# How do you solve (x+6)/(x^2-5x-24)>=0?

Oct 23, 2016

f(x)>=0 for x⋳[-6,-3]∪[8,+oo]

#### Explanation:

Start by factorising the denominator
${x}^{2} - 5 x - 24 = \left(x + 3\right) \left(x - 8\right)$
Then
$\frac{x + 6}{\left(x + 3\right) \left(x - 8\right)} \ge 0$
For this we make a sign table

$x$$\textcolor{w h i t e}{a a a a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- 6$$\textcolor{w h i t e}{a a a a}$$- 3$$\textcolor{w h i t e}{a a a a}$$+ 8$$\textcolor{w h i t e}{a a a a}$$+ \infty$
$x + 6$$\textcolor{w h i t e}{a a a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$
$x + 3$$\textcolor{w h i t e}{a a a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$
$x - 8$$\textcolor{w h i t e}{a a a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$
$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

As we need a result $\ge 0$, we keep the intervals where $f \left(x\right)$ os positive

So the result is x⋳[-6,-3]∪[8,+oo])