# What is an example of a quadratic equation with imaginary roots?

May 16, 2018

If we consider a general quadratic equation:

$a {x}^{2} + b x + c = 0$

And suppose that we denote roots by $\alpha$ and $\beta$, then

$x = \alpha , \beta \implies \left(x - \alpha\right) \left(x - \beta\right) = 0$
$\therefore {x}^{2} - \left(\alpha + \beta\right) x + \alpha \beta = 0$

Equivalently we can write as

$\therefore {x}^{2} - \left(\text{sum of roots")x+("product of roots}\right) = 0$

And comparing these identical equations we can readily derive the following important relationships:

$\text{sum of roots} = - \frac{b}{a}$ and $\text{product of roots} = \frac{c}{a}$

We also know that complex roots appear in conjugate pairs, so we can form some suitable equations.

Ex 1: $\alpha , \beta = 1 \pm 2 i$

$S = \left(1 - 2 i\right) + \left(1 + 2 i\right) = 2$
$P = \left(1 - 2 i\right) \left(1 + 2 i\right) = 1 + 4 = 5$

So the equation with these roots is:

${x}^{2} - 2 x + 5 = 0$

Ex 2: $\alpha , \beta = 2 \pm 1 i$

$S = \left(2 - i\right) + \left(2 + i\right) = 4$
$P = \left(2 - i\right) \left(2 + i\right) = 4 + 1 = 5$

So the equation with these roots is:

${x}^{2} - 4 x + 5 = 0$

If we strictly answer the question and require imaginary roots then we have no real component so:

Ex 3: $\alpha , \beta = \pm 3 i$

$S = \left(- 3 i\right) + \left(3 i\right) = 0$
$P = \left(- 3 i\right) \left(3 i\right) = 9$

So the equation with these roots is:

${x}^{2} - 0 x + 9 = 0$
$\therefore {x}^{2} + 9 = 0$