## Intermediate Algebra (12th Edition)

$4\sqrt 3$
We know from the radical form of $a^{\frac{m}{n}}$ that $a^{\frac{m}{n}}=\sqrt[n] a^{m}$ (where the indicated root is a real number). Therefore, $\sqrt[4] 48^{2}=48^{\frac{2}{4}}=48^{\frac{1}{2}}=\sqrt 48$ $\sqrt 48=\sqrt (16\times3)=\sqrt 16\times\sqrt 3=4\sqrt 3$