Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 10 - Exponents and Radicals - Review Exercises: Chapter 10: 42

Answer

$\dfrac{20}{\sqrt{10}+\sqrt{15}}$

Work Step by Step

Multiplying by an expression equal to $1$ which will make the numerator a perfect power of the index, then the rationalized-numerator form of the given expression is \begin{array}{l}\require{cancel} \dfrac{4\sqrt{5}}{\sqrt{2}+\sqrt{3}} \\\\= \dfrac{4\sqrt{5}}{\sqrt{2}+\sqrt{3}}\cdot\dfrac{4\sqrt{5}}{4\sqrt{5}} \\\\= \dfrac{(4\sqrt{5})^2}{4\sqrt{5}(\sqrt{2}+\sqrt{3})} \\\\= \dfrac{16(5)}{4\sqrt{5}(\sqrt{2}+\sqrt{3})} \\\\= \dfrac{80}{4\sqrt{5}(\sqrt{2}+\sqrt{3})} .\end{array} Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{80}{4\sqrt{5}(\sqrt{2}+\sqrt{3})} \\\\= \dfrac{80}{4\sqrt{5}(\sqrt{2})+4\sqrt{5}(\sqrt{3})} \\\\= \dfrac{80}{4\sqrt{10}+4\sqrt{15}} \\\\= \dfrac{\cancel4^{20}}{\cancel4\sqrt{10}+\cancel4\sqrt{15}} \\\\= \dfrac{20}{\sqrt{10}+\sqrt{15}} .\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.