# How do you solve –x^2 + 4 < 0?

Jul 1, 2018

The solution is $x \in \left(- \infty , - 2\right) \cup \left(2 , + \infty\right)$

#### Explanation:

The inequality is

$- {x}^{2} + 4 < 0$

$\implies$, ${x}^{2} - 4 > 0$

$\implies$, $\left(x + 2\right) \left(x - 2\right) > 0$

Let $y = \left(x + 2\right) \left(x - 2\right)$

Build a sign chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- 2$$\textcolor{w h i t e}{a a a a a}$$2$$\textcolor{w h i t e}{a a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x + 2$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$x - 2$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$y$$\textcolor{w h i t e}{a a a a a a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

Therefore,

$y > 0$ when $x \in \left(- \infty , - 2\right) \cup \left(2 , + \infty\right)$

graph{-x^2+4 [-7.9, 7.9, -3.95, 3.95]}