Intermediate Algebra (6th Edition)

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

Chapter 7 - Section 7.4 - Adding, Subtracting, and Multiplying Radical Expressions - Exercise Set - Page 439: 39

Answer

$\dfrac{5\sqrt{7}}{2x}$

Work Step by Step

Using the properties of radicals, then, \begin{array}{l} \sqrt{\dfrac{28}{x^2}}+\sqrt{\dfrac{7}{4x^2}} \\\\= \sqrt{\dfrac{4}{x^2}\cdot7}+\sqrt{\dfrac{1}{4x^2}\cdot7} \\\\= \dfrac{2\sqrt{7}}{x}+\dfrac{\sqrt{7}}{2x} \\\\= \dfrac{2(2\sqrt{7})+\sqrt{7}}{2x} \\\\= \dfrac{4\sqrt{7}+\sqrt{7}}{2x} \\\\= \dfrac{5\sqrt{7}}{2x} .\end{array}
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