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: 51


$x=\left\{ -3,1 \right\}$

Work Step by Step

Using the properties of equality, the given expression, $ 3x^2-2x=9-8x ,$ is equivalent to \begin{array}{l}\require{cancel} 3x^2-2x+8x-9=0 \\\\ 3x^2+6x-9=0 .\end{array} Factoring the above equation, $ 3x^2+6x-9=0 ,$ results to \begin{array}{l}\require{cancel} 3(x^2+2x-3)=0 \\\\ x^2+2x-3=0 \\\\ (x+3)(x-1)=0 .\end{array} Equating each factor to zero (Zero Product Principle), then the solutions to the equation, $ (x+3)(x-1)=0 ,$ are \begin{array}{l}\require{cancel} x+3=0 \\\\ x=-3 ,\\\\\text{OR}\\\\ x-1=0 \\\\ x=1 .\end{array} Hence, $ x=\left\{ -3,1 \right\} .$
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