Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 5 - Chapter 5 Test - Page 362: 17


$m=\left\{ -\dfrac{2}{5},1 \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given eqution, $ 5m(m-1)=2(1-m) ,$ express the equation in the form $ax^2+bx+c=0.$ Then express the equation in factored form. Next step is to equate each factor to zero (Zero Product Property). Finally, solve each equation. $\bf{\text{Solution Details:}}$ Using the Distributive Property which is given by $a(b+c)=ab+ac$ and then combining like terms, the expression above is equivalent to \begin{array}{l}\require{cancel} 5m^2-5m=2-2m \\\\ 5m^2+(-5m+2m)-2=0 \\\\ 5m^2-3m-2=0 .\end{array} Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $ 5(-2)=-10 $ and the value of $b$ is $ -3 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ 2,-5 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 5m^2+2m-5m-2=0 .\end{array} Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (5m^2+2m)-(5m+2)=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} m(5m+2)-(5m+2)=0 .\end{array} Factoring the $GCF= (5m+2) $ of the entire expression above results to \begin{array}{l}\require{cancel} (5m+2)(m-1)=0 .\end{array} Equating each factor to zero (Zero Product Property), the solutions to the equation above are \begin{array}{l}\require{cancel} 5m+2=0 \\\\\text{OR}\\\\ m-1=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 5m+2=0 \\\\ 5m=-2 \\\\ m=-\dfrac{2}{5} \\\\\text{OR}\\\\ m-1=0 \\\\ m=1 .\end{array} Hence, $ m=\left\{ -\dfrac{2}{5},1 \right\} .$
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