Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 10 - Section 10.5 - Rationalizing Numerators and Denominators of Radical Expressions - Exercise Set: 70

Answer

$\dfrac{5+\sqrt{2}}{\sqrt{2x}}=\dfrac{23}{5\sqrt{2x}-2\sqrt{x}}$

Work Step by Step

$\dfrac{5+\sqrt{2}}{\sqrt{2x}}$ Multiply the numerator and the denominator by the conjugate of the numerator and simplify if possible: $\dfrac{5+\sqrt{2}}{\sqrt{2x}}=\dfrac{5+\sqrt{2}}{\sqrt{2x}}\cdot\dfrac{5-\sqrt{2}}{5-\sqrt{2}}=\dfrac{5^{2}-(\sqrt{2})^{2}}{\sqrt{2x}(5-\sqrt{2})}=...$ $...=\dfrac{25-2}{5\sqrt{2x}-\sqrt{4x}}=\dfrac{23}{5\sqrt{2x}-2\sqrt{x}}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.