Discriminant on boundary of semi-algebraic surface Let
$P(t)$ be a polynomial in $t$ of degree $n$, 
with some contiguous coefficients (not the first or last) being $x_1,\dots,x_k$
and the rest of the coefficients are fixed.
(E.g. $p(t)=1+2t+x_1t^2+x_2t^3+(5i+7)t^6$ is ok).
Consider the set $S$ defined as $(x_1,\dots,x_k) \in \mathbb{C}^k$ 
with the property that the roots $t_1,\dots,t_{n}$ of $P(t)=0$
may be ordered increasingly w.r.t modulus, and such that
$$|t_j|=|t_{j+1}| = \dots = |t_{k+j}|,$$
for some fixed $j.$
This set is real $k$-dimensional.
Is it true, (in general), that the discriminant $D(x_1,\dots,x_n) = Discr_t(P)$ satisfy
$D(x_1,\dots,x_n)=0$ on the $k-1$-dimensional boundary of $S$?
 A: Let $P(t)=(t-t_1)(t-t_2)\dots(t-t_k)(t-a)$ where $(-1)^{k+1}t_1t_2\dots t_k a = b$
which has leading coefficient 1 and constant coefficient $b$. If you fix $b$, then $a$ is a rational function of the $t_i$. Treat the case where some $t_i=0$ with care.
The other coefficients are
$$x_i = (-1)^i \sigma_i(t_1,\dots,t_k,a) = (-1)^i\sum_{1\le j_1<\dots j_i\le k}t_{j_1}\dots t_{j_i}
+ (-1)^i a\sum_{1\le j_1<\dots j_{i-1}\le k}t_{j_1}\dots t_{j_{i-1}}$$
and the discriminant is 
$$ D=\prod_{i < j} (t_i-t_j)\cdot\prod_i (x_i-a).$$
Also let $$\sigma(t_1,\dots,t_k)=(\sigma_1(t_1,\dots,t_k,a),\dots\sigma_k(t_1,\dots,t_k,a)).$$
Denote by $s_i$ the Newton polynomials 
$\sum_{j}t_j^i+a^i$ which are related to the elementary symmetric 
function by 
$$ 
s_\ell-s_{\ell-1}\sigma_1+s_{k-2}\sigma_2+\dots+(-1)^{\ell-1}s_1\sigma_{\ell-1}+ 
(-1)^\ell\ell\sigma_\ell=0  \quad (\ell\leq k+1),
$$ 
where $\sigma_{k+1}=b$.
The corresponding mappings are related by a polynomial diffeomorphism 
$\psi$ (treating $a$ as a variable), given by:
$$
\sigma:=(\sigma_1,\dots,\sigma_k, b):\Bbb R^{k+1}\to \Bbb R^{k+1}$$
$$s:=(s_1,\dots,s_k, s_{k+1}):\Bbb R^{k+1}\to \Bbb R^{k+1}$$
$$s:=\psi^n\circ\sigma^n
$$ 
Note that the Jacobian (the determinant of the derivative) 
of $s$ is $(k+1)!$ times the Vandermonde determinant:
$$\det(ds(t,a))=(k+1)!\,\prod_{i>j}(t_i-t_j)\cdot\prod_i(t_i-a),$$ 
and even the derivative itself 
$d(s)(x)$ equals the 
Vandermonde matrix up to factors $i$ in the $i$-th row.
From this you should be able to decide your question. I do not have time to this now. 
