## Algebra 1

$\frac{25}{4}$
$k^2-5k+c$ Compare it with the standard form of quadratic equation $ax^2+bx+c$, we have $a=1, b=-5$ It will have discriminant, $d$ with formula : $d=b^2-4ac$ Since, $d=0$ for repeated root of the expression. Therefore, $b^2=4ac$ $\implies$ $c=\dfrac{b^2}{4a}$ To complete the square, plug in $a=1, b=-5$. Thus, $c=\dfrac{b^2}{4a}=\dfrac{(-5)^2}{4}=\frac{25}{4}$