How to simplify $-\sqrt {16x^2}$ when $x < 0$?

Question

What is $-\sqrt {16x^2}$ when $x < 0$?

Thanks

Answer 1

In any case, the expression $\sqrt {x^2}$ equals $|x|$. Then, we got that: $$-\sqrt{16x^2}=-\sqrt{16}\cdot\sqrt{x^2}=-4\cdot|x|.$$ Now, since $|x|=-x$ for all $x<0$, we conclude that: $$x<0\implies -\sqrt{16x^2}=\boxed{4x}$$