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

Dec 26, 2016

The answer is $x < - 2$ or $0 < x \le 4$
=x in ] -oo,-2 [ uu ] 0, 4]

#### Explanation:

Let's rewrite the inequality as

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

And

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

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{- 2 , 0\right\}$

Now, we can make 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}$$- 2$$\textcolor{w h i t e}{a a a a a a}$$0$$\textcolor{w h i t e}{a a a a a a}$$4$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x + 2$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a}$∥$\textcolor{w h i t e}{a a}$$+$∥$\textcolor{w h i t e}{a a}$$+$$\textcolor{w h i t e}{a a a}$$+$

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

$\textcolor{w h i t e}{a a a a}$$x - 4$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a}$∥$\textcolor{w h i t e}{a a}$$-$∥$\textcolor{w h i t e}{a a}$$-$$\textcolor{w h i t e}{a a a}$$+$

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

Therefore,

$f \left(x\right) \le 0$ when x in ] -oo,-2 [ uu ] 0, 4]

or, $x < - 2$ or $0 < x \le 4$