The cubic polynomial $f(x) = x^3+px+q\in K[x]$ has 3 roots $a_1,a_2,a_3\in \mathbb C$ Hence, the splitting field extension $L=K(a_1,a_2,a_3)$
$\delta=(a_1-a_2)(a_1-a_3)(a_2-a_3)\in L$ since $\delta^2=-4p^3-27q^2\in K$. MY QUESTION: why? what is the logic here? either $\delta \in K$or $K(\delta)$ is an extension field of dimension 2 over K. MY QUESTION: why $[K(\delta):K]=2$?
Now suppose $dim_K K(\delta)=2$ which leads to $dim_K L=6 ~~or~~ 3$. With the identity: $\sum_i a_i=0$, we have $a_3=-a_1-a_2$
$a_2^2=-a_1^2-a_1a_2-p$
$a_i^3=-pa_i-q~~(i=1,2)$
then $a_2={(\delta+2a_1p+3q)}/{2(3a_1^2+p)}$ MY QUESTION: what if $a_1^2= -p/3$ that makes $a_2$ undefined?? Please, help me understand it. Appreciate any tips.