<s>[INST] 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?
 [/INST] 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.
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