# How do you solve x^2<=8x using a sign chart?

##### 1 Answer
Dec 6, 2016

The answer is $x \in \left[0 , 8\right]$

#### Explanation:

Let's rearrange the equation

${x}^{2} \le 8 x$

${x}^{2} - 8 x \le 0$

$x \left(x - 8\right) \le 0$

Let $f \left(x\right) = x \left(x - 8\right)$

Let's do 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}$$0$$\textcolor{w h i t e}{a a a a}$$8$$\textcolor{w h i t e}{a a a a}$$+ \infty$

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

$\textcolor{w h i t e}{a a a a}$$x - 8$$\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}$$+$

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

So, $f \left(x\right) \le 0$, when $x \in \left[0 , 8\right]$

graph{x(x-8) [-41.1, 41.1, -20.56, 20.56]}