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: 41


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

Work Step by Step

Factoring the given equation, $ 4y^2=7y+15 ,$ results to \begin{array}{l}\require{cancel} 4y^2-7y-15=0 \\\\ (2y+3)(2y-5)=0 .\end{array} Equating each factor to zero (Zero Product Principle), then the solutions to the equation, $ (2y+3)(2y-5)=0 ,$ are \begin{array}{l}\require{cancel} 2y+3=0 \\\\ 2y=-3 \\\\ y=-\dfrac{3}{2} ,\\\\\text{OR}\\\\ 2y-5=0 \\\\ 2y=5 \\\\ y=\dfrac{5}{2} .\end{array} Hence, $ y=\left\{ -\dfrac{3}{2},\dfrac{5}{2} \right\} .$
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