In general, if n is even, $\displaystyle{\sqrt[{{n}}]{{{{x}}^{{n}}}}}={\left|{x}\right|}\usepackage{cmbright}\Huge\$ . When $\displaystyle{n}={2}\usepackage{cmbright}\Huge\$ , this means that $\displaystyle\sqrt{{{{x}}^{{2}}}}={\sqrt[{{2}}]{{{{x}}^{{2}}}}}={\left|{x}\right|}\usepackage{cmbright}\Huge\$ , so the simplified form of $\displaystyle\sqrt{{{16}{{x}}^{{2}}}}={4}{\left|{x}\right|}\usepackage{cmbright}\Huge\$ .