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 $S$?



Discriminant on boundary of semi-algebraic surface
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. 


