On a theorem of Galois I am currently teaching Galois theory and this week, I mentioned the following theorem of Galois :
Let $P(x) \in \mathbf{Q}[x]$ be an irreducible polynomial of prime degree. Then $P$ is solvable by radicals if and only if the splitting field of $P$ is generated by any two roots of $P$.
I was asked by a student whether this theorem can be generalized to polynomials whose degree is composite, maybe allowing the splitting field to be generated by more than two roots. I know that the proof of Galois's theorem relies on determining the solvable subgroups of $\mathfrak{S}_p$, but I don't know enough group theory to tell what can be proved in the case where the degree of the polynomial is composite, say $pq$ where $p$ and $q$ are (possibly equal) primes.
Does such a generalization of Galois's theorem exist? Or is there a conceptual reason why such a generalization cannot hold? In the latter case, do there already exist generalizations of Galois's theorem, possibly in different directions?
 A: The question asks about the relation of the properties 1. and 3., though possibly the intended meaning of 1. was 2.:
*

*The splitting field of $P$ is generated by two roots of $P$.

*The splitting field of $P$ is generated by any two roots of $P$.

*The Galois group $G$ of $P$ is solvable.

In the prime degree case, all three properties are equivalent.
For arbitrary degrees, 1. is a weak condition and so doesn't tell much about solvability of $G$. Condition 1. is also much weaker than 2. Despite its weakness, 3. does not imply 1. In fact, for each non-prime degree $n\ge6$, there is a solvable group $G$ of degree $n$ for which 1. does not hold. Indeed, it's easy to construct explicit examples for all such degrees: Let $n=rs$ with $r\ge3$, $s\ge2$. Then for suitable rational $a,b$, the polynomial $P(X)=(X^s-b)^r-a$ doesn't fulfill 1.
So let's forget about 1. Also, this example shows that 3. is far from implying 2.
The question remains whether 2. implies 3. Indeed, it comes close:
 If $n=\deg(P)\not\equiv1\pmod{120}$, and the splitting field of $P$ is not generated by a root (see Kevin Ventullo's comment above), then 2. implies 3.
The reason for this is as follows: 2. says that $G$ is a Frobenius group. By Frobenius' Theorem, $G$ is a semidirect product of a regular normal subgroup $N$ and a point stabilizer $H$, called the Frobenius complement. By Thompson's Theorem, $N$ is nilpotent, so in particular solvable. What about $H$? By an old result of Zassenhaus, $H$ is either solvable, or its series of derived subgroups terminates in $\text{SL}(2,5)$, a group of order $120$. As $n=\lvert N\rvert$, and $H$ has regular orbits on $N\setminus\{e\}$, we get what I claimed above.
As to the excluded degrees: The smallest candidate of degree $121$ exists group theoretically: $\text{SL}(2,5)$ has a regular action on the nonzero elements of $\mathbb F_{11}^2$, yielding a Frobenius group $\mathbb F_{11}^2\rtimes\text{SL}(2,5)$. By a result of Jack Sonn (see here), every finite Frobenius group is a Galois group over the rationals. Thus there is an irreducible $P(X)$ of degree $121$ for which 2. holds, but 3. does not.
Added (Answering François' question in his comment below): This minimal number of roots which generate the splitting field is the size of a so-called minimal base of the permutation group $G$. A base of a permutation group is a subset of the points the group acts on whose elementwise stabilizer is trivial. If you take the wreath product $G=C_2\rtimes C_m$, in its natural action on $2m$ points, this number is $m$. So even if the degree is a product of two primes, the minimal base size can be arbitrarily large.
Things are better if $G$ is primitive and solvable: Then the minimal base size is at most $4$. See here for more results on minimal bases. 
