Elementary Algebra

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

Chapter 9 - Roots and Radicals - 9.4 - Products and Quotients Involving Radicals - Problem Set 9.4: 62

Answer

$\dfrac{15\sqrt{2}+20}{2}$

Work Step by Step

Multiplying by the conjugate of the denominator and using $(a+b)(a-b)=a^2-b^2,$ the given expression, $ \dfrac{5}{3\sqrt{2}-4} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{5}{3\sqrt{2}-4}\cdot\dfrac{3\sqrt{2}+4}{3\sqrt{2}+4} \\\\= \dfrac{15\sqrt{2}+20}{(3\sqrt{2})^2-(4)^2} \\\\= \dfrac{15\sqrt{2}+20}{9\cdot2-16} \\\\= \dfrac{15\sqrt{2}+20}{18-16} \\\\= \dfrac{15\sqrt{2}+20}{2} .\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.