= Solution
The <existence and uniqueness of splitting fields> says that any two splitting fields of a polynomial $f\in F[X]$ are isomorphic by an isomorphism fixing $F$ pointwise. Inside a chosen algebraic closure, the subfield generated by all roots is unique.
For a polynomial $f\in\mathbb Q[X]$ with distinct roots, let $K$ be its splitting field. Its <Galois group of a polynomial> is
$$
\operatorname{Gal}_{\mathbb Q}(f)
=\operatorname{Gal}(K/\mathbb Q),
$$
the group of field automorphisms of $K$ fixing $\mathbb Q$. It acts faithfully on the roots.
If $f$ is irreducible and $\alpha,\beta$ are roots, the embedding $\mathbb Q(\alpha)\to K$ taking $\alpha$ to $\beta$ extends, because $K/\mathbb Q$ is normal and separable, to an element of the Galois group. Thus the action is transitive. Conversely, the roots belonging to one orbit have a product polynomial invariant under the whole Galois group, hence lying in $\mathbb Q[X]$. If there were more than one orbit, these orbit polynomials would give a nontrivial factorization of $f$. Therefore
$$
\boxed{f\text{ is irreducible over }\mathbb Q
\quad\Longleftrightarrow\quad
\operatorname{Gal}_{\mathbb Q}(f)\text{ acts transitively on its roots}.}
$$
Now let $g(X)=X^4+bX^2+c$ and let $\alpha$ be one root. The two roots of $Y^2+bY+c$ are $\alpha^2$ and $c/\alpha^2$. On choosing $\delta=\sqrt c$, the roots of $g$ are
$$
\alpha,\quad-\alpha,\quad\frac\delta\alpha,
\quad-\frac\delta\alpha.
$$
Consequently its splitting field is
$$
\boxed{K=\mathbb Q(\alpha,\sqrt c).}
$$
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