Why is this inequality regarding squares true?

inequality

Let, $x,b \in \mathbb{R}$. Then, $$x^2 < b \iff |x| < \sqrt{b}?$$

Any easy way to see why this is true? Thanks.


I imagine you are assuming that $b\geq 0$: $$x^2<b\Leftrightarrow x^2<\sqrt{b}^2\Leftrightarrow x^2-\sqrt{b}^2<0\Leftrightarrow (x-\sqrt b)(x+\sqrt b)<0$$

Hence $-\sqrt b<x<\sqrt b \Leftrightarrow |x|<\sqrt b.$

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