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