My Calculus II textbook shows the following example:
I do not understand how they got -2 from solving the inequality $\left|\left(\frac{x}{2}\right)^2\right| < 1$. Doesn't it result in an imaginary part when you square -4 in the last step?
Therefore, we have \begin{align*} \frac{x^2}{4-x^2}&=\frac{x^2}{4\left(1-\left(\frac{x}{2}\right)^2\right)}\\ &=\frac{\frac{x^2}{4}}{1-\left(\frac{x}{2}\right)^2}\\ &=\sum\limits_{n=0}^\infty\frac{x^2}{4}\left(\frac{x}{2}\right)^{2n} \end{align*} This series converges as long as $\left|\left(\frac{x}{2}\right)^2\right|<1$ (note that when $\left|\left(\frac{x}{2}\right)^2\right|=1$ the series does not converge). Solving this inequality, we conclude that the interval of convergence is $(-2,2)$ and
It sounds like you successfully went from $|(\frac x2)^2|<1$ to $-4<x^2<4$.
However, to solve this inequality for $x$, it is not correct to simply apply square roots to all three expressions. Mechanical algebraic steps (while great) need to be supplemented by knowledge about the functions involved.
Here, we know that $x^2$ is always nonnegative; therefore the inequalities $-4<x^2<4$ are equivlanent to the inequalities $0\le x^2<4$.
Now the square root function is defined and increasing on the entire range of values we care about, and so we can apply it to get $\sqrt 0\le\sqrt{x^2}<\sqrt4$, which evaluates to $0\le|x|<2$.