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: 41

Answer

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

Work Step by Step

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