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: 24


The solutions are $x=\frac{1}{27}$ and $x=-64$.

Work Step by Step

Subtract $4$ to both sides of the equation to obtain: $3x^{2/3}+11x^{1/3}-4=0 \\3(x^{1/3})^2+11x^{1/3}-4=0$ Let $u=x^{1/3}$ Rewrite the equation above using the variable $u$ to obtain: $3u^2+11u-4=0$ Factor the trinomial to obtain: $(3u-1)(u+4)=0$ Use the Zero-Factor Property by equating each unique factor to zero. then solve each equation: \begin{array}{ccc} &3u-1 = 0 &\text{ or } &u+4=0 \\&3u=1 &\text{ or } &u=-4 \\&u=\frac{1}{3} &\text{ or } &u=-4 \end{array} Since $u=x^{1/3}$, then \begin{array}{ccc} &u= \frac{1}{3} &\text{or} &u=-4 \\&x^{1/3}=\frac{1}{3} &\text{or} &x^{1/3}=-4 \\&(x^{1/3})^3=(\frac{1}{3})^3 &\text{or} &(x^{1/3})^3=(-4)^3 \\&x=\frac{1}{27} &\text{or} &x=-64 \end{array} Thus, the solutions are $x=\frac{1}{27}$ and $x=-64$.
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