Using the Quadratic Formula to Find Complex Roots

Back in Algebra 1, we learned about the quadratic formula. We used it to solve quadratic equations.

At first, we only used the formula to find real roots, which are real numbers. Now we are going to learn something new.

We will use the quadratic formula to find complex roots. These solutions go beyond real numbers and help us solve more types of equations.

The Quadratic Formula

The quadratic formula is:

$$x= \frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

for quadratic equations in standard form, $y=ax^2 +bx +c$. When the radicand of the square root is a positive integer, this formula makes perfect sense! We get two real roots. Now, think about what happens when the radicand is a negative number. We get complex roots!

Finding Complex Roots Using the Quadratic Formula

Quadratic equations have complex roots when the graph does not cross the $x$-axis. By looking at the graph, we can often tell whether a quadratic has real or complex roots.

In an earlier lesson, we learned how to use the discriminant to figure this out. The discriminant helps us know how many solutions a quadratic equation has.

• If the discriminant is positive, there are two real solutions.

• If the discriminant is zero, there is one real solution.

• If the discriminant is negative, there are two complex (imaginary) solutions.

Using the discriminant gives us a quick way to understand the solutions, even before solving the equation.

Algebraically, we can solve the quadratic equation using the quadratic formula.

Let's look at an example.

Say we are trying to solve the quadratic equation, $y=x^2+2x+5$. We know that $a=1$, $b=2$, and $c=5$. Let's plug in these values into the quadratic formula.

$$\begin{align} x &= \frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ x &= \frac{-2\pm \sqrt{(2)^2-4(1)(5)}}{2(1)} \\ x &= \frac{-2\pm \sqrt{4-20}}{5} \\ x &= \frac{-2\pm \sqrt{-16}}{5} \\ \end{align}$$

Obviously, we cannot find a real solution to $\sqrt{-16}$, but we can find a complex solution!

$$\begin{align} x &= \frac{-2\pm\sqrt{-16}}{5} \\ x &= \frac{-2\pm {4i}}{5} \\ x &= -\frac25 \pm \frac{4i}{5} \\ \end{align}$$

Thus, our two solutions are: $x = -\frac25 + \frac{4i}{5}$ and $x = -\frac25 - \frac{4i}{5}$.

Let's think about what we did. We used the quadratic formula to find the two complex roots to this quadratic equation!

Going Backward

As you practice more of these problems, you’ll notice a pattern. Complex roots always come in pairs called complex conjugates.

Complex conjugates look almost the same, except the sign in front of the $i$ is different. They work as a team.

When you multiply these two factors together, the $i$ terms cancel out. This brings you back to the original equation.

This is why complex roots always appear in pairs.

(Video) The Quadratic Formula - Solving for Complex Roots

by larryschmidt

This video by larryschmidt walks you through an example of how to use the quadratic formula to solve for complex roots.

Summary

Before we begin, it's important to review the quadratic formula:

$$x= \frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

The first example we walk through is using the quadratic formula to solve for the complex roots of: $x^2+8x+20=0$.

We plug in what we know into the quadratic formula.

$$\begin{align} x &= \frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ x &= \frac{-8\pm \sqrt{64-80}}{2} \\ \end{align}$$

Notice that there is a negative sign under the square root - that's our indicator that we are going to have complex roots!

$$\begin{align} x &= \frac{-8\pm \sqrt{64-80}}{2} \\ x &= \frac{-8\pm \sqrt{-16}}{2} \\ x &= \frac{-8\pm 4i}{2} \\ x &= -4 \pm 2i \end{align}$$

Try the next example on your own before watching the video!

Example

Solve $x^2 -2x +10 =0$.

Solution

$$\begin{align} x &= \frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ x &= \frac{2\pm \sqrt{4-40}}{2} \\ x &= \frac{2\pm \sqrt{-36}}{2} \\ x &= \frac{2\pm 6i}{2} \\ x &= 1 \pm 3i \\ \end{align}$$

Using the Quadratic Formula to Find Complex Roots

We are already familiar with the quadratic formula. You have used it to find solutions to quadratic equations.

Now we are going to take this idea one step further. We will use the quadratic formula to work with complex roots.

This will help us solve equations that do not have real-number solutions.

Sometimes the expression inside the square root is negative. This expression is called the discriminant, and it is written as $b^2−4ac$.

When the discriminant is negative, the equation has complex roots. This happens because you cannot take the square root of a negative number using real numbers.

Instead, we use complex numbers to find the solutions.

Instead, we use the imaginary unit $i$, where $i = \sqrt{-1}$. We've learned this concept before, but now we know that it also applies to the quadratic formula!