## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$\frac{1}{36}$, $(x+(1/6))^2$
To complete the square of $x^2 +\frac{1}{3}x$, we add $(b/2)^2$. In this case, the "$b$" is the second coefficient, $1/3$, so we add $((1/3)/2)^2=1/36$ and we have $x^2 +\frac{1}{3}x+\frac{1}{36}$. Then by factorization, we get $$x^2 +\frac{1}{3}x+\frac{1}{36}=(x+(1/6))^2 .$$