College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 1 - Equations and Inequalities - Exercise Set 1.6: 50

Answer

Solution set is $\{\frac{27}{8},-125\}$

Work Step by Step

$2x^{\frac{2}{3}}+7x^{\frac{1}{3}}-15=0$ This equation is equivalent to $2(x^{\frac{1}{3}})^{2}+7x^{\frac{1}{3}}-15=0$ Substituting $y = x^{\frac{1}{3}}$, $2y^{2}+7y-15=0$ By factoring, $(2y-3)(y+5)=0$ $2y-3 = 0$ or $y+5=0$ $y=\frac{3}{2}$ or $y=-5$ Let $y=\frac{3}{2}$ Replace $y = x^{\frac{1}{3}}$ $x^{\frac{1}{3}}=\frac{3}{2}$ Cubing both sides of the equations, $(x^{\frac{1}{3}})^{3}=(\frac{3}{2})^{3}$ $x=\frac{27}{8}$ Let $y=-5$ Replace $y = x^{\frac{1}{3}}$ $x^{\frac{1}{3}}=-5$ Cubing both sides of the equations, $(x^{\frac{1}{3}})^{3}=(-5)^{3}$ $x=-125$ Solution set is $\{\frac{27}{8},-125\}$
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