Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 7 - Section 7.5 - Rationalizing Denominators and Numerators of Radical Expressions - Exercise Set - Page 445: 26

Answer

$\dfrac{\sqrt{2x}}{8x}$

Work Step by Step

Rationalizing the denominator of $ \dfrac{1}{\sqrt{32x}} $ results to \begin{array}{l} \dfrac{1}{\sqrt{16\cdot2x}} \cdot \dfrac{\sqrt{2x}}{\sqrt{2x}} \\\\= \dfrac{\sqrt{2x}}{\sqrt{16\cdot4x^2}} \\\\= \dfrac{\sqrt{2x}}{4\cdot2x} \\\\= \dfrac{\sqrt{2x}}{8x} .\end{array}
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