# Explain the relationship between complex numbers and quadratic equations Does that mean that 2 and -3 are equations? No, that would be silly. The solutions to the equation [math]x^2+x+6 = 0[/math] are the complex. Demonstrates the relationship between complex roots of a quadratic and the intercepts of the related parabola. Also shows how to graph complex numbers. This Article is brought to you for free and open access by the Math in the Middle Institute beneficial to students to visualize a graphical connection. Figure 3 explain what we don't know graphically about quadratics with complex roots (Fig.

Roots that possess this pattern are called complex conjugates or conjugate pairs. This pattern of complex conjugates will occur in every set of complex roots that you will encounter when solving a quadratic equation.

When expressed as factors and multiplied, these complex conjugates will allow for the middle terms containing "i "s to cancel out. If the roots of a quadratic equation are imaginary, they always occur in conjugate pairs. Imaginary or complex roots will occur when the value under the radical portion of the quadratic formula is negative.

Notice that the value under the radical portion is represented by "b2 - 4ac". So, if b2 - 4ac is a negative value, the quadratic equation is going to have complex conjugate roots containing "i "s. If the discriminant is negative, you have a negative under the radical and the roots of the quadratic equation will be complex conjugates.

The discriminant, b2 - 4ac, offers valuable information about the "nature" of the roots of a quadratic equation where a, b and c are rational values. If you are trying to determine the "type" of roots of a quadratic equation not the actual roots themselvesyou need not complete the entire quadratic formula.

Simply look at the discriminant. So 2 times 2 is 4. So this is going to be equal to 6 plus or minus the square root of so let me just figure this out. This is 40 over here. So 36 minus 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. 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.

### Quadratic Equations with Complex Solutions - MathBitsNotebook(A2 - CCSS Math)

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. That's if I take the positive version of the i there.

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.

All I did-- you can see that this is just dividing both of these by 2. 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.

Introduction to i and imaginary numbers - Imaginary and complex numbers - Precalculus - Khan Academy

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. 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.

## Solving quadratic equations: complex roots

You'll get 3i twice. When you add them, you get 6i. I And then plus i squared, and i squared is negative 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. 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.