# How do you solve  x^2+5x+8=0?

Apr 30, 2016

$x = - \frac{5}{2} \pm \frac{\sqrt{7}}{2} i$

#### Explanation:

Pre-multiply by $4$ (to avoid arithmetic with fractions), complete the square then use the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

with $a = \left(2 x + 5\right)$ and $b = \sqrt{7} i$ as follows:

$0 = 4 \left({x}^{2} + 5 x + 8\right)$

$= 4 {x}^{2} + 20 x + 32$

$= {\left(2 x\right)}^{2} + 2 \left(2 x\right) \left(5\right) + 32$

$= {\left(2 x + 5\right)}^{2} - 25 + 32$

$= {\left(2 x + 5\right)}^{2} + 7$

$= {\left(2 x + 5\right)}^{2} - {\left(\sqrt{7} i\right)}^{2}$

$= \left(\left(2 x + 5\right) - \sqrt{7} i\right) \left(\left(2 x + 5\right) + \sqrt{7} i\right)$

$= \left(2 x + 5 - \sqrt{7} i\right) \left(2 x + 5 + \sqrt{7} i\right)$

So: $2 x = - 5 \pm \sqrt{7} i$

So: $x = - \frac{5}{2} \pm \frac{\sqrt{7}}{2} i$

$\textcolor{w h i t e}{}$

The equation ${x}^{2} + 5 x + 8 = 0$ is of the form $a {x}^{2} + b x + c = 0$ with $a = 1$, $b = 5$ and $c = 8$.

This has roots given by the quadratic formula:

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

$\frac{- 5 \pm \sqrt{{5}^{2} - \left(4 \cdot 1 \cdot 8\right)}}{2 \cdot 1}$

$= \frac{- 5 \pm \sqrt{25 - 32}}{2}$

$= \frac{- 5 \pm \sqrt{- 7}}{2}$

$= \frac{- 5 \pm \sqrt{7} i}{2}$

$= - \frac{5}{2} \pm \frac{\sqrt{7}}{2} i$