<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] $$|x^2 - 4| < 2 \implies -2 < x^2 - 4 < 2$$
$$\implies 2 < x^2 < 6$$
$$ \implies \sqrt{2} < |x| < \sqrt{6}$$
So 
$\sqrt{2} < -x < \sqrt{6} \implies -\sqrt{2} > x > -\sqrt{6}$ 
or
  $ \implies \sqrt{2} < x < \sqrt{6}$ and its done...
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