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: 27


The solutions are $x=-\frac{1}{125}$ and $x=\frac{1}{8}$.

Work Step by Step

The given equation can be written as: $20(x^{1/3})^2-6x^{1/3}-2=0$ Let $u=x^{1/3}$. Rewrite the equation above using $u$ to obtain: $20u^2-6u-2=0$ Factor out $2$ to obtain: $2(10u^2-3u-1)=0$ Factor the trinomial to obtain: $2(5u+1)(2u-1)=0$ Factor each factor with a variable to zero then solve each equation to obtain: \begin{array}{ccc} &5u+1=0 &\text{or} &2u-1=0 \\&5u=-1 &\text{or} &2u=1 \\&u=-\frac{1}{5} &\text{or} &u=\frac{1}{2} \end{array} Since $u=x^{1/3}$, then: \begin{array}{ccc} &u=-\frac{1}{5} &\text{or} &u=\frac{1}{2} \\&x^{1/3}=-\frac{1}{5} &\text{or} &x^{1//3}=\frac{1}{2} \\&(x^{1/3})^3=(-\frac{1}{5})^3 &\text{or} &(x^{1//3})^3=(\frac{1}{2})^3 \\&x=-\frac{1}{125} &\text{or} &x=\frac{1}{8} \end{array} Therefore, the solutions are $x=-\frac{1}{125}$ and $x=\frac{1}{8}$.
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