Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 8 - Radical Functions - 8.3 Multiplying and Dividing Radicals - 8.3 Exercises - Page 643: 62

Answer

$\frac{3\sqrt [4] {8x^3y^2}}{2xy}$

Work Step by Step

We don't want to leave radicals in the denominator, so to get rid of radicals in the denominator, we multiply both the numerator and denominator by as many factors that are needed to eliminate the radical. To figure this out, we need to factor the denominator completely: $\frac{3\sqrt [4] {9xy}}{\sqrt [4] {3^2 • 2 • x^2 • y^3}}$ In this case, to eliminate the radical in the denominator, we need to multiply the denominator by two more 3's, three more 2's, two more x's, and one more y. We multiply the numerator by the same factor: $\frac{3\sqrt [4] {3^2xy}}{\sqrt [4] {3^2 • 2 • x^2 • y^3}} • \frac{\sqrt [4] {3^2 • 2^3 • x^2 • y}}{\sqrt [4] {3^2 • 2^3 • x^2 • y}}$ Multiply to simplify: $\frac{3\sqrt [4] {3^4 • 2^3 • x^3 • y^2}}{\sqrt [4] {3^4 • 2^4 • x^4 • y^4}}$ Take the fourth roots: $\frac{9\sqrt [4] {8x^3y^2}}{6xy}$ Cancel any factors common in both numerator and denominator: $\frac{3\sqrt [4] {8x^3y^2}}{2xy}$
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