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 351: 34


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

Work Step by Step

Factoring the given equation, $ 9t^2=25 ,$ results to \begin{array}{l}\require{cancel} 9t^2-25=0 \\\\ (3t+5)(3t-5)=0 .\end{array} Equating each factor to zero (Zero Product Principle), then the solutions to the equation, $ (3t+5)(3t-5)=0 ,$ are \begin{array}{l}\require{cancel} 3t+5=0 \\\\ 3t=-5 \\\\ t=-\dfrac{5}{3} ,\\\\\text{OR}\\\\ 3t-5=0 \\\\ 3t=5 \\\\ t=\dfrac{5}{3} .\end{array} Hence, $ t=\left\{ -\dfrac{5}{3},\dfrac{5}{3} \right\} .$
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