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: 50

Answer

$6-2\sqrt{3}$

Work Step by Step

RECALL: (i) For non-negative real numbers $a$ and $b$, $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$ (ii) For all real numbers $a, b,$ and $c$, $a(b-c) = ab-ac$ Use rule (ii) above to obtain: $=\sqrt{6} \cdot \sqrt{6} -\sqrt{6} \cdot \sqrt{2}$ Use rule (i) above to obtain: $=\sqrt{36} -\sqrt{12} \\=\sqrt{6^2} -\sqrt{12} \\=6 -\sqrt{12}$ Factor the radicand so that one of the factors is a perfect square: $=6-\sqrt{4(3)} \\=6-\sqrt{2^2(3)}$ Simplify to obtain: $=6-2\sqrt{3}$
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