How do you solve and write the following in interval notation: x^2 - 2x - 8 > 0?

Nov 2, 2017

The solution is $x \in \left(- \infty , - 2\right) \cup \left(4 , + \infty\right)$

Explanation:

Factorise the inequality

${x}^{2} - 2 x - 8 = \left(x + 2\right) \left(x - 4\right) > 0$

Let $f \left(x\right) = \left(x + 2\right) \left(x - 4\right)$

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}$$4$$\textcolor{w h i t e}{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}$$x - 4$$\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}$$f \left(x\right)$$\textcolor{w h i t e}{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,

$f \left(x\right) > 0$ when $x \in \left(- \infty , - 2\right) \cup \left(4 , + \infty\right)$

graph{(x+2)(x-4) [-31.62, 26.12, -9.97, 18.9]}