm going to do the same thing over here . i 'm going to say what about the point 1.2 ? that would be equivalent . i 'm going to say blood pressure keeps rising . and let 's say we have our little atrial systole , something like that . so just to make it continuous instead of drawing two separate chunks , i think this will prove to you that it 's basically the same thing . so this is getting ready for the next heartbeat . but in terms of time , you would agree that that 's the same segment . so , then if i was actually to chunk it out , this part right here -- i 'm just going to draw it on the timeline axis -- is about 0.5 seconds . i 'm going to try to draw it big . i said 0.5 , sorry . 0.05 seconds . my mistake , sorry . so 0.05 seconds . the next chunk , this bit right here , is about 0.25 seconds . so you could say about a quarter of a second is right there . this is about a quarter of a second . and then you 've got the next chunk . this is about 0.15 seconds . and again , these numbers are not super important , but i just want you get a rough sense of simply the fact that this is actually not the same as the contraction bit . so it 's a little bit longer to relax . so just get that intuitive feel for that . and then finally , this last bit . this is obviously the longest bit . this is just going on and on and on . this is going to be about -- holy cow , it 's long -- 0.55 seconds . and of course , if you add up these four numbers , if you add them all up , they should add up to 1 second . because the whole point is that this is all happening in about a second . and that 's if we assume , of course , that our heart rate is 60 beats a minute . now , that is not always true , of course . certainly sometimes it 's much faster than that . but if we assume that , just for the sake of getting a sense or a feel for this stuff , then this might be a rough estimate of how it might look . now , one thing that i 've always thought is kind of interesting . when you look at this stuff , you think , ok , well , which parts are systole and which parts are diastole ? and these two , if you chunk them together , if you add them up , this makes up your systole . so that 's a nice way of thinking about it . and if you then add up the rest of it , this whole bit right here , this is your diastole . and you remember , we have talked about how the fact that diastole is about 2/3 of the time and systole is about 1/3 of the time . and you can see how that 's basically true here . and now , the final thing -- this is actually something that always threw me off , confused me a little bit -- is this chunk right here . i 've always wondered why this is n't part of systole . it certainly looks like it 's part of the lump , or the big mountain drawing . but the truth is that we have to remember that the left ventricle is relaxing during this time . and diastole is all about relaxation for the left ventricle . so because it 's part of relaxation , it is technically and truthfully part of diastole , even though it looks like it 's part of the lump . so just keep that in mind .
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let 's shut down , because pressure is going the other way . and then , our blood pressure is going to fall . it 's going to fall , and it 's going to fall over a bit of time .
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i am confused by the pressure in the graph , is it the force exerted by the ventricles or the pressure by the blood on the ventricles ?
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so the main thing i wanted to do in this video is just show you the timing of a heartbeat . and we 've been drawing the left ventricular pressure . and so far , i 've just sketched it out . but now , i want to try to be a little bit more careful with how i draw it so that you can get a real appreciation for how long everything seems to take . so these numbers i 'm going to write up are just estimates . they 're not exact numbers . and of course , you know that many things change how fast or slow a heartbeat can be . but they give you a real sense for the timing . so let 's get started . the left ventricle -- it begins , we know , with a pretty low pressure . let say this is about 50 millimeters mercury . i 'm just going to estimate this is about 10 . and we know that it 's going to begin contracting . and i have to pick a point somewhere , so i 'm just going to pick this point right here . it begins contracting at a pretty low pressure , about 10 millimeters of mercury , let 's say . and right before it contracts , the last thing that happens , remember , is atrial systole . and it 's a lump in the pressure . the pressure goes up , and then it slightly goes down . and that 's because of the atria contracting . and before that -- i 'm going backwards now , you see that ? before that , the atria and the ventricle are just slowly filling up with blood . and so the pressure 's just slowly creeping up . so that 's the first step in terms of what the left ventricle pressure looks like . it creeps up and then has that little bump in the end . now , it has to go from that point to a very high point . when the ventricle contracts , it 's going to skyrocket in pressure . and let 's say it 's going to get up to around 80 or so . and it does n't take much time . it actually does it all in about 0.05 seconds . it just shoots up like that . so it just skyrockets up . so that 's the next step . it rockets up . and really , let 's talk about these two points real quickly . these points here -- let 's call this b and i 'll call this a . and so between a and b , what 's happening exactly ? well , the big event at a is that the mitral valve closed . so i 'm going to write mitral closed . and then , of course , at b , the big event is that the aortic valve opened . so i 'm going to write aortic opened . and you know when i write aortic , i mean the valve , not the artery , because of course , the aorta is just an artery . it 's always open . but the aortic valve opened at that point . now , between those two -- and this is actually really a cool thing to think about -- between those two , what 's going on ? well here , when the mitral valve closes to the point where the other one opens , you 've got a chamber , the left ventricle , with kind of a room with two doors that are closed . there 's nothing open between these two spots where i 've drawn the red . and so because all of it 's closed , we actually have a special name for this because there 's contraction going on . the left ventricle is contracting . so we call that contraction . but because there 's a room with two doors closed , the blood has nowhere to go . so the volume of blood is not going to change . it 's going to be the same . and the medical word for that is `` iso . '' `` iso '' means same . so isovolumetric , same volume . so contraction . and all that means is that hey , the left ventricle 's contracting . and oh , by the way , the blood volume is not changing , because there 's nowhere for the blood to go . so a fancy word , but that 's what it means . so now the blood is going to start entering the aorta up here . and it 's going to really get a high pressure . and we know that it 's going to take a little bit of time for all that blood to go into the aorta . in fact , it takes about a quarter of a second . and at the end of it , we know that the pressure is going to be somewhere around 100 . it 's going to be around 100 , but my blood pressure is going to peak out somewhere higher than that , somewhere around 120 . in fact , i know that , because whenever i go to the doctor 's , they always check my blood pressure . and they tell me , hey , rishi , your blood pressure 's around 120/80 . so this is helping me draw my graph . so i can say , well , i know it has to get up to about 120 , because that 's what my doctor told me it was . and at some point , it 's going to dip down again to this point . and i 'm drawing it at 100 , because we know that the aortic valve is going to close at some point . and then , of course , you remember that whole dicrotic notch bit . but that 's roughly what it might look like . in fact , let me actually just draw that with a yellow . and the yellow just reminds us that now , at this point , blood is entering the aorta . this whole yellow bit is blood going into the aorta . and of course , you get to another important spot here . and you remember , let 's call this spot c. here , the aortic valve closes . so this is where the valve now says , hey , enough is enough . let 's shut down , because pressure is going the other way . and then , our blood pressure is going to fall . it 's going to fall , and it 's going to fall over a bit of time . it 's actually going to take about 0.15 seconds . and it 's going to go to about , let 's say -- i 'm just sketching it out -- let 's say about here . so about that point , let 's say , and it 's going to fall , fall , fall , fall , fall . and why did i choose this point ? well , that 's the point where we say , well , this is where what happens ? what happens at this point ? the mitral valve opens . the mitral valve opens at that spot . and then , of course , the pressure continues to fall . it gets pretty low , and then eventually it has to get back up to where we started . otherwise , the next heartbeat is not ready to go . so we have to creep our way back up as the blood fills in , and we 're done . so