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

Published by Pearson

# Chapter 10 - Exponents and Radicals - Review Exercises: Chapter 10 - Page 693: 33

#### Answer

$\sqrt{3}$

#### Work Step by Step

Extracting the factors that are perfect powers of the index, the given expression simplifies to \begin{array}{l}\require{cancel} 2\sqrt{75}-9\sqrt{3} \\\\= 2\sqrt{25\cdot3}-9\sqrt{3} \\\\= 2\sqrt{(5)^2\cdot3}-9\sqrt{3} \\\\= 2\cdot5\sqrt{3}-9\sqrt{3} \\\\= 10\sqrt{3}-9\sqrt{3} .\end{array} By combining like radicals, the given expression simplifies to \begin{array}{l}\require{cancel} 10\sqrt{3}-9\sqrt{3} \\\\= (10-9)\sqrt{3} \\\\= 1\sqrt{3} \\\\= \sqrt{3} .\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.