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 360: 48

Answer

$w=\frac{2}{3},w=\frac{5}{2}$

Work Step by Step

There is much work to be done here until the equation is simplified. The work speaks for itself! $$ \begin{aligned} 6 w^2-15 w&=4 w-10\\ 6 w^2-15 w-4 w&+10=0\\ (6 w^2-15 w)-(4 w&-10)=0\\ 3w(2 w-5)-2(2 w&-5)=0\\ (2 w-5)(3w-2)&= 0. \end{aligned} $$ Hence $$ \begin{aligned} 2 w-5& =0 \\ 2w & =5 \\ w & =\frac{5}{2} \\ & =2.5 \\ 3 w-2& =0 \\ 3w & =2 \\ w & =\frac{2}{3}. \end{aligned} $$ Check $$ \begin{aligned} 6 \cdot 2.5^2-15\cdot 2.5&\stackrel{?}{=}4\cdot 2.5-10 \\ 0& =0\checkmark\\ 6 \cdot \left(\frac{2}{3}\right)^2-15\cdot \frac{2}{3}&=4\cdot \frac{2}{3}-10 \\ -\frac{22}{3}& =-\frac{22}{3}\checkmark. \end{aligned} $$
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