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: 50


$x=\left\{ -1,\dfrac{6}{5} \right\}$

Work Step by Step

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