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 11 - Quadratic Functions and Equations - 11.5 Equations Reducible to Quadratic - 11.5 Exercise Set - Page 731: 10

Answer

Fill the blank with $w^{\frac{1}{6}}$.

Work Step by Step

With substitution $u=w^{\frac{1}{6}}$, then $u^{2}=w^{\frac{1}{3}}.$ The equation becomes $u^{2}-3u+8=0$
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