College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 1 - Equations and Inequalities - Exercise Set 1.5 - Page 160: 45


$\frac{1}{36}$; $(x-\frac{1}{6})^2$

Work Step by Step

To complete the square, add a term that is the square of half the coefficient of $x$ in the middle term. Then factor by trial and error. $x^2-\frac{1}{3}x$ $(\frac{-\frac{1}{3}}{2})^2 = (-\frac{1}{3}\times\frac{1}{2})^2=(-\frac{1}{6})^2 = \frac{1}{36}$ $x^2-\frac{1}{3}x+\frac{1}{36}$ $(x-\frac{1}{6})^2$
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.