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 7.6 - Radical Equations - Exercise Set - Page 562: 90

Answer

$\frac{54+43\sqrt2}{-46}$.

Work Step by Step

The given expression is $=\frac{7+4\sqrt2}{2-5\sqrt2}$ The conjugate of the denominator is $2+5\sqrt2$. Multiply the numerator and the denominator by $2+5\sqrt2$. $=\frac{7+4\sqrt2}{2-5\sqrt2}\cdot \frac{2+5\sqrt2}{2+5\sqrt2}$ $=\frac{7\cdot 2+7\cdot 5\sqrt2+4\sqrt2\cdot 2+4\sqrt2\cdot 5\sqrt2}{(2)^2-(5\sqrt2)^2}$ Use the product rule. $=\frac{14+35\sqrt2+8\sqrt2+40}{4-50}$ Add like terms. $=\frac{54+43\sqrt2}{-46}$.
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