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: 20

Answer

$\sqrt{\dfrac{26ab}{2b}}$

Work Step by Step

Rationalizing the denominator of $ \sqrt{\dfrac{13a}{2b}} $ results to \begin{array}{l} \sqrt{\dfrac{13a}{2b}} \cdot \sqrt{\dfrac{2b}{2b}} \\\\= \sqrt{\dfrac{26ab}{4b^2}} \\\\= \sqrt{\dfrac{26ab}{2b}} .\end{array}
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