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: 42

Answer

$-2\sqrt{3}$

Work Step by Step

Simplify each radical to obtain $4\sqrt{4(3)} -2\sqrt{25(3)} \\=4\sqrt{2^2(3)}-2\sqrt{5^2(3)} \\=4\cdot2\sqrt{3}-2\cdot5\sqrt{3} \\=8\sqrt{3}-10\sqrt{3}.$ 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 $(8-10)\sqrt{3} \\=-2\sqrt{3}.$
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