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: 42


$y=\left\{ -\dfrac{1}{4},\dfrac{2}{3} \right\}$

Work Step by Step

Factoring the given equation, $ 12y^2-5y=2 ,$ results to \begin{array}{l}\require{cancel} 12y^2-5y-2=0 \\\\ (3y-2)(4y+1)=0 .\end{array} Equating each factor to zero (Zero Product Principle), then the solutions to the equation, $ (3y-2)(4y+1)=0 ,$ are \begin{array}{l}\require{cancel} 3y-2=0 \\\\ 3y=2 \\\\ y=\dfrac{2}{3} ,\\\\\text{OR}\\\\ 4y+1=0 \\\\ 4y=-1 \\\\ y=-\dfrac{1}{4} .\end{array} Hence, $ y=\left\{ -\dfrac{1}{4},\dfrac{2}{3} \right\} .$
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