Help to solve a "Prove" that?

Question

Prove that if a and b are integers with a not equal to 0 and x is a real number such that ax(squared)+bx+b-a=0, then a|b.
Anyone got any thoughts on this one?
(the ax(squared), that is ax with the number 2 above the x)
(the "a not equal to 0" is "a with the equal sign with the line through it 0")

in "a|b", this means that "a divids b" or so I've been told....

Answer 1

Sounds to me like you're expected to use quadratic formula

ax^2 + bx + (b-a) = 0

This holds true where

x = (-b +/- SQRT(b^2 - 4a(b-a)))/2a {roots of a quadratic equation}

x = (-b +/- SQRT(b^2 - 4ab + 4a^2))/2a

x = (-b +/- SQRT((b-2a)^2))/2a

x = (-b +/- (b-2a))/2a

x = (-2b + 2a)/2a OR (-2a/2a)

x = 1 - 2(b/a) OR -1

Answer 2

hi. I think I am half way through.
If we solve the quadratic equation, a(x^2)+bx+(b-a)=0, we get the solution as
x=(-b+/-squareroot((b^2)-4ab+4a^2)/(2a))
=> x=1or x=(a-b)/a.
now if the condition were to be that x is also an integer, then it means that b is a multiple of a, the direct consequence of which is that a|b. Or is there any property of divisibility which might prove the result?