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: 34


The solutions are $2.5$ and $8$.

Work Step by Step

Subtract $5$ to both sides to obtain: $2(m-3)^2-9(m-3)-5=0$ Let $u=m-3$. Rewrite the equation above using $u$ to obtain: $2u^2-9u-5=0$ Factor the trinomial to obtain: $(2u+1)(u-5)=0$ 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-5=0 \\&2u=-1 &\text{or} & u=5 \\&u=-\frac{1}{2} &\text{or} & u=5 \end{array} Since $u=m-3$, then \begin{array}{ccc} \\&u=-\frac{1}{2} &\text{or} & u=5 \\&m-3=-\frac{1}{2} &\text{or} & m-3=5 \\&m=-\frac{1}{2}+3 &\text{or} & m=5+3 \\&m=2.5 &\text{or} & m=8 \end{array} Therefore, the solutions are $2.5$ and $8$.
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