# How do you solve the inequality (x^2-16)/(x^2-4x-5)>=0?

Apr 16, 2018

The solution is $x \in \left(- \infty , - 4\right] \cup \left(- 1 , 4\right] \cup \left(5 , + \infty\right)$

#### Explanation:

The numerator is

${x}^{2} - 16 = \left(x - 4\right) \left(x + 4\right)$

The denominator is

${x}^{2} - 4 x - 5 = \left(x - 5\right) \left(x + 1\right)$

Let

$f \left(x\right) = \frac{\left(x - 4\right) \left(x + 4\right)}{\left(x - 5\right) \left(x + 1\right)}$

Let's 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 a}$$- 4$$\textcolor{w h i t e}{a a a a}$$- 1$$\textcolor{w h i t e}{a a a a a}$$4$$\textcolor{w h i t e}{a a a a a}$$5$$\textcolor{w h i t e}{a a a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x + 4$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a}$$0$$\textcolor{w h i t e}{a a a}$$+$$\textcolor{w h i t e}{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}$$x + 1$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a}$color(white)(aaaa)-$\textcolor{w h i t e}{a}$$| |$$\textcolor{w h i t e}{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}$$x - 4$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a}$color(white)(aaaa)-$\textcolor{w h i t e}{a}$color(white)(aa)-$\textcolor{w h i t e}{a}$$0$$\textcolor{w h i t e}{a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$x - 5$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a}$color(white)(aaaa)-$\textcolor{w h i t e}{a}$color(white)(aa)-$\textcolor{w h i t e}{a}$color(white)(aaa)-$\textcolor{w h i t e}{a}$$| |$$\textcolor{w h i t e}{a a}$$+$

$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a}$$+$$\textcolor{w h i t e}{a a a}$$0$$\textcolor{w h i t e}{a a a}$$-$$\textcolor{w h i t e}{a}$$| |$$\textcolor{w h i t e}{a}$$+$$\textcolor{w h i t e}{a}$$0$$\textcolor{w h i t e}{a a}$$-$$\textcolor{w h i t e}{a}$$| |$$\textcolor{w h i t e}{a a}$$+$

graph{((x+4)(x-4))/((x-5)(x+1)) [-14.24, 14.23, -7.12, 7.12]}

Therefore,

$f \left(x\right) \ge 0$ when $x \in \left(- \infty , - 4\right] \cup \left(- 1 , 4\right] \cup \left(5 , + \infty\right)$