Given an arbitrary number field $K/\mathbb{Q}$, and let $\alpha$ an arbitrary primitive element, i.e. $K=\mathbb{Q}(\alpha)$.
What is the relation between $\text{Disc}(\text{min}_{\alpha}(x))$ and $\text{Disc}(K/\mathbb{Q})$?
For instance, in the case of quadratic extensions, it would be equal module ${\mathbb{Q}}/{\mathbb{Q}^{\times 2}}$.
If $\alpha$ is an algebraic integer, we have the well-known relation $disc(\min_\alpha)=disc(K/\mathbb{Q})[O_K:\mathbb{Z}[\alpha]]^2$, where $O_K$ is the ring of integers of $K$.
To prove it, choose an integral basis $(e_1,\ldots,e_n)$ of $O_K$ such that $(q_1e_1,\ldots, q_ne_n)$ is an integral basis of $\mathbb{Z}[\alpha]$. In particular $ [O_K:\mathbb{Z}[\alpha]]=q_1\cdots q_n$.
By definition, the discriminant of $\min_\alpha$ is $\prod_{i<j}(\alpha_i-\alpha_j)^2,$ where the $\alpha_i$'s are the roots of $\min_\alpha. $ This is also the discriminant of $(1,\alpha,\ldots,\alpha^{n-1})$, that is the discriminant of $\mathbb{Z}[\alpha]$. This discriminant does not depend on the integral basis of $\mathbb{Z}[\alpha]$, so we can also compute it using the basis $(q_1e_1,\ldots, q_ne_n)$. Easy calculations then yield that the discriminant of $\mathbb{Z}[\alpha]$ is equal to $(q_1\cdots q_n)^2disc(K/\mathbb{Q}),$ which is what we wanted to prove.