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# Solving quadratic equations: complex roots

CCSS.Math:

## Video transcript

we're asked to solve 2x squared plus 5 is equal to 6x and so we have a quadratic equation here but just to make it put it into a form that we're more familiar with let's try to put it into standard form in standard form of course is the form ax squared plus BX plus C is equal to 0 and to do that we essentially have to take the 6x and get rid of it from the right hand side so we just have a 0 on the right hand side and to do that let's just subtract 6x from both sides of this equation and so our left hand side becomes 2x squared minus 6x minus 6x plus 5 is equal to it on our right hand side these two characters cancel out and we just are left with 0 and there's many ways to solve this we could try to factor it and if I was trying to factor it I would divide both sides by 2 if I divide both sides by 2 I would get integer coefficients on the x squared and the X term but I would get 5 halves for the constant so it's not one of these easy things to factor we could complete the square or we could apply the quadratic formula which is really just to form the derived from completing the square so let's do that in this scenario and the quadratic formula tells us that if we have something in standard form like this that the roots of it are going to be negative B plus or minus so that gives us two roots right over there plus or minus square root of b squared minus 4ac over over 2a so let's apply that to this situation negative B this right here is B so negative B is negative negative 6 so that's going to be positive 6 plus or minus the square root of B squared negative 6 squared is 36 minus 4 minus 4 times a which is 2 times 2 times C which is 5 times 5 all of that over 2 times a a is 2 so 2 times 2 is 4 so this is going to be equal to 6 plus or minus the square root of 36 so let me just figure this out 36 minus so this is 4 times 2 times 5 this is 40 over here so 36 minus 40 and you already might be wondering what's going to happen here all of that over 4 or this is equal to 6 plus or minus the square root of negative 4 right 36 minus 40 is negative 4 over 4 and you might say hey wait Sal negative 4 if I take a square root I'm going to get an imaginary number and you would be right the only two roots of this quadratic equation right here are going to turn out to be complex because we're going to get when we evaluate this we're going to get an imaginary number so we're essentially going to get two complex numbers when we take the positive and negative version of this root so let's do that so the square root of negative 4 that is the same thing as 2i we know that's the same thing as 2i or if you want to think of it this way square root of negative 4 is the same thing as the square root of negative 1 times the square root of 4 which is the same well I could even take you one step that's the same thing as negative 1 times 4 under the radical which is the same thing as the square root of negative 1 times the square root of 4 and the principal square root of negative 1 is I times the principal square root of 4 is 2 so this is 2 I or I times 2 so this right over here is going to be 2i so we are left with X is equal to 6 plus or minus 2i over 4 and if we were to simplify it we can divide the numerator and the denominator by 2 and so that would be the same thing as 3 plus or minus I over 2 or if you want to write them as two distinct complex numbers you could write this as 3 plus I over 2 or 3 halves plus plus 1/2 I that's if I take the plot the positive version of the I there or we could view this as 3 halves minus minus 1/2 minus 1/2 I this and these two guys right here are equivalent those are the two roots now what I want to do is verify that these work verify that these two roots so this one I can rewrite as 3 plus I over 2 these are equivalent all I did you can see that this is just dividing both of these bikes - or if you would essentially factor out the 1/2 you would get either you could go either way on this expression and this over one over here is going to be 3 minus I over 2 or you could go directly from this this is 3 plus or minus I over 2 so 3 plus I over 2 or 3 minus I over 2 this and this or this and this or this these are all equal representations of both of the roots but let's see if they work so I'm first going to try I'm first going to try this character right over here it's going to get a little bit hairy because we're going to square it and all the rest but let's see if we can do it so what we want to do is we want to take two times this quantity squared so 2 times 3 plus I 3 plus I over 2 squared plus 5 and we want to verify that that's the same thing as 6 times this quantity as 6 times 3 plus I over 2 so what is 3 plus I squared so let me so this is 2 times let me just square this so 3 plus I that's going to be 3 squared which is 9 plus 2 times the product of 3 and I so 3 times I is 3 I times 2 is 6 I so plus 6 I and if that doesn't make sense to you I encourage you to kind of multiply it out either with the distributive property or foil it out and you'll get the middle term will be you'll get 3 times 3i twice when you add them you get 6 I and then plus I squared and I squared is negative 1 minus 1 all of that over 4 plus 5 is equal to well if you divide the numerator in the denominator by 2 you get a 3 here and you get a 1 here and 3 distribute it on 3 plus I is equal to 9 plus 3i 9 plus 3i and what we have over here we can simplify it before I just to save some screen real estate 9 minus 1 is 8 so if I get rid of this this is just 8 plus 6i 8 plus 6i we can divide the numerator and the denominator right here by 2 so the numerator would become 4 plus 3i if we divided it by 2 and the denominator here is just going to be 2 this 2 and this 2 are going to cancel out so on the left-hand side we're left with four plus three I plus five and this is needs to be equal to nine plus 3i well you could see we have a 3i on both sides of this equation and we have a four plus five which is exactly equal to nine so this solution three plus I definitely works now let's try three minus I 3 minus I so once again just looking at the original equation 2x squared plus 5 is equal to 6x so let me write it down over here so we have we rewrite the original equation we have 2x squared plus 5 is equal to 6x and now we're going to try this root verify that it works so we have 2 times 3 minus I over 2 squared plus 5 needs to be equal to 6 times this business 6 times 3 minus I over 2 once again a little hairy but as long as we do everything we put our head down and focus on it we should be able to get the right result so 3 minus I squared 3 minus I squared this 3 minus I times 3 minus I which is and you could get practice taking squares of of 2 termed expressions or complex numbers in this case actually it's going to be 9 that's 3 squared and then 3 times negative I is negative 3i and then you're going to have two of those so negative 6i and then I squared so negative I squared is also negative 1 that's negative 1 times negative 1 times I times I so that's also negative 1 you can negative I squared is also equal to negative 1 negative I is also another square root not the principal square root but one of the square roots of negative 1 so this is going to be so now we're going to have a plus 1 because oh sorry we're gonna have a minus 1 because this is negative I squared which is negative 1 negative 1 and all of that over 4 all of that over that's 2 squared is 4 times 2 over here times 2 plus 5 needs to be equal to well before I even multiply it out we can divide the numerator denominator by 2 so 6 divided by is 3 2 divided by 2 is 1 so 3 times 3 is 9 3 times negative I is negative 3i and if we simplify it a little bit more 9 minus 1 is going to be I'll do this in blue 9 minus 1 is going to be 8 we have 8 minus 6i and then if we divide 8 minus 6i by 2 and 4 by 2 in the numerator we're going to get 4 minus 3i and in the denominator over here in the denominator over here we're going to get a 2 if we divide we divided the numerator and the denominator by 2 then we have a 2 out here and we have a 2 in the denominator those two characters will cancel out and so this expression right over here cancels or simplifies to 4 minus 3i then we have a plus 5 needs to be equal to 9 minus 3i we have a negative 3 on the left negative 3i on the right we have a 4 plus 5 we could evaluate it this left hand side is 9 9 minus 3i which is the exact same complex number as we have on the right hand side 9 minus 3i so it also checks out it is also a root so we verified that both of these complex roots both of these complex roots satisfy this quadratic equation