# How do you solve  (x + 2)^2 = 36?

Apr 15, 2018

$x = - 8 \text{ or } x = 4$

#### Explanation:

$\textcolor{b l u e}{\text{take the square root of both sides}}$

$\sqrt{{\left(x + 2\right)}^{2}} = \pm \sqrt{36} \leftarrow \textcolor{b l u e}{\text{note plus or minus}}$

$\Rightarrow x + 2 = \pm 6$

$\text{subtract 2 from both sides}$

$\Rightarrow x = - 2 \pm 6$

$\Rightarrow x = - 2 - 6 = - 8 \text{ or } x = - 2 + 6 = 4$

$\textcolor{b l u e}{\text{As a check}}$

Substitute these values into the left side of the equation and if equal to the right side then they are the solutions.

$x = - 8 \to {\left(- 8 + 2\right)}^{2} = {\left(- 6\right)}^{2} = 36$

$x = 4 \to {\left(4 + 2\right)}^{2} = {6}^{2} = 36$

$\Rightarrow x = - 8 \text{ or "x=4" are the solutions}$