Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Section P.3 - Radicals and Rational Exponents - Exercise Set - Page 45: 54


$\displaystyle \frac{11(\sqrt{7}+\sqrt{3})}{4}$

Work Step by Step

Use $(a+b)(a-b)=a^{2}-b^{2}\quad$ when rationalizing. $[(a+\sqrt{b})(a-\sqrt{b})=a^{2}-b]$ $\displaystyle \frac{11}{\sqrt{7}-\sqrt{3}}\cdot\frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}+\sqrt{3}}=\frac{11(\sqrt{7}+\sqrt{3})}{7-3}$ $=\displaystyle \frac{11(\sqrt{7}+\sqrt{3})}{4}$
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.