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: 58



Work Step by Step

The given expression can be written as: $=\dfrac{1}{2}\cdot \dfrac{\sqrt{50xy}}{\sqrt{2}}$ 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: $\require{cancel}=\dfrac{1}{2} \cdot \sqrt{\dfrac{50xy}{2}} \\=\dfrac{1}{2} \cdot \sqrt{\dfrac{25\cancel{50}xy}{\cancel{2}}} \\=\dfrac{1}{2} \cdot \sqrt{25xy}$ Factor the radicand so that at least one factor is a perfect square to obtain: $=\dfrac{1}{2} \cdot \sqrt{25(xy)} \\=\dfrac{1}{2} \cdot \sqrt{5^2(xy)}$ Simplify to obtain: $=\frac{1}{2} \cdot 5\sqrt{xy} \\=\dfrac{5\sqrt{xy}}{2}$
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