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

Published by Pearson

# Chapter 10 - Exponents and Radicals - 10.5 Expressions Containing Several Radical Terms - 10.5 Exercise Set - Page 661: 63

#### Answer

$\dfrac{12-2\sqrt{3}+6\sqrt{5}-\sqrt{15}}{33}$

#### Work Step by Step

Multiplying by the conjugate of the denominator, the rationalized-denominator form of the given expression, $\dfrac{2+\sqrt{5}}{6+\sqrt{3}} ,$ is \begin{array}{l}\require{cancel} \dfrac{2+\sqrt{5}}{6+\sqrt{3}}\cdot\dfrac{6-\sqrt{3}}{6-\sqrt{3}} \\\\= \dfrac{2(6)+2(-\sqrt{3})+\sqrt{5}(6)+\sqrt{5}(-\sqrt{3})}{6^2-(\sqrt{3})^2} \\\\= \dfrac{12-2\sqrt{3}+6\sqrt{5}-\sqrt{5(3)}}{36-3} \\\\= \dfrac{12-2\sqrt{3}+6\sqrt{5}-\sqrt{15}}{33} .\end{array}

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.