Can discriminant polynomials become perfect powers on hyperplanes? Let
$$\displaystyle f(x) = a_d x^d + a_{d-1} x^{d-1} + \cdots + a_0.$$
Consider the discriminant of $f$, denoted by $\Delta(f)$, defined as
$$\displaystyle \Delta(f) = a_d^{2d-2} \prod_{i < j} (\theta_i - \theta_j)^2,$$
where $\theta_1, \cdots, \theta_d$ are the roots of $f(x) = 0$ (over some algebraic closure, say). 
It is well-known that $\Delta(f)$ is a homogeneous polynomial of degree $2d-2$ in the coefficients $a_d, \cdots, a_0$. 
We say that a homogeneous polynomial $F \in \mathbb{C}[x_0, \cdots, x_n]$ of degree $m$ ramifies completely on a hyperplane if there exists a hyperplane $P$ in $\mathbb{P}^n$ such that $F |_P$ is a perfect $k$-th power for some $k > 1$ dividing $m$ (as a polynomial). For example, the cubic polynomial $F(x,y,z) = x^3 + yz^2$ ramifies completely on the lines (hyperplanes in $\mathbb{P}^2$) $y = 0, z = 0$. 
For $d = 2$, we have that $\Delta(f) = a_1^2 - 4 a_2 a_0$ ramifies completely on $a_2 = 0, a_0 = 0$. Does this happen for $d > 2$? That is, does there exist $d > 2$ and a hyperplane $P \in \mathbb{P}^d$ such that $\Delta(f) |_P$ is a perfect $k$-th power, for some $k | 2d - 2$?
 A: For any $d \geq 2$, there are hyperplanes on which $\Delta_d$ ramifies, but for $d \geq 3$, it never ramifies completely. I guess there are many proofs of this fact, let me give one based on projective duality.
First note that $\Delta_d$ parametrizes polynomials of degree $d$ having a multiple root. Let me give a geometric interpretation of $\Delta_d$. Let $X = v_d(\mathbb{P}^1) \subset \mathbb{P}(S^d \mathbb{C}^2)$ be the $d$-th Vernoese embedding of $\mathbb{P}^1$. The equation $\Delta_d = 0$ gives in the dual projective space $\mathbb{P}(S^d \mathbb{C}^2)^*$ the variety which parametrizes singular hyperplane sections of $X$. This variety is known as the projective dual of $X$.
For any $Z \subset \mathbb{P}^N$, which is irreducible, the reflexivity Theorem tells you that $(Z^*)^* = Z$ and for any $z \in Z_{smooth}$, the tangency locus of $z^{\perp}$ with $Z^*$ is identified as a scheme to $\mathbb{P}(N_{Z/\mathbb{P}^N,z}^*)$.
Going back to your situation, we have $X = v_d(\mathbb{P}^1) \subset \mathbb{P}(S^d \mathbb{C}^2)$ is smooth, its projective dual $X^* \subset \mathbb{P}(S^d \mathbb{C}^2)^*$ is the hypersurface which equation is $\Delta_d = 0$. You want to know if there exists $x \in \mathbb{P}(S^d \mathbb{C}^2)$ such that $x^{\perp} \cap X^*$ is completely non-reduced. This is equivalent to saying that the reduced space underlying the singular locus of $x^{\perp} \cap X^*$ is  equal to the reduced space underlying $(x^{\perp} \cap X^*)$.
Two cases occcur:
1) if $x \notin X$, then $x^{\perp}$ is not tangent to $X^*$ (this is because $(X^*)^*= X$). Hence the singular locus of $x^{\perp} \cap X^*$ has codimension at least one in $x^{\perp} \cap X^*$ and the equation defining $x^{\perp} \cap X^*$ can't be a $k$-perfect power.
2) if $x \in X$, then $x^{\perp}$ is tangent to $X^*$ (this is again because $(X^*)^*= X$). Since $X$ is smooth, the reflexivity Theorem ensures that the tangency locus of $x^{\perp}$ with $X^*$ is scheme-theoretically a hyperplane in $x^{\perp}$. As a consequence, the singular locus of $x^{\perp} \cap X^*$ is generically scheme-theoretically a hyperplane. Now, if $x^{\perp} \cap X^*$ was a $k$-perfect power, then the fact that the singular locus of $x^{\perp} \cap X^*$ is generically scheme-theoretically a hyperplane in $x^{\perp}$ would imply that $k=2$ and $\deg \Delta_d =2$. This is only possible if $d=2$.
As far as non-complete ramification is concerned, the above argument shows that for any $x \in X$, the equation of $x^{\perp} \cap \{\Delta_d = 0\}$ can be written as $L^2Q$ where $L$ is a linear factor and $Q$ a polynomial of degree $2d-4$ which is not a power.
