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 - Review Exercises - Page 577: 72

Answer

$10\sqrt 5+15\sqrt2$

Work Step by Step

Simplify. $\dfrac{10}{2\sqrt5-3\sqrt2}$ Now, $\dfrac{10}{2\sqrt5-3\sqrt2}=\dfrac{10(2\sqrt5+3\sqrt2)}{(2\sqrt5+3\sqrt2)(2\sqrt5-3\sqrt2)}$ or, $=\dfrac{10(2\sqrt5)+10(3\sqrt2)}{(2\sqrt5)^2-(3\sqrt2)^2}$ or, $=\dfrac{10(2\sqrt{5}+3\sqrt{2})}{2}$ or, $=10\sqrt 5+15\sqrt2$
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