# How do solve (x^2+5x)/(x-3)>=0 and write the answer as a inequality and interval notation?

Nov 20, 2016

The answer is x in [-5,0 ] uu ]3, +oo[

#### Explanation:

Let $f \left(x\right) = \frac{{x}^{2} + 5 x}{x - 3} = \frac{x \left(x + 5\right)}{x - 3}$

The domain of $f \left(x\right)$ is ${D}_{f \left(x\right)} = \mathbb{R} - \left\{3\right\}$

Let's do a sign chart to solve this inequality

$\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}$$- 5$$\textcolor{w h i t e}{a a a a}$$0$$\textcolor{w h i t e}{a a a a}$$3$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x + 5$$\textcolor{w h i t e}{a a a a a}$$-$$\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 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 a a}$$-$$\textcolor{w h i t e}{a a 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}$$x - 3$$\textcolor{w h i t e}{a a a a a}$$-$$\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 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}$$-$$\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 a}$$+$

Therefore $f \left(x\right) \ge 0$ when x in [-5,0 ] uu ]3, +oo[

graph{(x^2+5x)/(x-3) [-72.6, 75.47, -19.23, 54.86]}