Answer
$k=1$
Work Step by Step
$kx^2-10x+25=0$ and $k\ne0$
$kx^2-5x-5x+25=0$
$x(kx-5)-5(x-5)=0$
The above formula is true when $k=1$, since we then have the following:
$x(kx-5)-5(x-5)=0$
$x(x-5)-5(x-5)=0$
$(x-5)(x-5)=0$
Since both of the equations within the problem are the same, we can determine the only solution is $x=5$.
