The vectors of a root-system were originally called "*roots*" because they are the zeros of a characteristic polynomial that comes from the connection of (crystallographic) root-systems to classifying semisimple Lie algbras. From an answer by José Carlos Santos,

It comes from the roots of the characteristic polynomial of an endomorphism. If $\mathfrak g$ is a complex semisimple Lie algbra, $\mathfrak h$ is a Cartan subalgebra and $\alpha\in\mathfrak{h}^*$, then $\alpha$ is a root if, for every $H\in\mathfrak h$, $\alpha(H)$ is an eigenvalue of the endomorphism of $\mathfrak g$ defined by $X\mapsto[H,X]$.

Is there a way to get at this same polynomial just from the definitions of a (crystallographic) root-system without having to talk about the connection to Lie theory?

To provide some specific motivation, I'm reading through a proof of Gabriel's theorem (classifying quivers of finite representation type) where you define the Tits form $q$ for the ADE Dynkin and Euclidean diagrams, and the set of *roots* associated to that diagram are the nonzero vectors $x \in \mathbb{Z}^n$ for which $q(x) \leq 1$. Since $q(x) \in \mathbb{Z}$ we could say these are the integer roots of the polynomial $q(x)(q(x)-1)$, but I doubt that this generalizes beyond the root systems you get from the ADE Dynkin diagrams. I was looking for a way to justify calling these "roots" without getting into Lie theory, and I thought there might be a justification for the word "root" in terms of general abstract root systems.

Reflection Groups and Coxeter Groupsby James Humphreys. I am flipping through it and cannot guarantee that it answers your concerns. On the other hand, I know of no book called Root Systems and Polynomials, so Humphreys is a good start. Plenty of roots, plenty of polynomials. $\endgroup$ – Will Jagy Mar 2 '18 at 23:17root" historically comes from Lie theory (section 1.2), and then makes the distinction between general root systems and crystallographic roots systems that relate to Lie theory (section 2.9). I haven't seen the answer to the question spelled out anywhere though. $\endgroup$ – Mike Pierce Mar 3 '18 at 0:37