# 3.4: Roots of Quadratic Equations

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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:

- \(z^2+2z+\frac 1 2=0\)
- \(z^2+2a−\frac 1 2=0\)
- \(z^2−\frac 1 2=0\)

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:

- \(z^2+2z+1=0\)
- \(z^2−2z+1=0\)
- \(z^2=0\)

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:

- \(z^2+2z+2=0\)
- \(z^2−2z+2=0\)
- \(z^2+2=0\)

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