# How do you solve (x^2+2x-15)/(x^2+7x)>0?

Nov 30, 2016

The answer is x in ] -oo,-7 [uu ] -5, 0[ uu ]3, oo[

#### Explanation:

Let $f \left(x\right) = \frac{{x}^{2} + 2 x - 15}{{x}^{2} + 7 x}$

$= \frac{\left(x - 3\right) \left(x + 5\right)}{x \left(x + 7\right)}$

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{0 , - 7\right\}$

To solve the equation $f \left(x\right) > 0$, we need a 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}$$- 7$$\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}$$x + 7$$\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 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 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 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 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 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 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 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 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 a}$$+$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a}$$+$

So,

$f \left(x\right) > 0$

when x in ] -oo,-7 [uu ] -5, 0[ uu ]3, oo[

graph{(x^2+2x-15)/(x^2+7x) [-16.02, 16.01, -8.01, 8.01]}