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: 48


$x=\left\{ -\dfrac{5}{9},\dfrac{5}{9} \right\}$

Work Step by Step

Using the properties of equality, the given expression, $ 81x^2-5=20 ,$ is equivalent to \begin{array}{l}\require{cancel} 81x^2-5-20=0 \\\\ 81x^2-25=0 .\end{array} Factoring the above equation, $ 81x^2-25=0 ,$ results to \begin{array}{l}\require{cancel} (9x+5)(9x-5)=0 .\end{array} Equating each factor to zero (Zero Product Principle), then the solutions to the equation, $ (9x+5)(9x-5)=0 ,$ are \begin{array}{l}\require{cancel} 9x+5=0 \\\\ 9x=-5 \\\\ x=-\dfrac{5}{9} ,\\\\\text{OR}\\\\ 9x-5=0 \\\\ 9x=5 \\\\ x=\dfrac{5}{9} .\end{array} Hence, $ x=\left\{ -\dfrac{5}{9},\dfrac{5}{9} \right\} .$
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