回复：You totally misunderstanding me, f(x)=a(X-5)^2+b,why b=2? you

by writting f(x)=a(x-5)^2+b, and f(2) = 0

You got f(x)=-b/9(x-5)^2+b

and after a useless calcul ff=-b/9(-b/9(x-5)^2+b-5)^2+b

you get f(f(5))=-b/9(b-5)^2+b=0, which is (b-5)^2=9

*** if you had noticed f(5)=b, you would have noticed also that b is a root of f(x), that is (x-5)^2=9, or (b-5)^2=9, since f(b)=0

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since you do not undestand Kang's explanation, you have no right to say b-5 < 0.

(logically, one get first the conclusion f(5)=2, by reasoning, not by guessing)

Do you understand why Kang put f(x)=a(X-5)^2+b, while this is not a general form of a quadratic polynomial?