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: 81



Work Step by Step

Multiplying by the conjugate of the numerator, then the rationalized-numerator form of the given expression, $ \dfrac{\sqrt{x}+1}{\sqrt{x}-1} ,$ is \begin{array}{l}\require{cancel} \dfrac{\sqrt{x}+1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}-1} \\\\= \dfrac{(\sqrt{x})^2-1^2}{(\sqrt{x}-1)^2} \\\\= \dfrac{x-1}{(\sqrt{x})^2+2(\sqrt{x})(-1)+(-1)^2} \\\\= \dfrac{x-1}{x-2\sqrt{x}+1} .\end{array}
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.