# 3.4: Roots of Quadratic Equations

You probably first encountered complex numbers when you studied values of $$z$$ (called roots or zeros) for which the following equation is satisfied

$az^2 + bz + c = 0$

For $$a≠0$$ (as we will assume), this equation may be written as

$z^2 + \frac b a z + \frac c a = 0$

Let's denote the second-degree polynomial on the left-hand side of this equation by $$p(z)$$:

$p(z)=z^2 + \frac b a z + \frac c a$

This is called a monic polynomial because the coefficient of the highest-power term $$(z^2)$$ is 1. When looking for solutions to the quadratic equation $$z^2 + \frac b a z + \frac c a = 0$$, we are really looking for roots (or zeros) of the polynomial $$p(z)$$. The fundamental theorem of algebra says that there are two such roots. When we have found them, we may factor the polynomial $$p(z)$$ as follows:

$p(z) = z^2 + \frac b a z + \frac c a = (z−z1)(z−z2)$

In this equation, $$z_1$$ and $$z_2$$ are the roots we seek. The factored form $$p(z)=(z−z_1)(z−z_2)$$ shows clearly that $$p(z_1) = p(z_2) = 0$$, meaning that the quadratic equation $$p(z) = 0$$ is solved for $$z=z_1$$ and $$z=z_2$$. In the process of factoring the polynomial $$p(z)$$, we solve the quadratic equation and vice versa.

By equating the coefficients of $$z^2,z^1,$$ and $$z^0$$ on the left-and right-hand sides of Equation, we find that the sum and the product of the roots $$z_1$$ and $$z_2$$ obey the equations

$z_1 + z_2 = −\frac b a$

$z_1z_2 = \frac c a$

You should always check your solutions with these equations.

## Completing the Square

In order to solve the quadratic equation $$z^2 + \frac b a z + \frac c a = 0$$ (or, equivalently, to find the roots of the polynomial $$z^2 + \frac b a z + \frac c a$$), we “complete the square” on the left-hand side of the equation:

$\left(z + \frac b {2a}\right)^2 − \left(\frac b {2a} \right)^2 + \frac c a = 0$

This equation may be rewritten as

$\left(z + \frac b {2a}\right)^2 = \left(\frac 1 {2a} \right)^2 + \left(b^2 - 4ac\right)$

We may take the square root of each side to find the solutions

$z_{1,2} = −\frac b {2a} ± \frac 1 {2a} \sqrt {b^2−4ac}$

Exercise $$\PageIndex{1}$$

With the roots $$z_1$$ and $$z_2$$ defined in Equation 1.29, prove that $$(z−z_1)(z−z_2)$$ is, indeed, equal to the polynomial $$z^2 + \frac b a z + \frac c a$$. Check that $$z_1+z_2=−\frac b a$$ and $$z_1z_2=\frac c a$$

In the equation that defines the roots $$z_1$$ and $$z_2$$, the term $$b^2−4ac$$ is critical because it determines the nature of the solutions for $$z_1$$ and $$z_2$$. In fact, we may define three classes of solutions depending on $$b^2−4ac$$.

### Overdamped

$$(b^2 − 4ac > 0)$$. In this case, the roots $$z_1$$ and $$z_2$$ are

$z_{1,2} = −\frac b {2a} ± \frac 1 {2a} \sqrt {b^2−4ac}$

These two roots are real, and they are located symmetrically about the point $$−\frac b {2a}$$. When $$b=0$$, they are located symmetrically about 0 at the points $$±\frac 1 {2a} \sqrt {−4ac}$$. (In this case, $$−4ac>0$$.) Typical solutions are illustrated in Figure.

Exercise $$\PageIndex{2}$$

Compute and plot the roots of the following quadratic equations:

1. $$z^2+2z+\frac 1 2=0$$
2. $$z^2+2a−\frac 1 2=0$$
3. $$z^2−\frac 1 2=0$$

For each equation, check that $$z_1+z_2=−\frac b a$$ and $$z_1z_2=\frac c a$$

### Critically Damped

$$(b^2 − 4ac = 0)$$. In this case, the roots $$z_1$$ and $$z_2$$ are equal (we say they are repeated):

$z_1 = z_2 = −\frac b {2a}$

These solutions are illustrated in Figure.

Exercise $$\PageIndex{3}$$

Compute and plot the roots of the following quadratic equations:

1. $$z^2+2z+1=0$$
2. $$z^2−2z+1=0$$
3. $$z^2=0$$

For each equation, check that $$z_1 + z_2 = −\frac b a$$ and $$z_1z_2=\frac c a$$

### Underdamped

$$(b^2 − 4ac < 0)$$. The underdamped case is, by far, the most fascinating case. When $$b^2 − 4ac < 0$$, then the square root in the solutions for $$z_1$$ and $$z_2$$ $$\left(z_{1,2} = −\frac b {2a} ± \frac 1 {2a} \sqrt {b^2−4ac}\right)$$ produces an imaginary number. We may write $$b^2 − 4ac$$ as $$−(4ac - b^2)$$ and write $$z_{1,2}$$ as

$z_{1,2} = -\frac b {2a} ± -\frac 1 {2a} \sqrt {-(4ac−b^2)} = -\frac b {2a} ± -j\frac 1 {2a} \sqrt {4ac−b^2}$

These complex roots are illustrated in Figure. Note that the roots are purely imaginary when $$b=0$$, producing the result

$z_{1,2} = ±j \sqrt {\frac c a}$

In this underdamped case, the roots $$z_1$$ and $$z_2$$ are complex conjugates:

$z_2 = z^∗_1$

Thus the polynomial $$p(z) = z^2 + \frac b a z + \frac c a = (z−z_1)(z−z_2)$$ also takes the form

$p(z)=(z−z_1)(z−z^∗_1) = z^2 − 2 \mathrm {Re}[z_1]z + |z_1|^2$

$$\mathrm {Re}[z_1]$$  and $$|z_1|^2$$ are related to the original coefficients of the polynomial as follows:

$2\mathrm {Re}[z_1] = −\frac b a$

$∣z_1∣^2 = \frac c a$

Always check these equations.

Let's explore these connections further by using the polar representations for $$z_1$$ and (z_2\):

$z_{1,2} = re^{±jθ}$

Then Equation for the polynomial $$p(z)$$ may be written in the “standard form”

$p(z)=(z−re^{jθ})(z−re^{−jθ}) = z^2 − 2r \cosθz +r^2$

Equation is now

$2r\cosθ = −\frac b a$

$r^2 = \frac c a$

These equations may be used to locate $$z_{1,2} = re^{±jθ}$$
$r = \sqrt{\frac c a}$

$θ = ±\cos^{−1}{\left(\frac {-b} {\sqrt{4ac}}\right)}$

Exercise $$\PageIndex{4}$$

Prove that $$p(z)$$ may be written as $$p(z)=z^2−2r\cosθz+r^2 in the underdamped case. Exercise \(\PageIndex{5}$$

Prove the relations in Equation. Outline a graphical procedure for locating $$z_1 = re^{jθ}$$ and $$z_2 = re^{−jθ}$$ from the polynomial $$z^2+\frac b a z + \frac c a$$

Exercise $$\PageIndex{6}$$

Compute and plot the roots of the following quadratic equations:

1. $$z^2+2z+2=0$$
2. $$z^2−2z+2=0$$
3. $$z^2+2=0$$

For each equation, check that $$2 \mathrm {Re} [z_{1,2}] = −\frac b a$$ and $$∣z_{1,2}∣^2 = \frac c a$$.