Quadratic Equation

Answer
Facts:

If f(x)-x has roots d,e
then
f(d)-d=0
f(e)-e=0

i.e
f(d) = d ......... EQ I
f(e) = e ......... EQ II

Proof by contradiction

We assume if d,e are roots of f(x)-x=0 then they cannot be roots of f(f(x))-x=0 i.e. f(f(d))-d cannot be 0


Proof:

We have f(f(x))-x = L

putting x as d (or e)

f(f(d))-d=L

f(d)-d = L ... from EQ I

d-d = L .... from EQ I

0 = L

which seesntially proves that the assumption was wrong

So By proof of contradiction d,e are also the roots of f(f(x))-x=0

d,e are roots of f(x)-x=0
so if x=d,e
f(d)-d=0 & f(d)=d
and
f(e)-e=0 & f(e)=e
so

f[f(d)]=f(d)=d and f[f(e)]=f(e)=e

so if f[f(x)]-x=L
and x=d,e
then
f[f(d)]-d=d-d=L=0
and
f[f(e)]-e=e-e=L=0
and d & e are roots of f[f(x)]-x=0

Contradiction would be to assume that L<>0 and then prove L=0, but if you can prove L=O from deduction you can use the definition of a equation root directly and save an extra step.

Rap

You remind me of my Maths Prof... She used to pinpoint things that were not so important

Anyways your approach is proper ... I agree Sir!

Re: Quadratic Equation


This looks a bit wrongly stated to me, because we are given

f(x) = ax^2 + bx + c and f(x) - x = 0 or stated otherwise f(x) = x, implying a = 0, b = 1 and c = 0, a contradiction!

So if f(x) equals x of course f[f(x)] - x = 0 because f(x) = x so f(f(f(f(x)))) = x so f(f(x)) - x must equal 0.

Knowledge of the roots D and E is superflous when the question is stated so!

The only correct way of asking what you meant would be to say

f(x) = ax^2 + bx + c where a, b, c are real and a is not equal to zero

If d, e are the roots of f(x) - x, prove d, e are also the roots of f[f(x)] - x

* * *

Adding the = 0 as you did technically introduced a contradiction that a was not equal to zero!