For what values of a does the quadratic equation ax^2 + 20x + 4 = 0 have equal roots?

To solve for what values the equation ax^2+20x+4  0 has
zero roots.

Given that ax^2+20x+4  =  0 has equal
roots.

We divide both sides by a. This is  possible as
a cannot be be zero. Further, if a = 0, then it is a contradicion for ax^2+bx+c to be
quadratic.

So x^2+20x/a +4/a is = 0 has equal
roots.

Let the equal roots be m. Then x^2+20x/a +c/a =
(x-m)^2.

So x^2+20x/a+4/a =
x^2-2m+m^2.

So this has to be an
identiy.

Sa 20/a = -2m,

 or
10/a  = -m..(1)

 4/a =
m^2...(2)

We eliminate m from (1) and (2):  m^2 = 4/a =
(10/a)^2.

So 4a = 100 . Or a = 100/4
.

a = 25.

Therefore a =
25.