College Algebra (10th Edition)

Published by Pearson

Chapter 1 - Section 1.4 - Radical Equations; Equations Quadratic in Form; Factorable Equations - 1.4 Assess Your Understanding - Page 118: 89

Answer

$x=0$ or $x=3$ or $x=\frac{5}{2}$

Work Step by Step

We solve by factoring: $x(x^{2}-3x)^{1/3}+2(x^{2}-3x)^{4/3}=0$ $(x^{2}-3x)^{1/3}[x+2(x^{2}-3x)]=0$ $(x^{2}-3x)^{1/3}(x+2x^{2}-6x)=0$ $(x^{2}-3x)^{1/3}(2x^{2}-5x)=0$ $(x^{2}-3x)^{1/3}=0$ or $2x^{2}-5x=0$ $x^{2}-3x=0^3$ or $2x^{2}-5x=0$ $x(x-3)=0$ or $x(2x-5)=0$ $x=0$ or $x=3$ or $x=0$ or $2x-5=0$ $x=0$ or $x=3$ or $2x=5$ $x=0$ or $x=3$ or $x=\frac{5}{2}$

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