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: 44

Answer

$\sqrt{6}+12\sqrt{7}$

Work Step by Step

Simplify each radical to obtain $3\sqrt{9(6)} -2\sqrt{4(6)}-\sqrt{16(6)}+4\sqrt{9(7)} \\=3\sqrt{3^2(6)} -2\sqrt{2^2(6)}-\sqrt{4^2(6)}+4\sqrt{3^2(7)} \\=3\cdot 3\sqrt{6} -2\cdot 2\sqrt{6}-4\sqrt{6}+4\cdot3\sqrt{7} \\=9\sqrt{6}-4\sqrt{6}-4\sqrt{6}+12\sqrt{7}.$ RECALL: The distributive property states that for any real numbers a, b, and c, $(ac-bc)=(a-b)c \\(ac+bc) = (a+b)c.$ Use the rule above to obtain $(9-4-4)\sqrt{6}+12\sqrt{7} \\=\sqrt{6}+12\sqrt{7}.$
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