this part right here , this third segment , where i 'm going to use green , again you have the aortic valve is closed , but the mitral has not yet opened . and so does blood have anywhere to go ? nope . again , it 's stuck in a room . so if the blood is stuck in the left ventricle and the left ventricle is relaxing , then you better believe we 're going to have a fancy word for it . we 're going to call that relaxation . i guess not that fancy , but the first part of it is `` iso , '' same , volumetric . so same volume . isovolumetric relaxation , and that 's this part right here . because again , the blood has nowhere to go , and the left ventricle is relaxing . and the last chunk out here , i 'm going to do in a different color -- blue , let 's say -- is where blood is slowly just filling back into the left atrium . and obviously , since the mitral valve is open , also the left ventricle . so these are the four segments . and you might think well , wait a second . what about that first segment ? i did n't color that part in . and what i 'm going to do instead is i 'm going to say , well , let 's say this is 0.2 . so i 'm going to do the same thing over here . i 'm going to say what about the point 1.2 ? that would be equivalent . i 'm going to say blood pressure keeps rising . and let 's say we have our little atrial systole , something like that . so just to make it continuous instead of drawing two separate chunks , i think this will prove to you that it 's basically the same thing . so this is getting ready for the next heartbeat . but in terms of time , you would agree that that 's the same segment . so , then if i was actually to chunk it out , this part right here -- i 'm just going to draw it on the timeline axis -- is about 0.5 seconds . i 'm going to try to draw it big . i said 0.5 , sorry . 0.05 seconds . my mistake , sorry . so 0.05 seconds . the next chunk , this bit right here , is about 0.25 seconds . so you could say about a quarter of a second is right there . this is about a quarter of a second . and then you 've got the next chunk . this is about 0.15 seconds . and again , these numbers are not super important , but i just want you get a rough sense of simply the fact that this is actually not the same as the contraction bit . so it 's a little bit longer to relax . so just get that intuitive feel for that . and then finally , this last bit . this is obviously the longest bit . this is just going on and on and on . this is going to be about -- holy cow , it 's long -- 0.55 seconds . and of course , if you add up these four numbers , if you add them all up , they should add up to 1 second . because the whole point is that this is all happening in about a second . and that 's if we assume , of course , that our heart rate is 60 beats a minute . now , that is not always true , of course . certainly sometimes it 's much faster than that . but if we assume that , just for the sake of getting a sense or a feel for this stuff , then this might be a rough estimate of how it might look . now , one thing that i 've always thought is kind of interesting . when you look at this stuff , you think , ok , well , which parts are systole and which parts are diastole ? and these two , if you chunk them together , if you add them up , this makes up your systole . so that 's a nice way of thinking about it . and if you then add up the rest of it , this whole bit right here , this is your diastole . and you remember , we have talked about how the fact that diastole is about 2/3 of the time and systole is about 1/3 of the time . and you can see how that 's basically true here . and now , the final thing -- this is actually something that always threw me off , confused me a little bit -- is this chunk right here . i 've always wondered why this is n't part of systole . it certainly looks like it 's part of the lump , or the big mountain drawing . but the truth is that we have to remember that the left ventricle is relaxing during this time . and diastole is all about relaxation for the left ventricle . so because it 's part of relaxation , it is technically and truthfully part of diastole , even though it looks like it 's part of the lump . so just keep that in mind .
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so this part right here , this third segment , where i 'm going to use green , again you have the aortic valve is closed , but the mitral has not yet opened . and so does blood have anywhere to go ? nope .
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in the part of isovolumetric relaxation u said that the blood has no where to go ... now where does that blood come from if the ventricle has just finished contracting and pushing blood into the aorta ?
