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: 54


$t=\left\{ -\dfrac{3}{5},\dfrac{5}{4} \right\}$

Work Step by Step

Using $(a+b)(c+d)=ac+ad+bc+bd$ or the Distributive Property, the given expression, $ (2t+1)(4t-1)=14 ,$ is equivalent to \begin{array}{l}\require{cancel} 2t(4t)+2t(-1)+1(4t)+1(-1)=14 \\\\ 8t^2-2t+4t-1=14 \\\\ 8t^2-2t+4t-1-14=0 \\\\ 8t^2+2t-15=0 .\end{array} Factoring the above equation, $ 8t^2+2t-15=0 ,$ results to \begin{array}{l}\require{cancel} (2t+3)(4t-5)=0 .\end{array} Equating each factor to zero (Zero Product Principle), then the solutions to the equation, $ (2t+3)(4t-5)=0 ,$ are \begin{array}{l}\require{cancel} 2t+3=0 \\\\ 2t=-3 \\\\ t=-\dfrac{3}{5} ,\\\\\text{OR}\\\\ 4t-5=0 \\\\ 4t=5 \\\\ t=\dfrac{5}{4} .\end{array} Hence, $ t=\left\{ -\dfrac{3}{5},\dfrac{5}{4} \right\} .$
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