Intermediate Algebra (6th Edition)

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

Chapter 7 - Review: 92



Work Step by Step

Using the properties of radicals, the expression $ \sqrt{\dfrac{8}{x^2}}-\sqrt{\dfrac{50}{16x^2}} $ simplifies to \begin{array}{l} \sqrt{\dfrac{4}{x^2}\cdot2}-\sqrt{\dfrac{25}{16x^2}\cdot2} \\\\= \dfrac{2\sqrt{2}}{x}-\dfrac{5\sqrt{2}}{4x} \\\\= \dfrac{4(2\sqrt{2})-1(5\sqrt{2})}{4x} \\\\= \dfrac{8\sqrt{2}-5\sqrt{2}}{4x} \\\\= \dfrac{3\sqrt{2}}{4x} .\end{array}
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