= Solution
The elementary symmetric functions are
$$
e_r(x_1,\ldots,x_n)=\sum_{1\leq i_1<\cdots<i_r\leq n}x_{i_1}\cdots x_{i_r}.
$$
The <Fundamental theorem of symmetric polynomials> says that every symmetric polynomial is uniquely a polynomial in $e_1,\ldots,e_n$.
If $r_1,\ldots,r_n$ are the roots of a monic polynomial, its <polynomial discriminant>[discriminant] is
$$
\operatorname{Disc}(f)=\prod_{i<j}(r_i-r_j)^2.
$$
This is symmetric in the roots, hence the theorem and Vieta's formulas express it polynomially in the coefficients. Equivalently,
$$
\operatorname{Disc}(f)=(-1)^{n(n-1)/2}\operatorname{Res}(f,f').
$$
For $x^5+q$ this gives
$$
\boxed{\operatorname{Disc}(x^5+q)=5^5q^4=3125q^4}.
$$
For $f=x^5+px^2+q$, a repeated root solves $f=f'=0$. Elimination gives
$$
\boxed{\operatorname{Disc}(f)=q(108p^5+3125q^3)}.
$$
\b[Thus the discriminant vanishes exactly when $q=0$ or $108p^5+3125q^3=0$.]
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