Solve this

$|x|<4$

Square

$\Rightarrow x^{2}<16$

$\Rightarrow x^{2}-16<0$

$\Rightarrow x^{2}-4^{2}<0$

$\Rightarrow(x+4)(x-4)<0$

Observe that when x is greater than 4, (x + 4)(x – 4) is positive

And for each root the sign changes hence

We want less than 0 that is negative part

Hence $x$ should be between $-4$ and 4 for $(x+4)(x-4)$ to be negative

Hence $x \in(-4,4)$

Hence the solution set of $|x|<4$ is $(-4,4)$