Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 7 - Section 7.3 - Simplifying Radicals, the Distance Formula, and Circles - 7.3 Exercises - Page 459: 8


$\sqrt[4]{x}\cdot\sqrt[4]{x}=\sqrt{x}$ for $x\ge0$

Work Step by Step

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 $x\ge0$.
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