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: 51


{$-1 - \dfrac{\sqrt{6}}{3},-1 + \dfrac{\sqrt{6}}{3}$}

Work Step by Step

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

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.