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: 10


$x=\left\{ 1,\dfrac{5}{8} \right\}$

Work Step by Step

Equating each factor to zero (Zero Product Principle), then the solutions to the given equation, $ (5t-8)(t-1)=0 ,$ is \begin{array}{l} 5t-8=0 \\\\ 5t=8 \\\\ t=\dfrac{8}{5} ,\\\\\text{OR}\\\\ t-1=0 \\\\ t=1 .\end{array} Hence, $ x=\left\{ 1,\dfrac{5}{8} \right\} .$
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