The quadratic equation $ax^2+32x+c=0$ has exactly one solution. If a+c=130, and a>c, find the ordered pair (a,c).

If we have only one solution, the discriminant must  = 0

So

32^2  - 4ac  = 0             ⇒  c  = 130 - a

1024 - 4  (a) (130 - a)  = 0

1024 - 520a + 4a^2  = 0

4a^2 -520a + 1024  = 0       divide through by 4

a^2 - 130a + 256  = 0      factor as

(a -128) (a - 2)  = 0

Set each factor to 0  and solve for a

a - 128  = 0           a - 2 =0

a = 128                  a = 2

But...since  a > c

Then a must = 128    and c = 130 - 128  = 2  

(a, c )  =  (128, 2)