Incorrect. Factor out as many squares as possible. Remember that if $\displaystyle{x}\ge{0}\usepackage{cmbright}\Huge\$ , then $\displaystyle\sqrt{{{{x}}^{{2}}}}={x}\usepackage{cmbright}\Huge\$ . In this case, $\displaystyle\sqrt{{{12}{{a}}^{{5}}}}=\sqrt{{{{2}}^{{2}}\cdot{3}\cdot{{a}}^{{4}}\cdot{a}}}\usepackage{cmbright}\Huge\$ . Does this help you simplify the expression?