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: 26


The solutions are $x=-1$ and $x=-7.5$.

Work Step by Step

Multiply $(x+6)^2$ to both sides of the equation to obtain: $(x+6)^2\left(2-\dfrac{7}{x+6}\right)=\dfrac{15}{(x+6)^2} \cdot (x+6)^2 \\2(x+6)^2-(x+6)(7)=15 \\2(x+6)^2-7(x+6)-15=0$ Let $u=x+6$ Rewrite the equation above using the variable $u$ to obtain: $2u^2-7u-15=0$ Factor the trinomial to obtain: $(2u+3)(u-5)=0$ Use the Zero-Factor Property by equating each unique factor to zero. then solve each equation: \begin{array}{ccc} &2u+3 = 0 &\text{ or } &u-5=0 \\&2u=-3 &\text{ or } &u=5 \\&u=-\frac{3}{2} &\text{ or } &u=5 \end{array} Since $u=x+6$, then \begin{array}{ccc} &u= -\frac{3}{2} &\text{or} &u=5 \\&x+6=-\frac{3}{2} &\text{or} &x+6=5 \\&x=-\frac{3}{2}]-6 &\text{or} &x=5-6 \\&x=-7.5 &\text{or} &x=-1 \end{array} Thus, the solutions are $x=-1$ and $x=-7.5$.
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