# How do you find all the real and complex roots of x^2 + 8x + 25 = 0?

Jan 10, 2016

$x = - 4 \pm 3 i$

#### Explanation:

Complete the square.

$\left({x}^{2} + 8 x + 16\right) + 25 - 16 = 0$

${\left(x + 4\right)}^{2} + 9 = 0$

${\left(x + 4\right)}^{2} = - 9$

$\sqrt{{\left(x + 4\right)}^{2}} = \sqrt{- 9}$

$x + 4 = \pm 3 i$

$x = - 4 \pm 3 i$

Jan 10, 2016

Roots (both complex) are at $x = - 4 + 3 i$ and $x = - 4 - 3 i$

#### Explanation:

The quadratic formula tells us that for a quadratic equation in the form:
$\textcolor{w h i t e}{\text{XXX}} a {x}^{2} + b x + c = 0$
the roots are given by:
$\textcolor{w h i t e}{\text{XXX}} x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

For the given example: ${x}^{2} + 8 x + 25 = 0$
$\textcolor{w h i t e}{\text{XXX}} a = 1 , b = 8 , \mathmr{and} c = 25$

So the roots are:
$\textcolor{w h i t e}{\text{XXX}} x = \frac{- 8 \pm \sqrt{{8}^{2} - 4 \left(1\right) \left(25\right)}}{2 \left(1\right)}$

$\textcolor{w h i t e}{\text{XXXX}} = \frac{- 8 \pm \sqrt{- 36}}{2}$

$\textcolor{w h i t e}{\text{XXXX}} = \frac{- 8 \pm \sqrt{36} \cdot \sqrt{- 1}}{2}$

$\textcolor{w h i t e}{\text{XXXX}} = - 4 \pm 3 i$