# How do you solve (x^2-x-12)/(x^2+4)<=0 using a sign chart?

Jun 30, 2018

The solution is $x \in \left[- 3 , 4\right]$

#### Explanation:

Factorise the numerator

${x}^{2} - x - 12 = \left(x + 3\right) \left(x - 4\right)$

Therefore,

$\frac{{x}^{2} - x - 12}{{x}^{2} + 4} \le 0$

$\iff$, $\frac{\left(x + 3\right) \left(x - 4\right)}{{x}^{2} + 4} \le 0$

$\forall x \in \mathbb{R} , \left({x}^{2} + 4\right) > 0$

Let $f \left(x\right) = \frac{\left(x + 3\right) \left(x - 4\right)}{{x}^{2} + 4}$

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 a}$$- 3$$\textcolor{w h i t e}{a a a a a a a}$$4$$\textcolor{w h i t e}{a a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x + 3$$\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 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}$color(white)(aaaa)-$\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}$$f \left(x\right)$$\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 a}$$0$$\textcolor{w h i t e}{a a a}$$+$

Therefore,

$f \left(x\right) \le 0$ when $x \in \left[- 3 , 4\right]$

graph{(x^2-x-12)/(x^2+4) [-10, 10, -5, 5]}