# How do you solve x/(x^2 - 16)>0?

Jun 17, 2016

$\frac{x}{{x}^{2} - 16} > 0 \iff x \in \left(- 4 , 0\right) \cup \left(4 , \infty\right)$

#### Explanation:

An expression $f \frac{x}{g} \left(x\right)$ will be greater than $0$ when either both $f \left(x\right)$ and $g \left(x\right)$ are positive or both $f \left(x\right)$ and $g \left(x\right)$ are negative.

Looking at the signs of the given functions, we have

$x < 0 \mathmr{if} x \in \left(- \infty , 0\right)$
$x > 0 \mathmr{if} x \in \left(0 , \infty\right)$

and

${x}^{2} - 16 < 0 \mathmr{if} x \in \left(- 4 , 4\right)$
${x}^{2} - 16 > 0 \mathmr{if} x \in \left(- \infty , - 4\right) \cup \left(4 , \infty\right)$

If we compare the signs, then, we have both being negative on $\left(- 4 , 0\right)$ and both being positive on $\left(4 , \infty\right)$.

Thus, we have $\frac{x}{{x}^{2} - 16} > 0 \iff x \in \left(- 4 , 0\right) \cup \left(4 , \infty\right)$