Answer
$c=\displaystyle \frac{81}{4}$
$(x-\displaystyle \frac{9}{2})^{2}$
Work Step by Step
$ x^{2}-9x+c\qquad$ ... find half the coefficient of $x$, which is $\displaystyle \frac{-9}{2}$.
$\qquad$ ...square the result.
$(\displaystyle \frac{-9}{2})^{2}=\frac{81}{4}\qquad$ ...substitute $c$ with $\displaystyle \frac{81}{4}$ in the original expression
$x^{2}-9x+\displaystyle \frac{81}{4}=(x-\frac{9}{2})^{2}$
The trinomial $x^{2}-9x+c$ is a perfect square when $c=\displaystyle \frac{81}{4}.$
Then $x^{2}-9x+\displaystyle \frac{81}{4}=(x-\frac{9}{2})(x-\frac{9}{2})=(x-\frac{9}{2})^{2}$.
