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: 33

Answer

$t=\left\{ -\dfrac{7}{2},\dfrac{7}{2} \right\}$

Work Step by Step

Factoring the given equation, $ 4t^2=49 ,$ results to \begin{array}{l}\require{cancel} 4t^2-49=0 \\\\ (2t+7)(2t-7)=0 .\end{array} Equating each factor to zero (Zero Product Principle), then the solutions to the equation, $ (2t+7)(2t-7)=0 ,$ are \begin{array}{l}\require{cancel} 2t+7=0 \\\\ 2t=-7 \\\\ t=-\dfrac{7}{2} ,\\\\\text{OR}\\\\ 2t-7=0 \\\\ 2t=7 \\\\ t=\dfrac{7}{2} .\end{array} Hence, $ t=\left\{ -\dfrac{7}{2},\dfrac{7}{2} \right\} .$
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