Let $f(X)=X^5+aX+b \in \mathbb{Q}[X]$. I want to know the Galois group of $f(X)$ if all the following conditions are met:
1) $f(X)$ is irreducible in $\mathbb{Q}[X]$.
2) The discriminant $D(f)$ of polynomial $f(X)$ is a square in $\mathbb{Q}.$
3)The equation $f(X)=0$ is soluble by radicals.
So by 1) the Galois group is a transitive subgroup of $S_5$ and by 2) is contained in $A_5$, my question is the following by the 3) I know the Galois group is a solvable group? If this is the case what is the Galois group?
Edit 2: Ok, some news. I know $f(X)$ can have 1 or 5 real roots, because $f(X)$ can't have two real roots and neither four real roots because this is a contradiction with the number of possible complex roots. Futhermore $f(X)$ can't have 3 real roots because if $f(X)$ have exactly two complex roots tha Galois group will be $S_5$, a contradiction with 2). Now the question is how can i prove that $f(X)$ can't have five real roots?
The irreducibility condition implies that the Galois group $G$ acts transitively on the 5 roots.
There are exactly 5 such groups: $S_5$, $A_5$, the Frobenius group of order $20$ (which is the normalizer of the p-Sylow and also isomorphic to the group of linear affine transformations of $\mathbf{F}_5$), the dihedral group of order 10 (a subgroup of the Frobenius group), and the cyclic group of order $5$.
The discriminant condition implies that $G$ is a subgroup of $A_5$, which rules out $S_5$ and the Frobenius group, since the linear map $x \mapsto 2x$ defines a 4-cycle which is odd.
The solvability condition rules out $A_5$, leaving the cyclic group and the dihedral group.
The form of the polynomial implies (by Descartes rule of signs, for example) that not all the roots are real. This implies that complex conjugation is non-trivial, and hence that the Galois group has even order. This rules out the cyclic group of order $5$, and thus by elimination the Galois group must be the dihedral group of order $10$.