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 - Section 8.1 - The Square Root Property and Completing the Square - Exercise Set - Page 593: 52


{$1 - \dfrac{\sqrt{3}}{3},1 + \dfrac{\sqrt{3}}{3}$}

Work Step by Step

Given: $3x^2-6x+2=0$ This can be re-written as: $x^2-2x=\dfrac{-2}{3}$ We will have to add both sides $(-1)^2$ to complete the square. Thus, $x^2-2x+(1)^2=\dfrac{-2}{3}+(-1)^2$ or, $(x-1)^2=\dfrac{1}{3}$ or, $(x-1)=\pm \dfrac{\sqrt{3}}{3}$ or, $x=1 \pm \dfrac{\sqrt{3}}{3}$ Hence, our desired solution set is {$1 - \dfrac{\sqrt{3}}{3},1 + \dfrac{\sqrt{3}}{3}$}
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