How to prove $x^2<4 \implies|x|<2$? 
How to prove $x^2<4 \implies|x|<2$?

I don't know exactly there is a proof for this or we take this as an axiom. Please help.
What I did so far is, $$x^2-4<0$$
$$(x-2)(x+2)<0$$
$$-2<x<2$$
From this step can we directly say? $|x|<2$

Saying $-2\lt x \lt 2$ is exactly the same thing as saying $|x|\lt 2$. Here's why: from the definition of absolute value we have that $$|x|=\begin{cases}x&x\gt 0\\-x&x\leq0\end{cases}$$ so the inequality $|x|\lt 2$ becomes $$\begin{cases}x\lt 2&x>0\\-x\lt 2& x\leq0\end{cases}$$ the second inequality can be simplified changing the sing, which changes the direction of the inequality, so $$x\gt -2\;\;\;x\leq 0$$ In the end you can take the union of the solutions to have a solution so that the won't be disjoint $$-2\lt x\leq 0\;\; \cup0\lt x\lt 2\;\; \implies -2\lt x\lt2$$
