f(x)=a(x-2)(x-b) a is not 0.
case 1: f(5) = 2 => b=(15a-2)/3a => f(0)=2ab=10a - 4/3.
case 2: f(5) = b => b=15a/(3a+1) => f(0)=2ab=30a/(3a+1).
Discussion:
let f(f(x)) = a^3(x-5)(x-p)(x-q)(x-r).
case 1: we have f(x) - 2 = a(x-5)(x-p) and f(x) - b = a(x-q)(x-r)
case 2: we have f(x) - b= a(x-5)(x-p) and f(x) - 2 = a(x-q)(x-r).
If a is real, in either case, b should be real, and we already know the unique answer.
If a is unreal, in either case, b is unreal too. Easy to see p, q and r are all unreal.
That is, a could be any unreal and f(0) could be any unreal.
