## Algebra 1

$k=1$
$kx^2-10x+25=0$ 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 $x=5$.