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
$\frac 92 + 2 = -\frac 92 + 11\implies 2\frac 92 = 9$.
 A: Perhaps simpler (fewer cases to distinguish):
Note that $|a|=|b|$ (with real numbers $a,b$) means that $a=b$ or $a=-b$.
So here either $x^2+2=x^2-11$, which is absurd, or $x^2+2=-(x^2-11)$, which leads to $2x^2=9$ and so $$x=\pm\frac 32\sqrt 2.$$
Verification: If $x=\pm\frac32\sqrt 2$, then $x^2=\frac 92$ and $|x^2+2|=|\frac{13}2|=\frac{13}2$ and $|x^2-11|=|\frac92-11|=|-\frac{13}2|=\frac{13}2$ as well.
A: 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 $2=-11$.
A: $\mathrm{dist}(x^2,-2)=|x^2+2|=|x^2-11|=\mathrm{dist}(x^2,11)$. Therefore you want $x^2$ to be equidistant from $-2$ and $11$, i.e. $x^2=11-\frac{\mathrm{dist}(-2,11)}{2}=-2+\frac{\mathrm{dist}(-2,11)}{2}=\frac{9}{2}$. It follows that $x=\pm \frac{3}{\sqrt{2}}$
A: Hint:
For real $x,$ $x^2\ge0$, so $x^2+2>0$, so $|x^2+2|=x^2+2$.  $|x^2-11|=x^2-11$ or $-(x^2-11).$  Equate them and see what you get.  
A: Maybe another approach might also be interesting:
$$|x^2 +2| = |x^2-11| \Leftrightarrow (x^2+2)^2 = (x^2-11)^2\Leftrightarrow (x^2+2)^2 - (x^2-11)^2 = 0$$
$$\stackrel{a^2-b^2 = (a-b)(a+b)}{\Leftrightarrow} (x^2+2 - (x^2-11))(x^2 + 2 + x^2-11)= 13(2x^2 - 9)= 0$$
$$\Leftrightarrow x^2= \frac{9}{2}\Leftrightarrow x= \pm\frac{3}{\sqrt{2}}$$
