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 R - Elementary Algebra Review - R.5 Polynomials and Factoring - R.5 Exercise Set - Page 970: 40


$x=\left\{ -\dfrac{3}{10},\dfrac{3}{10} \right\}$

Work Step by Step

Dividing both sides by $100,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 100x^2=9 \\\\ x^2=\dfrac{9}{100} .\end{array} Taking the square root of both sides (Square Root Property), the solutions are \begin{array}{l}\require{cancel} x=\pm\sqrt{\dfrac{9}{100}} \\\\ x=\pm\sqrt{\left(\dfrac{3}{10}\right)^2} \\\\ x=\pm\dfrac{3}{10} .\end{array} Hence, the solutions are $ x=\left\{ -\dfrac{3}{10},\dfrac{3}{10} \right\} .$
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