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: 46


$\frac{1}{64}$; $(x-\frac{1}{8})^2$

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. $x^2-\frac{1}{4}x$ $(\frac{-\frac{1}{4}}{2})^2 = (-\frac{1}{4}\times\frac{1}{2})^2=(-\frac{1}{8})^2 = \frac{1}{64}$ $x^2-\frac{1}{4}x+\frac{1}{64}$ $(x-\frac{1}{8})^2$
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