Answer
$\left(-3, -\frac{5}{2}\right] \cup \left[\frac{5}{2}, 3\right)$
Work Step by Step
1. Express the inequality in the form $f(x)<0, f(x)>0, f(x)\leq 0$, or $f(x)\geq 0,$
where $f$ is a Rational function.
$\displaystyle \frac{4x^2-25}{x^2-9} \leq 0$
$f(x)=\displaystyle \frac{(2x-5)(2x+5)}{(x-3)(x+3)} \leq 0$
2. The cut points are:
$\displaystyle \frac{(2x-5)(2x+5)}{(x-3)(x+3)} = 0$
$x=-3$ or $x=\frac{-5}{2}$ or $x=\frac{5}{2}$ or $x=3$
3. Make a table or diagram: use the test values to make a table or diagram of the sign of each factor in each interval.
4. Test each interval's sign of $f(x)$ with a test value,
$\begin{array}{llll}
Intervals: & a=test.v. & f(a),signs & f(a) \leq 0 ? \\
& & \displaystyle \frac{(2a-5)(2a+5)}{(a-3)(a+3)} & \\
(-\infty,-3) & -5 & \frac{(-)(-)}{(-)(-)}=(+) & F\\
(-3,-\frac{5}{2}) & -\frac{11}{4} & \frac{(-)(-)}{(-)(+)}=(-) & T\\
(-\frac{5}{2}, \frac{5}{2}) & 0 & \frac{(-)(+)}{(-)(+)}=(+) & F\\
(\frac{5}{2}, 3) & \frac{11}{4} & \frac{(+)(+)}{(-)(+)}=(-) & T\\
(3,\infty) & 10 & \frac{(+)(+)}{(+)(+)}=(+) & F\\
\end{array}$
5. Write the solution set, selecting the interval or intervals that satisfy the given inequality. If the inequality involves $\leq$ or $\geq$, include the boundary points.
Solution set: $\left(-3, -\frac{5}{2}\right] \cup \left[\frac{5}{2}, 3\right)$
