Intermediate Algebra (12th Edition)

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

Chapter 5 - Review Exercises - Page 361: 41


$x=\left\{ -\dfrac{5}{2},\dfrac{10}{3} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given eqution, $ 6x^2=5x+50 ,$ 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 properties of equality, the equation above is equivalent to \begin{array}{l}\require{cancel} 6x^2-5x-50=0 .\end{array} Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $ 6(-50)=-300 $ and the value of $b$ is $ -5 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ 15,-20 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 6x^2+15x-20x-50=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} (6x^2+15x)-(20x+50)=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 3x(2x+5)-10(2x+5)=0 .\end{array} Factoring the $GCF= (2x+5) $ of the entire expression above results to \begin{array}{l}\require{cancel} (2x+5)(3x-10)=0 .\end{array} Equating each factor to zero (Zero Product Property), the solutions to the equation above are \begin{array}{l}\require{cancel} 2x+5=0 \\\\\text{OR}\\\\ 3x-10=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 2x+5=0 \\\\ 2x=-5 \\\\ x=-\dfrac{5}{2} \\\\\text{OR}\\\\ 3x-10=0 \\\\ 3x=10 \\\\ x=\dfrac{10}{3} .\end{array} Hence, $ x=\left\{ -\dfrac{5}{2},\dfrac{10}{3} \right\} .$
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