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: 42


The solutions are $-\frac{125}{216}$ and $125$.

Work Step by Step

The given equation can be written as: $6(x^{1/3})^2-25x^{1/3}-25=0$ Let $u=x^{1/3}$. Rewrite the equation above using $u$ to obtain: $6u^2-25u-25=0$ Factor the trinomial to obtain: $(6u+5)(u-5)=0$ Use the Zero Factor Property by equating each factor to zero, then solve each equation to obtain: \begin{array}{ccc} \\&6u+5=0 &\text{or} &u-5=0 \\&6u=-5 &\text{or} &u=5 \\&u=-\frac{5}{6} &\text{or} &u=5 \end{array} Since $u=x^{1/3}$, then: \begin{array}{ccc} \\&u=-\frac{5}{6} &\text{or} &u=5 \\&x^{1/3}=-\frac{5}{6} &\text{or} &x^{1/3}=5 \\&(x^{1/3})^3=(-\frac{5}{6})^3 &\text{or} &(x^{1/3})^3=5^3 \\&x=-\frac{125}{216} &\text{or} &x=125 \end{array} Therefore, the solutions are $-\frac{125}{216}$ and $125$.
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