3. Roots of Quadratics

The solutions of a quadratic equation, may be real or complex. In this case complex.

Let

$$\; az^2 + bz + c = 0 \;$$

be a quadratic where

$$\quad a, b, c \in \mathbb{R}$$

but the roots of the quadratic are non-real complex numbers.

How to solve

Case 1

If $\;z_1\;$ is a root, then the other root is $\; z_2 = \overline z$.

 

Case 2

If the quadratic is in the form

$$(z - \alpha)(z - \beta) = 0$$

then

$$z^2 - (\alpha + \beta)z + \alpha\beta$$

and if we write

$$az^2 + bz + c = 0$$

as

$$z^2 + \dfrac{b}{a}z + \dfrac{c}{a} = 0$$

then

$$\alpha + \beta = -\dfrac{b}{a}, \quad \alpha\beta = \dfrac{c}{a}$$

these results can be used to find the quadratic equation given a single root.

 

Example

Given that $\; \alpha = 7 + 2i \;$ is one of the roots of a quadratic with real coefficients

$$\begin{aligned} \text{a)} \quad & \text{State the other root of } \beta \\[1mm] \text{b)} \quad & \text{Find the quadratic} \\[1mm] \text{c)} \quad & \text{Find the values of } \alpha, \beta, \; \alpha \beta \text{ and interpret your results} \end{aligned}$$

Answer

$$\begin{aligned} \text{a)} \quad & \beta = \overline{\alpha} = 7 - 2i \\[1mm] \text{b)} \quad & \text{Quadratic: } z^2 - (\alpha + \beta)z + (\alpha \beta) = z^2 - 14z + 53 \\[1mm] \text{c)} \quad & \alpha = 7 + 2i, \quad \beta = 7 - 2i, \quad \alpha \beta = 53, \quad \text{sum } \alpha + \beta = 14 \end{aligned}$$