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# Solving quadratic equations: complex roots

CCSS Math: HSN.CN.C

## 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 put it into a form that we're more familiar with, let's try to put it into standard form. And 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 plus 5 is equal to-- and then 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 in the x term, but I would get 5/2 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 a formula 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 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 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. 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 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. And 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. I could even do it 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 2i, 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 could 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 3/2 plus 1/2i. That's if I take the positive version of the i there. Or we could view this as 3/2 minus 1/2i. This and these two guys right here are equivalent. Those are the two roots. Now what I want to do is a verify that these work. Verify 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 by 2. Or if you were to essentially factor out the 1/2, you could go either way on this expression. And this 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 this character right over here. It's going to get a little bit hairy, because we're going to have 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 2 times this quantity squared. So 2 times 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 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 three and i. So 3 times i is 3i, times 2 is 6i. So plus 6i. 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. You'll get 3i twice. When you add them, you get 6i. 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 and the denominator by 2, you get a 3 here and you get a 1 here. And 3 distributed on 3 plus i is equal to 9 plus 3i. And what we have over here, we can simplify it 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. 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 4 plus 3i plus 5. And this needs to be equal to 9 plus 3i. Well, you can see we have a 3i on both sides of this equation. And we have a 4 plus 5, which is exactly equal to 9. So this solution, 3 plus i, definitely works. Now let's try 3 minus i. So once again, just looking at the original equation, 2x squared plus 5 is equal to 6x. Let me write it down over here. Let me 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 times 3 minus i, which is-- and you could get practice taking squares of two 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. 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. 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 now we're going to have a plus 1, because-- oh, sorry, we're going to have a minus 1. Because this is negative i squared, which is negative 1. And all of that over 4. All of that over-- that's 2 squared is 4. Times 2 over here, plus 5, needs to be equal to-- well, before I even multiply it out, we could divide the numerator and the denominator by 2. So 6 divided by 2 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, we're going to get a 2. 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. I We have a negative 3i on the left, a negative 3i on the right. We have a 4 plus 5. We could evaluate it. This left hand side is 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, satisfy this quadratic equation.