## College Algebra (6th Edition)

$\frac{1}{36}$; $(x-\frac{1}{6})^2$
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. $x^2-\frac{1}{3}x$ $(\frac{-\frac{1}{3}}{2})^2 = (-\frac{1}{3}\times\frac{1}{2})^2=(-\frac{1}{6})^2 = \frac{1}{36}$ $x^2-\frac{1}{3}x+\frac{1}{36}$ $(x-\frac{1}{6})^2$