Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Section P.3 - Radicals and Rational Exponents - Exercise Set - Page 45: 43

Answer

$20\sqrt{2}-5\sqrt{3}$

Work Step by Step

Simplify each radical to obtain $3\sqrt{4(2)} -\sqrt{16(2)}+3\sqrt{36(2)}-\sqrt{25(3)} \\=3\sqrt{2^2(2)} -\sqrt{4^2(2)}+3\sqrt{6^2(2)}-\sqrt{5^2(3)} \\=3\cdot2\sqrt{2} -4\sqrt{2}+3\cdot6\sqrt{2}-5\sqrt{3)} \\=6\sqrt{2}-4\sqrt{2}+18\sqrt{2}-5\sqrt{3}.$ RECALL: The distributive property states that for any real numbers a, b, and c, $(ac-bc)=(a-b)c.$ Use the rule above to obtain $(6-4+18)\sqrt{2}-5\sqrt{3} \\=20\sqrt{2}-5\sqrt{3}.$
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