 Solution

First assume $\mathrm{char}F\ne 2$. Set $d=\det V(\alpha )\ne 0$. If ${\Delta }_f=b^2$ with $b\in F$, then $(d-b)(d+b)=0$ in the field $K$, so $d=\pm b\in F$. For an automorphism inducing the root permutation $\pi$,Since $\sigma$ fixes $d$ and $1\ne -1$ in $F$, every induced permutation has positive sign. The root action is faithful because the roots generate $K$, givingThis is the discriminant criterion for an alternating Galois group.

As printed for arbitrary fields, the assertion is false in characteristic two. Over ${\mathbb{F}}_2$, the polynomial $T^2+T+1$ is irreducible and separable, with roots $\alpha ,\alpha +1$ in ${\mathbb{F}}_4$. Their difference is one, so the discriminant is the square $1$. Nevertheless its Galois group is $C_2$, whereas $A_2$ is trivial. In characteristic two the permutation signs become indistinguishable as field elements; square discriminants in characteristic two therefore cannot detect the alternating subgroup in this way.