# How do you solve ((x+7)(x-3))/(x-5)^2>0?

Dec 20, 2016

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

#### Explanation:

Let $f \left(x\right) = \frac{\left(x + 7\right) \left(x - 3\right)}{x - 5} ^ 2$

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{5\right\}$

$\forall x \in {D}_{f} \left(x\right)$, the denominator is $> 0$

We can now do 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}$$- 7$$\textcolor{w h i t e}{a a a a}$$3$$\textcolor{w h i t e}{a a a a}$$5$$\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}$$+$$\textcolor{w h i t e}{}$∥$\textcolor{w h i t e}{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}{}$∥$\textcolor{w h i t e}{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}{}$∥$\textcolor{w h i t e}{a a}$$+$

Therefore,

$f \left(x\right) > 0$ when x in ] -oo,-7 [ uu ] 3,5 [uu ] 5,+oo [

graph{((x+7)(x-3))/(x-5)^2 [-23.16, 34.55, -11.88, 16.99]}