This question came to me while trying to determine whether $\mathbb{Q}(a)$, is a splitting field, if $a^7 = 5$. I thought that since $\\{ 1, a, ..., a^6 \\} $ is a basis for $\mathbb{Q}(a)$, and we can easily see that some roots of the minimal polynomial of $a$, namely $m(t)=t^7 - 5$, are not inside $\mathbb{Q}(a)$, that would be sufficient to show that it is not a splitting field.
Then I thought, wait, what if $\mathbb{Q}(a)$ is a splitting field for a different polynomial? By considering that an extension is normal and finite iff it is the splitting field for some polynomial, if $\mathbb{Q}(a)$ was really the splitting field for some other polynomial, it would be normal, so $m$ would also split in it, which is a contradiction.
Is that sufficient to prove the claim that $\mathbb{Q}(a)$ is not a splitting field?
EDIT: My question is whether the second argument is really needed or not.
It is a theorem in field theory that a finite field extension $E/F$ of the form $E=F(a_1,...,a_n)$ is normal iff each generator $a_i$ is normal over $F$ (meaning that the minimal polynomial of each $a_i$ splits into linear factors in $E[X]$). So your argument by just considering the minimal polynomial of $a$ is sufficient.