## Introductory Algebra for College Students (7th Edition)

Published by Pearson

# Chapter 8 - Section 8.3 - Operations with Radicals - Exercise Set - Page 591: 40

#### Answer

$7\sqrt{6}+8\sqrt{5}$

#### Work Step by Step

Factor the radicands so that one of the factors is a perfect square: $=3\sqrt{9(6)}-2\sqrt{4(5)}+4\sqrt{9(5)}-\sqrt{4(6)} \\=3\sqrt{3^2(6)}-2\sqrt{2^2(5)}+4\sqrt{3^2(5)}-\sqrt{2^2(6)}$ Simplify to obtain: $\\=3\cdot 3\sqrt{6}-2\cdot 2\sqrt{5}+4\cdot 3\sqrt{5}-2\sqrt{6} \\=9\sqrt{6}-4\sqrt{5}+12\sqrt{5}-2\sqrt{6}$ Combine like terms: $=(9\sqrt{6} - 2\sqrt{6}) + (-4\sqrt{5}+12\sqrt{5}) \\=(9-2)\sqrt{6} + (-4+12)\sqrt{5} \\=7\sqrt{6}+8\sqrt{5}$

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.