# How do you solve the inequality x^2-4x<=5?

Jan 28, 2017

The answer is $x \in \left[- 1 , 5\right]$

#### Explanation:

Let's rewrite the inequality

${x}^{2} - 4 x - 5 \le 0$
Let's factorise the LHS

$\left(x + 1\right) \left(x - 5\right) \le 0$

Let $f \left(x\right) = \left(x + 1\right) \left(x - 5\right)$

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

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

Therefore,

$f \left(x\right) \le 0$ when $x \in \left[- 1 , 5\right]$