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so the main thing i wanted to do in this video is just show you the timing of a heartbeat . and we 've been drawing the left ventricular pressure . and so far , i 've just sketched it out . but now , i want to try to be a little bit more careful with how i draw it so that you can get a real appreciation for how long everything seems to take . so these numbers i 'm going to write up are just estimates . they 're not exact numbers . and of course , you know that many things change how fast or slow a heartbeat can be . but they give you a real sense for the timing . so let 's get started . the left ventricle -- it begins , we know , with a pretty low pressure . let say this is about 50 millimeters mercury . i 'm just going to estimate this is about 10 . and we know that it 's going to begin contracting . and i have to pick a point somewhere , so i 'm just going to pick this point right here . it begins contracting at a pretty low pressure , about 10 millimeters of mercury , let 's say . and right before it contracts , the last thing that happens , remember , is atrial systole . and it 's a lump in the pressure . the pressure goes up , and then it slightly goes down . and that 's because of the atria contracting . and before that -- i 'm going backwards now , you see that ? before that , the atria and the ventricle are just slowly filling up with blood . and so the pressure 's just slowly creeping up . so that 's the first step in terms of what the left ventricle pressure looks like . it creeps up and then has that little bump in the end . now , it has to go from that point to a very high point . when the ventricle contracts , it 's going to skyrocket in pressure . and let 's say it 's going to get up to around 80 or so . and it does n't take much time . it actually does it all in about 0.05 seconds . it just shoots up like that . so it just skyrockets up . so that 's the next step . it rockets up . and really , let 's talk about these two points real quickly . these points here -- let 's call this b and i 'll call this a . and so between a and b , what 's happening exactly ? well , the big event at a is that the mitral valve closed . so i 'm going to write mitral closed . and then , of course , at b , the big event is that the aortic valve opened . so i 'm going to write aortic opened . and you know when i write aortic , i mean the valve , not the artery , because of course , the aorta is just an artery . it 's always open . but the aortic valve opened at that point . now , between those two -- and this is actually really a cool thing to think about -- between those two , what 's going on ? well here , when the mitral valve closes to the point where the other one opens , you 've got a chamber , the left ventricle , with kind of a room with two doors that are closed . there 's nothing open between these two spots where i 've drawn the red . and so because all of it 's closed , we actually have a special name for this because there 's contraction going on . the left ventricle is contracting . so we call that contraction . but because there 's a room with two doors closed , the blood has nowhere to go . so the volume of blood is not going to change . it 's going to be the same . and the medical word for that is `` iso . '' `` iso '' means same . so isovolumetric , same volume . so contraction . and all that means is that hey , the left ventricle 's contracting . and oh , by the way , the blood volume is not changing , because there 's nowhere for the blood to go . so a fancy word , but that 's what it means . so now the blood is going to start entering the aorta up here . and it 's going to really get a high pressure . and we know that it 's going to take a little bit of time for all that blood to go into the aorta . in fact , it takes about a quarter of a second . and at the end of it , we know that the pressure is going to be somewhere around 100 . it 's going to be around 100 , but my blood pressure is going to peak out somewhere higher than that , somewhere around 120 . in fact , i know that , because whenever i go to the doctor 's , they always check my blood pressure . and they tell me , hey , rishi , your blood pressure 's around 120/80 . so this is helping me draw my graph . so i can say , well , i know it has to get up to about 120 , because that 's what my doctor told me it was . and at some point , it 's going to dip down again to this point . and i 'm drawing it at 100 , because we know that the aortic valve is going to close at some point . and then , of course , you remember that whole dicrotic notch bit . but that 's roughly what it might look like . in fact , let me actually just draw that with a yellow . and the yellow just reminds us that now , at this point , blood is entering the aorta . this whole yellow bit is blood going into the aorta . and of course , you get to another important spot here . and you remember , let 's call this spot c. here , the aortic valve closes . so this is where the valve now says , hey , enough is enough . let 's shut down , because pressure is going the other way . and then , our blood pressure is going to fall . it 's going to fall , and it 's going to fall over a bit of time . it 's actually going to take about 0.15 seconds . and it 's going to go to about , let 's say -- i 'm just sketching it out -- let 's say about here . so about that point , let 's say , and it 's going to fall , fall , fall , fall , fall . and why did i choose this point ? well , that 's the point where we say , well , this is where what happens ? what happens at this point ? the mitral valve opens . the mitral valve opens at that spot . and then , of course , the pressure continues to fall . it gets pretty low , and then eventually it has to get back up to where we started . otherwise , the next heartbeat is not ready to go . so we have to creep our way back up as the blood fills in , and we 're done . so this part right here , this third segment , where i 'm going to use green , again you have the aortic valve is closed , but the mitral has not yet opened . and so does blood have anywhere to go ? nope . again , it 's stuck in a room . so if the blood is stuck in the left ventricle and the left ventricle is relaxing , then you better believe we 're going to have a fancy word for it . we 're going to call that relaxation . i guess not that fancy , but the first part of it is `` iso , '' same , volumetric . so same volume . isovolumetric relaxation , and that 's this part right here . because again , the blood has nowhere to go , and the left ventricle is relaxing . and the last chunk out here , i 'm going to do in a different color -- blue , let 's say -- is where blood is slowly just filling back into the left atrium . and obviously , since the mitral valve is open , also the left ventricle . so these are the four segments . and you might think well , wait a second . what about that first segment ? i did n't color that part in . and what i 'm going to do instead is i 'm going to say , well , let 's say this is 0.2 . so i 'm going to do the same thing over here . i 'm going to say what about the point 1.2 ? that would be equivalent . i 'm going to say blood pressure keeps rising . and let 's say we have our little atrial systole , something like that . so just to make it continuous instead of drawing two separate chunks , i think this will prove to you that it 's basically the same thing . so this is getting ready for the next heartbeat . but in terms of time , you would agree that that 's the same segment . so , then if i was actually to chunk it out , this part right here -- i 'm just going to draw it on the timeline axis -- is about 0.5 seconds . i 'm going to try to draw it big . i said 0.5 , sorry . 0.05 seconds . my mistake , sorry . so 0.05 seconds . the next chunk , this bit right here , is about 0.25 seconds . so you could say about a quarter of a second is right there . this is about a quarter of a second . and then you 've got the next chunk . this is about 0.15 seconds . and again , these numbers are not super important , but i just want you get a rough sense of simply the fact that this is actually not the same as the contraction bit . so it 's a little bit longer to relax . so just get that intuitive feel for that . and then finally , this last bit . this is obviously the longest bit . this is just going on and on and on . this is going to be about -- holy cow , it 's long -- 0.55 seconds . and of course , if you add up these four numbers , if you add them all up , they should add up to 1 second . because the whole point is that this is all happening in about a second . and that 's if we assume , of course , that our heart rate is 60 beats a minute . now , that is not always true , of course . certainly sometimes it 's much faster than that . but if we assume that , just for the sake of getting a sense or a feel for this stuff , then this might be a rough estimate of how it might look . now , one thing that i 've always thought is kind of interesting . when you look at this stuff , you think , ok , well , which parts are systole and which parts are diastole ? and these two , if you chunk them together , if you add them up , this makes up your systole . so that 's a nice way of thinking about it . and if you then add up the rest of it , this whole bit right here , this is your diastole . and you remember , we have talked about how the fact that diastole is about 2/3 of the time and systole is about 1/3 of the time . and you can see how that 's basically true here . and now , the final thing -- this is actually something that always threw me off , confused me a little bit -- is this chunk right here . i 've always wondered why this is n't part of systole . it certainly looks like it 's part of the lump , or the big mountain drawing . but the truth is that we have to remember that the left ventricle is relaxing during this time . and diastole is all about relaxation for the left ventricle . so because it 's part of relaxation , it is technically and truthfully part of diastole , even though it looks like it 's part of the lump . so just keep that in mind .
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it 's always open . but the aortic valve opened at that point . now , between those two -- and this is actually really a cool thing to think about -- between those two , what 's going on ?
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or there is some blood that remains in the ventricle even after contarction is over and aortic valve shuts ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color .
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why does n't sal shade in the yellow circle ( which marks the beginning of the red line ) 4 ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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is the tangent line the derivative of the function at the certain point ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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is there a case in which we have a curve in the derivative , or are there only lines ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here .
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whats the difference between the shaded and not-shaded circles ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color .
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around sal draws a open dot to start the blue line - should n't it be closed because we know it 's zero ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about .
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0 , should the graph of the derivative resemble something like the graph of tangent ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 .
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why is the derivative of second and third function a horizontal line ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there .
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so does the y-axis on the lower graph represent the slope of the tangent lines ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive .
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why does sal not shade in the second blue circle ?
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