quadratic equ. problem no.4
No I dont think I understand the second step in your proof.
You say f(x) - x = 0
Then you say f(f(x)) - x = 0 ???
The idea is functions obey "axiom of substitution",(AOS)
that is if
So if f(x) = x for some x,
Then by AOS, f(f(x)) = f(x) = x for those x
if i'll just follow the argument, it would be like this..
now, from your AOS,
But if I say it obeys axiom of substitution, it looks like functions follow some kind of law. Makes it look more important
P.S:LOL, I recently learned that term from Terence Tao's Book on Real Analysis. He builds the number system and introduces this as a basic axiom . That funny name stuck me as something important, so i use it repeatedly to demonstrate wisdom
oic.. i have not seen his book yet.. anyhow, thanks for that info..
if you say that  equals , why don't you apply your result to [b]?
i don't know if this is the proper proof, but i'll write it anyway..
substitute to and show that it is equal to 0..
here it is..
since is a root of , we have .
therefore is a root of .. (similarly, is a root.)
for [b], this is a hint.. ..
for example, and
got it? so, your [b] should be
just expand it and simplify..
Since we already know roots of f(x)=x are also roots of f(f(x)) = x,
Solve f(x) = x, i.e. x^2 - 3x + 2 = x. That is .
f(f(x)) - x = (x^2 - 3x + 2)^2 - 3(x^2 - 3x + 2) + 2 - x
And it is a quartic equation. But we already know two roots of it by (i), so divide the quartic by , you will get a quadratic quotient. Solve that quadratic to get the remaining 2 roots