Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 5 - Polynomials and Factoring - 5.7 Solving Quadratic Equations by Factoring - 5.7 Exercise Set - Page 352: 53


$a=\left\{ -\dfrac{5}{2},\dfrac{4}{3} \right\}$

Work Step by Step

Using $(a+b)(c+d)=ac+ad+bc+bd$ or the Distributive Property, the given expression, $ (6a+1)(a+1)=21 ,$ is equivalent to \begin{array}{l}\require{cancel} 6a(a)+6a(1)+1(a)+1(1)=21 \\\\ 6a^2+6a+a+1=21 \\\\ 6a^2+6a+a+1-21=0 \\\\ 6a^2+7a-20=0 .\end{array} Factoring the above equation, $ 6a^2+7a-20=0 ,$ results to \begin{array}{l}\require{cancel} (2a+5)(3a-4)=0 .\end{array} Equating each factor to zero (Zero Product Principle), then the solutions to the equation, $ (2a+5)(3a-4)=0 ,$ are \begin{array}{l}\require{cancel} 2a+5=0 \\\\ 2a=-5 \\\\ a=-\dfrac{5}{2} ,\\\\\text{OR}\\\\ 3a-4=0 \\\\ 3a=4 \\\\ a=\dfrac{4}{3} .\end{array} Hence, $ a=\left\{ -\dfrac{5}{2},\dfrac{4}{3} \right\} .$
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