Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 8 - Section 8.3 - Operations with Radicals - Exercise Set: 42

Answer

$9\sqrt{2}+8\sqrt{3}$

Work Step by Step

Factor the radicands so that one of the factors is a perfect square: $=4\sqrt{4(2)}-\sqrt{64(2)}+2\sqrt{16(3)}+3\sqrt{9(2)} \\=4\sqrt{2^2(2)}-\sqrt{8^2(2)}+2\sqrt{4^2(3)}+3\sqrt{3^2(2)}$ Simplify to obtain: $\\=4\cdot 2\sqrt{2}-8\sqrt{2}+2\cdot 2\sqrt{3}+3\cdot 3\sqrt{2} \\=8\sqrt{2}-8\sqrt{2}+8\sqrt{3}+9\sqrt{2}$ Combine like terms: $=(8\sqrt{2} - 8\sqrt{2}+9\sqrt{2}) + 8\sqrt{3} \\=(8-8+9)\sqrt{2} + 8\sqrt{3} \\=9\sqrt{2}+8\sqrt{3}$
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