Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 9 - Roots and Radicals - Chapter 9 Review Problem Set: 49

Answer

$\dfrac{3\sqrt{2}+\sqrt{6}}{6}$

Work Step by Step

Multiplying by the conjugate of the denominator and using $(a+b)(a-b)=a^2-b^2$, the simplifed form of given expression, $ \dfrac{2}{3\sqrt{2}-\sqrt{6}} ,$ is \begin{array}{l}\require{cancel} \dfrac{2}{3\sqrt{2}-\sqrt{6}}\cdot\dfrac{3\sqrt{2}+\sqrt{6}}{3\sqrt{2}+\sqrt{6}} \\\\= \dfrac{2(3\sqrt{2}+\sqrt{6})}{(3\sqrt{2})^2-(\sqrt{6})^2} \\\\= \dfrac{2(3\sqrt{2}+\sqrt{6})}{9(2)-6} \\\\= \dfrac{2(3\sqrt{2}+\sqrt{6})}{12} \\\\= \dfrac{\cancel{2}(3\sqrt{2}+\sqrt{6})}{\cancel{2}(6)} \\\\= \dfrac{3\sqrt{2}+\sqrt{6}}{6} .\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.