Find all numbers x which satisfy $|x^2 + 2| = |x^2 − 11|$. This question is taken from book: Exercises and Problems in Calculus, by
John M. Erdman, available online, from chapter 1.1, question $4$. 
Request help, as not clear if my approach is correct.

(4) Find all numbers x which satisfy $|x^2 + 2| = |x^2 − 11|$.

Have two conditions, leading to three intervals;
(i) $x \lt \sqrt{-2}$
(ii) $\sqrt{-2} \le x \lt \sqrt{11}$
(iii) $x \ge \sqrt{11}$
(i) no soln.
(ii) $2x^2 = 9 \implies x = \pm\sqrt{\frac 92}$
(iii) no soln.
Verifying:
First, the two solutions must satisfy that they lie in the given interval,
this means $\sqrt{-2} \le x \lt \sqrt{11}\implies \sqrt{2}i \le x \lt \sqrt{11}$.
Am not clear, as the lower bound is not a real one. So, given the two real values of $x = \pm\sqrt{\frac 92}$ need only check with the upper bound.
For further verification, substitute in values of $x$, for interval (ii):
a) For $x = \sqrt{\frac 92}$, $|x^2 + 2| = |x^2 − 11|$ leads to
$\frac 92 + 2 = -\frac 92 + 11\implies 2\frac 92 = 9$.
b) For $x = -\sqrt{\frac 92}$, $|x^2 + 2| = |x^2 − 11|$ leads to
$\frac 92 + 2 = -\frac 92 + 11\implies 2\frac 92 = 9$.

Maybe another approach might also be interesting:
$$|x^2 +2| = |x^2-11| \Leftrightarrow (x^2+2)^2 = (x^2-11)^2\Leftrightarrow (x^2+2)^2 - (x^2-11)^2 = 0$$
$$\stackrel{a^2-b^2 = (a-b)(a+b)}{\Leftrightarrow} (x^2+2 - (x^2-11))(x^2 + 2 + x^2-11)= 13(2x^2 - 9)= 0$$
$$\Leftrightarrow x^2= \frac{9}{2}\Leftrightarrow x= \pm\frac{3}{\sqrt{2}}$$
