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: 38


The solutions are $-8$ and $64$.

Work Step by Step

The given equation can be written as: $(x^{1/3})^2-2x^{1/3}-8=0$ Let $u=x^{1/3}$. Rewrite the equation above using $u$ to obtain: $u^2-2u-8=0$ Factor the trinomial to obtain: $(u-4)(u+2)=0$ Use the Zero Factor Property by equating each factor to zero, then solve each equation to obtain: \begin{array}{ccc} \\&u-4=0 &\text{or} &u+2=0 \\&u=4 &\text{or} &u=-2 \end{array} Since $u=x^{1/3}$, then: \begin{array}{ccc} \\&u=4 &\text{or} &u=-2 \\&x^{1/3}=4 &\text{or} &x^{1/3}=-2 \\&(x^{1/3})^3=4^3 &\text{or} &(x^{1/3})^3=(-2)^3 \\&x=64 &\text{or} &x=-8 \end{array} Therefore, the solutions are $-8$ and $64$.
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