## Intermediate Algebra (12th Edition)

$\sqrt[4]{x}\cdot\sqrt[4]{x}=\sqrt{x}$ for $x\ge0$
Using $\sqrt[n]{a}\cdot\sqrt[n]{b}=\sqrt[n]{ab}$, the expression $\sqrt[4]{x}\cdot\sqrt[4]{x}$ is equivalent to \begin{align*} & \sqrt[4]{x\cdot x} \\&= \sqrt[4]{x^2} \\&= \sqrt{\sqrt{x^2}} \\&= \sqrt{|x|} &(\text{use }\sqrt{x^2}=|x|) \\&= \sqrt{x} &(\text{since }x\ge0) .\end{align*} Hence, $\sqrt[4]{x}\cdot\sqrt[4]{x}=\sqrt{x}$ for $x\ge0$.