Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 8 - Section 8.3 - Solving Equations by Using Quadratic Methods - Exercise Set - Page 501: 23


The solutions are $x=-\frac{1}{8}$ and $x=27$.

Work Step by Step

Subtract $3$ to both sides of the equation to obtain: $2x^{2/3}-5x^{1/3}-3=0 \\2(x^{1/3})^2-5x^{1/3}-3=0$ Let $u=x^{1/3}$ Rewrite the equation above using the variable $u$ to obtain: $2u^2-5u-3=0$ Factor the trinomial to obtain: $(2u+1)(u-3)=0$ Use the Zero-Factor Property by equating each unique factor to zero. then solve each equation: \begin{array}{ccc} &2u+1 = 0 &\text{ or } &u-3=0 \\&2u=-1 &\text{ or } &u=3 \\&u=-\frac{1}{2} &\text{ or } &u=3 \end{array} Since $u=x^{1/3}$, then \begin{array}{ccc} &u= -\frac{1}{2} &\text{or} &u=3 \\&x^{1/3}=-\frac{1}{2} &\text{or} &x^{1/3}=3 \\&(x^{1/3})^3=(-\frac{1}{2})^3 &\text{or} &(x^{1/3})^3=3^3 \\&x=-\frac{1}{8} &\text{or} &x=27 \end{array} Thus, the solutions are $x=-\frac{1}{8}$ and $x=27$.
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