Is there an algebraic way to prove this relationship between the roots of a real polynomial and the roots of its derivative?

Question

Let $$\sum_{0 \leq i\leq n} a_ix^i$$

be a polynomial (real coefficients) with at least two real roots. Is there an algebraic way to show that for any two roots $k_1, k_2$ of this polynomial, the polynomial

$$\sum_{1 \leq i\leq n} i \cdot a_ix^{i-1} $$

admits at least one root $c$ satisfying $k_1 <c < k_2$?

Analytically, this is of course a consequence of Rolle's theorem.

Edit: "Algebra" is as broad as you want it to be. Elementary or abstract. The completeness of $\mathbb{R}$ is essential, so it won't be purely algebraic. I was mainly hoping for something without derivatives.

Answer 1

Probably not. Note that in $\mathbb Q(\sqrt{6})$, the polynomial $x^3 - 6x$ has three roots. However, its derivative has no roots. These would be $\pm \sqrt{2}$. Thus, whatever you mean by purely algebraic needs to be stronger than the theory of ordered fields.

If, however, you are willing to stretch your definition of "algebraic" somewhat, and add to the theory of ordered fields even the axiom "each positive number has a square root", and an axiom scheme containing for each polynomial of odd degree an axiom "this polynomial has a root", then you can prove any first order property of the reals. See for instance this Wikipedia page.