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: 25


The solutions are $t=-\frac{2}{3}$ and $t=-\frac{4}{3}$.

Work Step by Step

Multiply $(3t-2)^2$ to both sides of the equation to obtain: $(3t-2)^2\left(1+\dfrac{2}{3t-2}\right)=\dfrac{8}{(3t-2)^2} \cdot (3t-2)^2 \\(3t-2)^2+(3t-2)(2)=8 \\(3t-2)^2+2(3t-2)-8=0$ Let $u=3t-2$ Rewrite the equation above using the variable $u$ 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 unique factor to zero. then solve each equation: \begin{array}{ccc} &u+4 = 0 &\text{ or } &u-2=0 \\&u=-4 &\text{ or } &u=2 \end{array} Since $u=3t-2$, then \begin{array}{ccc} &u= -4 &\text{or} &u=2 \\&3t-2=-4 &\text{or} &3t-2=2 \\&3t=-4+2 &\text{or} &3t=2+2 \\&3t=-2 &\text{or} &3t=4 \\&t=-\frac{2}{3} &\text{or} &t=\frac{4}{3} \end{array} Thus, the solutions are $t=-\frac{2}{3}$ and $t=-\frac{4}{3}$.
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