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 7.4 - Adding, Subtracting, and Dividing Radical Expressions - Exercise Set - Page 539: 72

Answer

$47x\sqrt {2x}$.

Work Step by Step

The given expression is $=7\sqrt {2x^3}+\frac{40x^3\sqrt {150x^2}}{5x^2\sqrt {3x}}$ Divide the radicands and retain the common index. $=7\sqrt {2x^3}+\frac{40x^3}{5x^2}\cdot \sqrt {\frac{150x^2}{3x}}$ Factor the radicands into square terms. $=7\sqrt {2x^2\cdot x}+\frac{40x^3}{5x^2}\cdot \sqrt {\frac{5^2\cdot 2\cdot 3\cdot x^2}{3x}}$ Cancel out similar terms. $=7\sqrt {2x^2\cdot x}+8x\cdot \sqrt {5^2\cdot 2\cdot x}$ Simplify. $=7x\sqrt {2x}+8x\cdot 5\cdot \sqrt { 2 x}$ $=7x\sqrt {2x}+40x\sqrt { 2x}$ Apply the distributive property. $=(7+40)x\sqrt {2x}$ Simplify. $=47x\sqrt {2x}$.
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