Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 1-14 - Cumulative Review - Final Exam - Page 930: 14

Answer

$|y|\sqrt[12]{x^5y^2}$

Work Step by Step

Converting the radicals to like indices, the given expression, $ \dfrac{\sqrt[3]{x^2y^5}}{\sqrt[4]{xy^2}} ,$ is equivalent to \begin{align*} & \dfrac{\left(\sqrt[3(4)]{x^2y^5}\right)^4}{\left(\sqrt[4(3)]{xy^2}\right)^3} \\\\&= \dfrac{\sqrt[12]{\left(x^2y^5\right)^4}}{\sqrt[12]{\left(xy^2\right)^3}} &\left(\left(\sqrt[n]{x}\right)^m=\sqrt[n]{x^m}\right) \\\\&= \dfrac{\sqrt[12]{x^{2(4)}y^{5(4)}}}{\sqrt[12]{x^{1(3)}y^{2(3)}}} &\left((x^my^n)^p=x^{mp}y^{np}\right) \\\\&= \dfrac{\sqrt[12]{x^{8}y^{20}}}{\sqrt[12]{x^{3}y^{6}}} .\end{align*} Using $\dfrac{\sqrt[n]{x^m}}{\sqrt[n]{x^n}}=\sqrt[n]{\dfrac{x^m}{x^n}},$ the expression above is equivalent to \begin{align*} & \sqrt[12]{\dfrac{x^{8}y^{20}}{x^{3}y^{6}}} \\\\&= \sqrt[12]{x^{8-3}y^{20-6}} &\left(\dfrac{x^m}{x^n}=x^{m-n}\right) \\\\&= \sqrt[12]{x^{5}y^{14}} \\\\&= \sqrt[12]{y^{12}\cdot x^5y^2} \\\\&= \sqrt[12]{y^{12}}\cdot\sqrt[12]{x^5y^2} \\\\&= |y|\sqrt[12]{x^5y^2} &\left(\sqrt[n]{x^n}=|x|\text{, if $n$ is even}\right) .\end{align*} Hence, the expression $\dfrac{\sqrt[3]{x^2y^5}}{\sqrt[4]{xy^2}}$ simplifies to $|y|\sqrt[12]{x^5y^2}$.
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