## Algebra 2 (1st Edition)

The trinomial $x^{2}-13x+c$ is a perfect square when $c=\displaystyle \frac{169}{4}.$ Then $x^{2}-13x+\displaystyle \frac{169}{4}=(x-\frac{13}{2})(x-\frac{13}{2})=(x-\frac{13}{2})^{2}$.
$x^{2}-13x+c\qquad$ ... find half the coefficient of $x$, which is $\displaystyle \frac{-13}{2}$. $\qquad$ ...square the result. $(\displaystyle \frac{-13}{2})^{2}=\frac{169}{4}\qquad$ ...substitute $c$ with $\displaystyle \frac{169}{4}$ in the original expression $x^{2}-13x+\displaystyle \frac{169}{4}=(x-\frac{13}{2})^{2}$