Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 4 - Quadratic Functions - 4.5 Solving Equations by Factoring - 4.5 Exercises - Page 362: 91

Answer

The solution set is $\left\{-3, 0\right\}$.

Work Step by Step

Factor out $6b$ to obtain: $$6b(b+3)=0$$ Use the Zero-Product Property by equating each factor to zero, then solving each equation to obtain: \begin{align*} 6b&=0 &\text{or}& &b+3=0\\ b&=0 &\text{or}& &b=-3\\ \end{align*} Checking: \begin{align*} 6(0^2)+18(0)&=0\\ 6(0)+0&=0\\ 0&\stackrel{\checkmark}=0 \end{align*} \begin{align*} 6(-3)^2+18(-3)&=0\\ 6(9)+(-54)&=0\\ 0&\stackrel{\checkmark}=0 \end{align*} Therefore, the solution set is $\left\{-3, 0\right\}$.
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