<s>[INST] 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
$\frac 92 + 2 = -\frac 92 + 11\implies 2\frac 92 = 9$.
 [/INST] Your solution is not correct. But contains good ideas. 
You write $\sqrt{2}i\leq x\leq \sqrt{11}$. 
You can not do that, because you can not order complex numbers! So that is a mistake.
Also it is not clear, how you get your three conditions, which are also the wrong intervals.
It is $x^2+2>0$ for all $x\in\mathbb{R}$.
So $|x^2+2|=x^2+2$.
Now: $|x^2-11|=\begin{cases} -x^2+11, \sqrt{11}\geq x\geq-\sqrt{11}\\ x^2-11, x\in\mathbb{R}\setminus (-\sqrt{11},\sqrt{11})\end{cases}$
Now seperate the cases and check, if 

$x^2+2=-x^2+11$ has solutions in $(-\sqrt{11},\sqrt{11})$

or

$x^2+2=x^2-11$ has solutions in $\mathbb{R}\setminus (-\sqrt{11},\sqrt{11})$

For the first equation, you get $2x^2=9\Leftrightarrow x_{1,2}=\sqrt{\frac92}\in (-\sqrt{11},\sqrt{11})$
The second equation has obviously no solutions, because it is equivalent to $2=-11$.
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