# How do you solve X^2>16?

Aug 8, 2015

$x > 4$ or $x < - 4$

#### Explanation:

If $x \ge 0$ and ${x}^{2} > 16$
then $x > \sqrt{16} = 4$

If $x > 0$ and ${x}^{2} > 16$
then $- x > \sqrt{16} = 4$
and (multiplying by $\left(- 1\right)$ which reverses the inequality sign)
$\textcolor{w h i t e}{\text{XXXX}}$$x < - 4$