Multiple irrational roots of irreducible polynomial over Q Is it possible for irreducible polynomial $f \in Q[x]$ to have irrational multiple root?

I guess not, so I supposed to assume that my polynomial has factor $(x - a)^k$ in some extention, where $a$ is irrational, $k&gt;1$ and come to contradiction, but I don't know what  could it be.
No.  The field $\mathbb{Q}$ is a perfect field: no irreducible polynomial over it has multiple roots.

To see it, recall that a polynomial $P$ has multiple roots if and only if it has a non-constant common factor with its formal derivative $P'$.  Now, since $P$ is irreducible, its only factors are itself and constant polynomials.  Since $P'$ has degree smaller than that of $P$, the only factors it can share with $P$ are constant polynomials.
No, let $G$ be the Galois group of $Q[X]/P$, $(X-a)^k$ divides $P$ implies that  $(\Pi_{g\in G}(X-g(a)))^k$ divides $P$ which implies that $P$ is not irreducible if $k&gt;1$ since $\Pi_{g\in G}(X-g(a))\in Q[X]$.