Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 8 - Review Exercises - Page 657: 3


{$\pm \dfrac{\sqrt{6}}{3}$}

Work Step by Step

Given: $3x^2-2=0$ Take the common terms out. $(x^2-\dfrac{2}{3})=0$ This can be written as: $[x^2-(\sqrt{\dfrac{2}{3}})^2]=0$ $(x-\sqrt{\dfrac{2}{3}})(x+\sqrt{\dfrac{2}{3}})=0$ or,$(x-\dfrac{\sqrt{6}}{3})(x+\dfrac{\sqrt{6}}{3})=0$ Thus, $x=\pm \dfrac{\sqrt{6}}{3} $ Hence, $x=${$\pm \dfrac{\sqrt{6}}{3}$}
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