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: 14

Answer

$25 - 2\sqrt{5}$

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, $\sqrt{a} \cdot \sqrt{b}=\sqrt{ab}$ (3) $(a+b)(c-d) = a(c-d) + b(c-d)$ Use rule (3) above to obtain: $=4(10 - 3\sqrt{5})+ \sqrt{5}(10 - 3\sqrt{5})$ Use rules (1) and (2) above then simplify to obtain: $=4(10) - 4(3\sqrt{5}) + \sqrt{5}(10) - \sqrt{5}(3\sqrt{5}) \\=40 - 12\sqrt{5} + 10\sqrt{5}-3\sqrt{5(5)} \\=40 + (-12+10)\sqrt{5} - 3\sqrt{25} \\=40 + (-2\sqrt{5}) - 3(5) \\=40 - 2\sqrt{5}-15 \\=25 - 2\sqrt{5}$
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