Let $p(x)$ be a quadratic polynomial such that for distinct reals $\alpha$ and $\beta$, $$p(\alpha)=\alpha\ \&\ p(\beta)=\beta$$
Show that $\alpha$ and $\beta$ are the roots of the following equation $$p(p(x))-x=0$$
Also find the remaining roots.
The first part was very simple to prove, in order to find the remaining roots, I assumed $t$ to be a root of the second equation with $p(t)=u.$ Hence it immediately follows that $u$ is also the root of the second equation with $p(u)=t$. Now the task is to find such $u$ and $t$. We have
$$at^2+bt+c=u \ \ \ \ (1)$$ and $$au^2+bu+c=t\ \ \ (2)$$
Thus, taking (1) – (2) and cancelling $u-t$ we get
$$u+t=\frac{-(1+b)}{a}$$ Now, taking $u^2*(1) – t^2*(2)$ and cancelling $u-t$ again, we get $$ut=\frac{1+b+ac}{a^2}$$
With this, we see that $u$ and $t$ are the roots of the following equation
$$a^2x^2+a(1+b)x+(1+b+ac)=0$$
And thus the roots can be computed using the quadratic formula.
First of all, I want to know whether my answer is correct or not, as the book that I use does not provide any answer to this problem, and if it is incorrect, I would like to know the correct answer.
If my answer is correct, can the answer be better in anyway? (as I only came up with an equation for the roots… and writing down the final answer using the quadratic formula looks crazy!!)
Thanks for any answers!!
Edit: Here I have assumed $p(x)=ax^2+bx+c$
Best Answer
Write $q(x)= p(x)-x$, then given equation is equivalent to $$q(q(x)+x)+q(x)=0$$
Since $\alpha $ and $\beta $ are roots for $q$ we have $$q(x)=c(x-\alpha)(x-\beta)$$ where $c\ne 0$, so $$ c(q(x)+x-\alpha )(q(x)+x-\beta)+c(x-\alpha)(x-\beta)=0$$
so $$(x-\alpha)(x-\beta)\Big(\color{\red}{(cx-c\alpha+1)(cx-c\beta+1)+1}\Big)=0$$
So you need to solve the red equation...