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: 44


$\frac{4}{25}$; $(x+\frac{2}{5})^2$

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. $x^2+\frac{4}{5}x$ $(\frac{\frac{4}{5}}{2})^2 = (\frac{4}{5}\times\frac{1}{2})^2=(\frac{2}{5})^2 = \frac{4}{25}$ $x^2+\frac{4}{5}x+\frac{4}{25}$ $(x+\frac{2}{5})^2$
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