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: 53

Answer

$2x^2\sqrt{5}$

Work Step by Step

RECALL: (1) The quotient rule: $\dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}=\sqrt[n]{\dfrac{a}{b}}$ where $\sqrt[n]{a}$ and $\sqrt[n]{b}$ are real numbers and $b\ne0$ (2) $\dfrac{a^m}{a^n} = a^{m-n}, a \ne =0$ Use the quotient rule above to obtain: $\require{cancel}=\sqrt{\dfrac{200x^3}{10x^{-1}}} \\=\sqrt{\dfrac{20\cancel{200}x^3}{\cancel{10}x^{-1}}} \\=\sqrt{\dfrac{20x^3}{x^{-1}}}$ Use rule (2) above to obtain: $=\sqrt{20x^{3-(-1)}} \\=\sqrt{20x^{3+1}} \\=\sqrt{20x^4}$ Factor the radicand so that at least one factor is a perfect square to obtain: $=\sqrt{4x^4(5)} \\=\sqrt{(2x^2)^2(5)}$ Simplify to obtain: $=2x^2\sqrt{5}$
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