Degree 2 Recurring monic polynomials Consider a monic polynomial $x^2+ax+b=0$, with real coefficients. If it has real roots $p$ and $q$, such that $p\leq q$, then you construct a new monic polynomial as $x^2+px+q=0$. 
If this polynomial has real roots $p_1$ and $q_1$, $p_1\leq q_1$ you again construct a monic polynomial as $x^2+p_1x+q_1=0$. You go on doing this until you get complex roots. Once you get complex roots, you stop and count the number of polynomials constructed so far. 
So my question here is if you begin with any arbitrary monic polynomial with real coefficients, how many such polynomials can be created, will this process go on forever or will it give a finite number of polynomials? How do you go on to start with the solution?

If you start with the polynomial $a=b=0$, then the polynomial will always be $x^2$ and so the process goes on forever.  However, in all other cases, it will terminate in finitely many steps.
First, suppose $b=0$ and $a\neq 0$.  If $a<0$, the next polynomial is $x^2-a$ which does not have real roots.  If $a>0$, the next polynomial is $x^2-ax$ and then the next polynomial is $x^2+a$ which does not have real roots.
Now suppose $b\neq0$.  Note then that the initial polynomial does not have $0$ as a root, and consequently neither will any of the others (since the constant coefficient is never $0$).  Let us suppose the process goes on forever and say the polynomials are $x^2+ax+b$, $x^2+p_0x+q_0$, $x^2+p_1x+q_1$, $x^2+p_2x+q_2$, and so on.
Note first that the signs of the coefficients $(p,q)$ go in a cycle $$(-,-)\to(-,+)\to (+,+)\to (-,-).$$  For instance, if $p_n$ and $q_n$ are both negative, then the roots of $x^2+p_nx+q_n$ have opposite sign (since their product $q_n$ is negative), so $p_{n+1}$ is negative and $q_{n+1}$ is positive.
Let us now analyze how the absolute value of the linear coefficient changes in this cycle.  If $p_n$ and $q_n$ are negative, then $p_{n+1}<0<q_{n+1}$ and $p_{n+1}+q_{n+1}=-p_n>0$, so $|p_{n+1}|<|q_{n+1}|$.  Thus $$|p_{n+1}|<\sqrt{|p_{n+1}q_{n+1}|}=\sqrt{|q_n|}\leq\sqrt{|p_n|}.$$
Now suppose $p_n$ is negative and $q_n$ is positive.  Then $0<p_{n+1}\leq q_{n+1}$ and so $$|p_{n+1}|=p_{n+1}\leq \frac{p_{n+1}+q_{n+1}}{2}=\frac{-p_n}{2}=\frac{|p_n|}{2}.$$
Finally, suppose $p_n$ and $q_n$ are positive.  Then $p_{n+1}\leq q_{n+1}<0$ so $$|p_{n+1}|<|p_{n+1}+q_{n+1}|=|-p_n|=|p_n|.$$
We thus see that in all cases, $|p_{n+1}|\leq\max(|p_n|,1)$.  Moreover, in every third step, $|p_{n+1}|\leq\frac{|p_n|}{2}$.  It follows that for all sufficiently large $n$, $|p_n|\leq 1$.
Let us now pick $n$ such that $|p_n|\leq 1$ and $p_n$ and $q_n$ are positive.  We then have $$p_n^2-4q_n\leq p_n^2-4p_n<0.$$  Thus the discriminant of $x^2+p_nx+q_n$ is negative and it does not have real roots.  This is a contradiction.
