Answer
$x\in(-3,-2)\cup(2,3)$
Work Step by Step
The square root function is defined for nonnegative real numbers, and is an increasing function.
This means that if a and b are nonnegative,
$a \lt b\Rightarrow\sqrt{a} \lt \sqrt{b}$
An alternate definition of the absolute value is $|x|=\sqrt{x^{2}}$.
All parts of the compound inequality sign are nonnegative, so we may take the square root of all parts, with the direction of inequality being unchanged:
$\sqrt{4} \lt \sqrt{x^{2}} \lt \sqrt{9}$
$2 \lt |x| \lt 3$
Now we have two inequalities:
$\left[\begin{array}{lll}
& 2 \lt |x| & \\
& \swarrow\searrow & \\
x \lt -2 & ..or.. & x \gt 2
\end{array}\right]\qquad$and$\qquad \left[\begin{array}{l}
|x| \lt 3\\
-3 \lt x \lt 3\\
\end{array}\right]$
$x $ is in the intersection of $(-3,3)$ with $(-\infty,-2)\cup(2,\infty)$
From the image below, this solution set in interval form is
$x\in(-3,-2)\cup(2,3)$
