Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 8 - Review - Page 532: 60

Answer

$n=\left\{ -\dfrac{4}{9}, \dfrac{2}{9} \right\}$

Work Step by Step

Taking the square root of both sides, the solution/s of the given quadratic equation, $ (9n+1)^2=9 ,$ is/are \begin{array}{l}\require{cancel} 9n+1=\pm\sqrt{9} \\\\ 9n+1=\pm3 \\\\ 9n=-1\pm3 \\\\ n=\dfrac{-1\pm3}{9} \\\\ n=\left\{ -\dfrac{4}{9}, \dfrac{2}{9} \right\} .\end{array}
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