Solve the inequality: $|x^2 − 4| < 2$ This is a question on a calculus assignment our class received, I am a little confused on a few parts to the solution, can someone clear a few things up with it?
Since $x^2-4 = 0$ that means $x = 2$ and $x = -2$ are turning points.
When $x < -2$ :
$x^2-4<2$
$x^2<6$
$x < \sqrt{6}$ and $x < - \sqrt{6}$
But the solution says that $x > -\sqrt{6}$, what am I not getting?
 A: $$|x^2-4|<2 \\ \implies -2<x^2-4<2 \\ \implies 2<x^2<6 \\ \implies \sqrt{2}<\pm x<\sqrt{6} \\ \implies \sqrt{2}<x<\sqrt{6},\ \ -\sqrt{2}>x>-\sqrt{6}$$
A: $$|x^2 - 4| < 2 \implies -2 < x^2 - 4 < 2 \implies 2 < x^2 < 6 \implies \sqrt{2} < |x| < \sqrt{6}$$
So, the solutions are $x$ such that $-\sqrt{6} < x < -\sqrt{2}$ or $\sqrt{2} < x < \sqrt{6}$. Drawing the number line is very helpful for these exercises.
A: $$|x^2 - 4| < 2 \implies -2 < x^2 - 4 < 2$$
$$\implies 2 < x^2 < 6$$
$$ \implies \sqrt{2} < |x| < \sqrt{6}$$
So 
$\sqrt{2} < -x < \sqrt{6} \implies -\sqrt{2} > x > -\sqrt{6}$ 
or
  $ \implies \sqrt{2} < x < \sqrt{6}$ and its done...
A: In this case, we have $ |x^2 - 4| < 2 $, which leads to $ (x^2 - 4)^2 < 4 $, because $ |a| = \sqrt{a^2} $ where the root is the positive result. 
Continuing, $ (x^2 - 4)^2 - 4 < 0$ and $ x^4 - 8x^2 + 12 < 0 $ or $ (x^2 - 6)(x^2 - 2) < 0 $.
This is only satisfied when $ x^2 > 2 $ and $ x^2 < 6 $. This is possible when $ \sqrt{2} < x < \sqrt{6} $ or $ -\sqrt{2} > x > -\sqrt{6} $.
To figure this out, break it into cases. When is $x^2 > 2$? One case is when $ x > \sqrt{2} $. Then you compare this with the two possibilities of the other inequality, namely $x < \sqrt{6}$ and $x > -\sqrt{6}$. You will notice that one result will be eliminated as impossible. Then do the two cases for when $ x < -\sqrt{2} $. Same thing will happen.
