# How do you solve (16-x^2)/(x^2-9)>=0?

Jan 24, 2016

Identify points and intervals at which this is satisfied to find solution:

$x \in \left[- 4 , - 3\right) \cup \left(3 , 4\right]$

#### Explanation:

Let $f \left(x\right) = \frac{16 - {x}^{2}}{{x}^{2} - 9} = \frac{\left(4 - x\right) \left(4 + x\right)}{\left(x - 3\right) \left(x + 3\right)}$

First note that if $x = \pm 4$ then $16 - {x}^{2} = 0$ and ${x}^{2} - 9 \ne 0$, so these values of $x$ are part of the solution set.

Next note that if $x = \pm 3$ then ${x}^{2} - 9 = 0$ and $16 - {x}^{2} \ne 0$, so $f \left(x\right)$ is not even defined for these values of $x$, which are therefore not part of the solution set.

Apart from these values of $x$ the rational expression is non-zero and continuous, so is consistently positive or negative throughout each of the individual intervals: $\left(- \infty , - 4\right)$, $\left(- 4 , - 3\right)$, $\left(- 3 , 3\right)$, $\left(3 , 4\right)$, $\left(4 , \infty\right)$

If $x < - 4$ or $x > 4$ then $16 - {x}^{2} < 0$ and ${x}^{2} - 9 > 0$, so $f \left(x\right) < 0$.

If $- 4 < x < - 3$ or $3 < x < 4$ then $16 - {x}^{2} > 0$ and ${x}^{2} - 9 > 0$, so $f \left(x\right) > 0$.

If $- 3 < x < 3$ then $16 - {x}^{2} > 0$ and ${x}^{2} - 9 < 0$, so $f \left(x\right) < 0$

So $f \left(x\right) \ge 0$ when $x \in \left[- 4 , - 3\right) \cup \left(3 , 4\right]$