How do solve (x^2-x-12)/(x^2+4)<=0 and write the answer as a inequality and interval notation?

Jan 21, 2017

The answer is $x \in \left[- 3 , 4\right]$ or $- 3 \le x \le 4$

Explanation:

Let's factorise the numerator

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

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

The domain of $f \left(x\right)$ is $\mathbb{R}$

The denominator is $> 0$

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

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

$- 3 \le x \le 4$