College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 1 - Review Exercises - Page 145: 20

Answer

$\left\{-1, \dfrac{1}{2}\right\}$

Work Step by Step

Let $u=x^{-3}$. Then, the given equation becomes: $$u^2-7u-8=0$$ Factor the trinomial to obtain: $$(u-8)(u+1)=0$$ Use the Zero-Product Property by equating each factor to zero. Then, solve each equation to obtain: \begin{align*} u-8&=0 &\text{or}& &u+1=0\\ u&=8 &\text{or}& &u=-1 \end{align*} With $u=x^{-3}$, substitute $x^{-3}$ to $u$ to obtain: \begin{align*} x^{-3}&=8 &\text{or}& &x^{-3}=-1\\ \frac{1}{x^3}&=8 &\text{or}& &\frac{1}{x^3}=-1\\ \frac{1}{x^3}&=8 &\text{or}& &\frac{1}{x^3}=-1\\ 8x^3&=1 &\text{or}& &-x^3=1\\ x^3&=\frac{1}{8} &\text{or}& &x^3=-1\\ \sqrt[3]{x^3}&=\sqrt[3]{\frac{1}{8}} &\text{or}& &\sqrt[3]{x^3}=\sqrt[3]{-1}\\ x&=\frac{1}{2} &\text{or}& &x=-1 \end{align*}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.