Does a polynomial $P(X,Y)$ that specializes to a polynomial $P(x_0,Y)$ with distincts roots in $\overline{k}$ has distincts roots in $\overline{k(X)}$

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Let $k$ be a field, $P\in k[X][Y]$ be a monic polynomial of degree $n$ in $Y$.

I would like a simple proof of the following fact.

"If there exists $x_0\in k$ such that $P(x_0,Y) \in k[Y]$ has (at least) two distincts roots in an algebraic closure $\overline{k}$ then $P(X,Y)$ has (at least) two distincts roots in an algebraic closure $\overline{k(X)}$."

Intuitively, if there were a specialization map $\overline{k(X)} \to \overline{k}$, that would be obvious, but unfortunately I don't think this is the case.

So the idea would be to find an algebraic characterization of the polynomials that have only one root in an algebraic closure only in terms of its coefficients. I kwow how to do it in characteristic zero (say writing down the condition implied by the fact that the image of the linear map $U,V\mapsto PU+P'V$ is "small"), but it does not work quite well in characteristic $p$, for we can have $P'=0$ for instance.

I feel that I maybe need more hypothesis on $k$ (probably perfect ?), although I have not been able to find any counter-example.

I am sure this should be well-known to someone here...