# Function and root

• March 23rd 2011, 11:40 AM
mathfun
Function and root
Let $a \in {{\Cal R}^*}$ and $b,c \in {\Cal R}$ such that $4ac < {\left( {b - 1} \right)^2}$. Let $f:{\Cal R} \to {\Cal R}$ be a function such that $f\left( {a{x^2} + bx + c} \right) = a{f^2}\left( x \right) + bf\left( x \right) + c,\forall x \in {\Cal R}$. Prove that $f\left( {f\left( x \right)} \right) = x$ has at least one root
• March 23rd 2011, 11:57 AM
TheChaz
What have you tried so far?
"Prove that..." is your responsibility, and while many of us (myself included) are willing and able to help, you have to show some effort besides registering for a forum and copying the question down in LaTex markup.
• March 23rd 2011, 12:02 PM
mathfun
Ok, seeing the hypothesis i said: there are roots $d,e$ of the equation $a{x^2} + bx + c = x$ because its equivalent to $a{x^2} + \left( {b - 1} \right)x + c = 0$. I set d and e in the given relation. and then i get stuck.
• March 25th 2011, 01:38 AM
mathfun
Let ${x_0}$ be such that it is the solution of the equation $a{x^2} + bx + c = x$ where ${x_0} = \tfrac{{1 - b + \sqrt {{{\left( {b - 1} \right)}^2} - 4ac} }}{{2a}}$.
I set x0 in the given relation $f\left( {a{x_0}^2 + b{x_0} + c} \right) = a{f^2}\left( {{x_0}} \right) + bf\left( {{x_0}} \right) + c \Rightarrow a{f^2}\left( {{x_0}} \right) + \left( {b - 1} \right)f\left( {{x_0}} \right) + c = 0$

Now one of the roots of the last equation are $f\left( {{x_0}} \right) = \tfrac{{1 - b + \sqrt {{{\left( {b - 1} \right)}^2} - 4ac} }}{{2a}} = {x_0}$ and we get the wanted.

Is it correct?
• March 25th 2011, 08:27 PM
TheChaz
Quote:

Originally Posted by mathfun
Ok, seeing the hypothesis i said: there are roots $d,e$ of the equation $a{x^2} + bx + c = x$ because its equivalent to $a{x^2} + \left( {b - 1} \right)x + c = 0$. I set d and e in the given relation. and then i get stuck.

You're almost there. What is the discriminant of this quadratic? If you subtract 4ac from the given inequality, you will conclude that
("the discriminant") >= 0
and this implies that the quadratic has >= 1 real root.
• March 26th 2011, 01:32 AM
mathfun
Actually it has 2 roots, right? Plaese see the solution i gave in post No. 5
• March 26th 2011, 06:20 AM
TheChaz
I don't believe so. If (b - 1)^2 = 4ac, then your "two" solutions are actually the same.
• March 26th 2011, 06:26 AM
mathfun
but ${\left( {b - 1} \right)^2} > 4ac$
• March 26th 2011, 07:08 AM
TheChaz
Hah! All this time I thought it was >=!!
ok you win :)
• March 26th 2011, 07:10 AM
mathfun
Quote:

Originally Posted by mathfun
Let ${x_0}$ be such that it is the solution of the equation $a{x^2} + bx + c = x$ where ${x_0} = \tfrac{{1 - b + \sqrt {{{\left( {b - 1} \right)}^2} - 4ac} }}{{2a}}$.
I set x0 in the given relation $f\left( {a{x_0}^2 + b{x_0} + c} \right) = a{f^2}\left( {{x_0}} \right) + bf\left( {{x_0}} \right) + c \Rightarrow a{f^2}\left( {{x_0}} \right) + \left( {b - 1} \right)f\left( {{x_0}} \right) + c = 0$

Now one of the roots of the last equation are $f\left( {{x_0}} \right) = \tfrac{{1 - b + \sqrt {{{\left( {b - 1} \right)}^2} - 4ac} }}{{2a}} = {x_0}$ and we get the wanted.

Is it correct?

Could you please tell me if this solution is correct?
• March 26th 2011, 08:34 AM
worc3247
Looks fine to me.