Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 7 - Section 7.5 - Multiplying with More Than One Term and Rationalizing Denominators - Exercise Set: 6

Answer

$24 - 6\sqrt{3}$

Work Step by Step

RECALL: (1) Distributive Property: For any real numbers a, b, and c, $a(b−c)=ab−ac$ (2) For any real numbers real numbers a and b within the domain, $\sqrt[n]{a} \cdot \sqrt[n]{b}=\sqrt[n]{ab}$ (3) $\sqrt{a} \cdot \sqrt{a} = a, a\ge 0$ Use rule (1) above to obtain: $=\sqrt{6}(4\sqrt{6}) -\sqrt{6} \cdot 3\sqrt{2}$ Use rule (2) above to obtain: $=4\sqrt{36} - 3\sqrt{12} \\=4\sqrt{6^2} - 3\sqrt{4(3)} \\=4\sqrt{6^2} - 3\sqrt{2^2(3)}$ Simplify each radical to obtain: $=4(6) - 3(2)\sqrt{3} \\=24 - 6\sqrt{3}$
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