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? $|x|<2$
 A: Hint:


*

*Square root monotonic grows (proof is trivial)

*$\sqrt{x^2}=|x|$ (follows directly from the definition of $\sqrt{}$)



Alternative way, continuing what you started:


*$a\cdot b \lt 0 \leftrightarrow (a \lt 0 \land b \gt 0) \lor (a \gt 0 \land b \lt 0)$

*So you have to find the interval where $(x-2)(x+2)\lt 0$

A: 
What I did so far is, $$x^2-4<0$$
  $$(x-2)(x+2)<0$$

Since $\,x^2=|x|^2\,$, you could also factor it as:
$$\left(|x|-2\right)\left(|x|+2\right) \lt 0$$
Given that $\,|x|+2 \gt 0\,$ it follows that $\,|x| - 2\lt 0\,$.
A: 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 $$-2\lt x\leq 0\;\; \cup0\lt x\lt 2\;\; \implies -2\lt x\lt2$$
A: The implication is stronger, indeed we have
$$x^2<4 \iff x^2-4<0 \iff (x+2)(x-2)<0 \iff -2< x < 2 \iff |x|<2$$
indeed 


*

*for $x\ge 0$ we have $|x|<2 \iff x<2\iff 0\le x<2$

*for $x< 0$ we have $|x|<2 \iff -x<2\iff x>-2\iff -2<x<0$
therefore the last implication follows.
