The quadratic equation ax^2 + 8x + x = 0 has exactly one solution. If a + c = 10, and a < c, find the ordered pair (a, c).

My original answer was (-8, -2), but it was incorrect so I must have done something wrong.

Thanks in advance.

Guest Feb 13, 2018

#1**+1 **

Do you mean this ???

ax^2 + 8x + c = 0 ??? if so

If we have only one solution, the discriminant will = 0 .....so.....

8^2 - 4ac = 0 and a + c = 10 ⇒ c = 10 - a

So....substituting, we have

8^2 - 4a(10 - a) = 0

8^2 - 40a + 4a^2 = 0

4a^2 - 40a + 64 = 0 divide through by 4

a^2 - 10a + 16 = 0 factor

(a - 8) ( a - 2) = 0

The solutions are a = 8 or a = 2

Note that if a = 8, then c = 2 or

If a = 2 then c = 8

But a > c so (a, c) = ( 8, 2 )

Check the ggraph here that only one zero exists at x = -1/2 :

https://www.desmos.com/calculator/idsuikzats

CPhill Feb 13, 2018