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).
Thanks!
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)