# How do you solve x^2 + 6x +8= 0  by completing the square?

Feb 4, 2016

$x = - 4 \mathmr{and} x = - 2$

(see below for method)

#### Explanation:

Given:
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + 6 x + 8 = 0$

It will be convenient (although not necessary) to move the constant to the right side as:
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b l u e}{{x}^{2} + 6 x = - 8}$

Completing the square

For the general squared binomial: ${\left(x + a\right)}^{2} = {x}^{2} + 2 a + {a}^{2}$

So if ${x}^{2} + 6 x$
$\textcolor{w h i t e}{\text{XXX}}$are the first two terms of an expanded squared binomial
then $a = 3 \mathmr{and} {a}^{2} = 9$
and we need to add $9$ (to both sides) to complete the square

$\textcolor{b l u e}{{x}^{2} + 6 x} \textcolor{red}{+ 9} = \textcolor{b l u e}{- 8} \textcolor{red}{+ 9}$

$\textcolor{g r e e n}{{\left(x + 3\right)}^{2} = 1}$

If ${\left(x + 3\right)}^{2} = 1$
then
$\textcolor{w h i t e}{\text{XXX}} x + 3 = \pm 1$

$\textcolor{w h i t e}{\text{XXX}} x = - 3 \pm 1$

$\textcolor{w h i t e}{\text{XXX}} x = - 4 \mathmr{and} x = - 2$