# A sequence of quadratic equations

Let $$p\le q$$ be roots of the (real) quadratic equation $$x^2+ax+b=0$$, $$|p|+|q|\ne 0.$$ Form the new equation $$x^2+px+q=0$$, find its real roots (if exist), etc. For example, if $$a=3, b=2$$, then $$p=-2,q=-1$$, $$x^2-2x-1=0$$ has two real roots $$p_1= 1 - \sqrt{2}$$ and $$q_1= 1 + \sqrt{2}$$ (note that $$p_1\le q_1$$) but the equation $$x^2+p_1x+q_1$$ does not have real roots, the process ends.

Question: What is the longest possible sequence of quadratic equations we can get?

• If it's possible to make something like $p$ and $q$ keep switching back and forth between 2 values, does that count as $2$ or $\infty$? Jun 9, 2020 at 19:49
• @user6247850 I think it would be infinity (OP please confirm). Did you find such an example? Jun 9, 2020 at 19:52
• Infinity is not possible. The process always terminats after finitely many steps. Jun 9, 2020 at 23:15
• @user158834 how do you see it always terminates? Jun 9, 2020 at 23:50
• @doetoe: It is basically obvious. The key is that $p_i\le q_i$. Jun 10, 2020 at 0:12

Let's work backwards: given a final state, how many steps would it take to reach the beginning? Let the final state be represented by $$r_1, r_2$$ with $$r_1 \le r_2$$.

We can see that the quadratic before the final one must have been $$(x-r_1)(x-r_2) = x^2+(-r_1-r_2)x + r_1r_2$$

In order to have a sequence longer than $$1$$, it must be true that $$-r_1-r_2 \le r_1r_2 \tag 1$$

To get that quadratic, the previous quadratic must have been $$(x+r_1+r_2)(x-r_1r_2) = x^2 + (r_1+r_2-r_1r_2)x - r_1r_2(r_1+r_2)$$

In order to get a sequence longer than $$2$$, it must be true that $$r_1+r_2-r_1r_2 \le - r_1r_2(r_1+r_2) \tag 2$$

Taking it one more step, the following condition must also hold in order to get a sequence longer than $$3$$: $$(r_1r_2-r_1-r_2+r_1r_2(r_1+r_2)) \le (r_1r_2-r_1-r_2)(r_1r_2(r_1+r_2)) \tag 3$$

Finally, this last condition must also hold in order to get a sequence longer than $$4$$: $$-\left(r_{1}r_{2}-r_{1}-r_{2}+r_{1}r_{2}\left(r_{1}+r_{2}\right)\right)-\left(r_{1}r_{2}-r_{1}-r_{2}\right)\left(r_{1}r_{2}\left(r_{1}+r_{2}\right)\right)\le\left(r_{1}r_{2}-r_{1}-r_{2}+r_{1}r_{2}\left(r_{1}+r_{2}\right)\right)\left(r_{1}r_{2}-r_{1}-r_{2}\right)\left(r_{1}r_{2}\left(r_{1}+r_{2}\right)\right) \tag 4$$

In order to satisfy $$(2)$$, it must be true that $$r_1 < 0$$ or that $$r_2 < 0$$. Letting the $$x$$-axis be $$r_1$$ and the $$y$$-axid be $$r_2$$, this eliminates the first quadrant. In order to satisfy $$(3)$$, it must be true that $$r_1 > 0$$ or that $$r_2 > 0$$, eliminating the third quadrant. However, in order to satisfy both $$(1)$$ and $$(4)$$, $$(r_1, r_2)$$ can only be in the second and fourth quadrants. Therefore, there are no real $$r_1, r_2$$ that satisfy $$(1), (2), (3), (4)$$.

This then means that the maximum length of a sequence of the quadratic equations is $$4$$, obtained by setting $$a = r_1r_2-r_1-r_2+r_1r_2(r_1+r_2), b = (r_1r_2-r_1-r_2)(r_1r_2(r_1+r_2)$$

for any $$r_1, r_2$$ that satisfy $$(1), (2), (3)$$, and $$r_1 \le r_2$$.

Edit: The conditions $$(1), (2), (3), r_1 \le r_2$$ can be rewritten as $$-\frac{r_{2}}{r_{2}+1}\le r_{1}\le\frac{-\left(r_{2}^{2}+1-r_{2}\right)+\sqrt{\left(r_{2}^{2}+1-r_{2}\right)^{2}-4r_{2}^{2}}}{2r_{2}}$$ with $$r_1 \le 0 \le r_2$$

• That is correct. Jun 10, 2020 at 4:57

The equation $$y=x^2 + px + q$$ can be rewritten as $$y-\left(q-\frac{p^2}{4}\right)=\left(x-\left(\frac{-p}{2}\right)\right)^2$$ And so it is clear that this parabola's vertex is at the point $$(\frac{-p}{2},q-\frac{p^2}{4})$$. Therefore the process will terminate at step $$n$$ if $$q_n-\frac{{p_n}^2}{4} > 0$$.

• This is just a good first step. I suspect there is no limit to the length of a sequence we can create. Jun 9, 2020 at 23:36
• This is equivalent to the discriminant being non-negative, no? Jun 10, 2020 at 0:13
• This is not a good first step. See the correct sollution. Jun 10, 2020 at 5:41