The quadratic equation $ax^2+20x+c=4x$ has exactly one solution. If $a+c=20$, and $a<c$ find the ordered pair $(a,c)$.

Answer 1

Subtracting $4x$ from both sides gives us $ax^2 +16x +c = 0.$ Since there is only one solution, the discriminant of this quadratic must be $0$. This gives us
\[16^2 - 4ac = 0 \implies 256 - 4ac = 0 \implies 64 = ac\]
Given $a+c=20$, we solve to get $a=4$ and $c=16$. Thus $(a,c) = \boxed{(4,16)}$.