# How do you solve 1/2x^2<=x+12 using a sign chart?

Aug 9, 2017

The solution is $x \in \left[- 4 , 6\right]$

#### Explanation:

Let's rearrange the inequality

$\frac{1}{2} {x}^{2} \le x + 12$

$\frac{1}{2} {x}^{2} - x - 12 \le 0$

${x}^{2} - 2 x - 24 \le 0$

Factorising, we get

$\left(x + 4\right) \left(x - 6\right) \le 0$

Let, $f \left(x\right) = \left(x + 4\right) \left(x - 6\right)$

We can 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 a}$$- 4$$\textcolor{w h i t e}{a a a a a a}$$6$$\textcolor{w h i t e}{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 a}$$-$$\textcolor{w h i t e}{a a}$$0$$\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}$$x - 6$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a}$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a}$$0$$\textcolor{w h i t e}{a a}$$+$

$\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 a}$$+$$\textcolor{w h i t e}{a a}$$0$$\textcolor{w h i t e}{a a a}$$-$$\textcolor{w h i t e}{a}$$0$$\textcolor{w h i t e}{a a}$$+$

Therefore,

$f \left(x\right) \le 0$ when $x \in \left[- 4 , 6\right]$

graph{1/2x^2-x-12 [-16.22, 15.8, -12.55, 3.47]}