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 - Concept and Vocabulary Check: 7

Answer

$5$; $6\sqrt3$

Work Step by Step

Simplifying the radical gives $\sqrt{3} + \sqrt{25\cdot 3} \\=\sqrt{3} + \sqrt{5^2 \cdot 3} \\=\sqrt{3} + \sqrt{5^2} \cdot \sqrt{3} \\=\sqrt{3} + \underline{5}\sqrt{3}.$ Use the rule $ac+bc=(a+b)c$ to obtain $=(1+5)\sqrt{3} \\=\underline{6\sqrt{3}}.$ Thus, the missing expressions are: $5$ and $6\sqrt{3}.$
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