Show that $\displaystyle f(x)=x^4+bx^2+d$ is irreducible over
$\displaystyle \mathbb Q(\sqrt{d(b^2-4d)})[x]$
the question doesn't make sense! for example, choose d = 1 and b = 3. then $\displaystyle \sqrt{d(b^2 - 4d)}=\sqrt{5}.$ now in $\displaystyle \mathbb{Q}(\sqrt{5})[x]$ we have:
$\displaystyle x^4+3x^2 + 1 = \left(x^2 + \frac{3 + \sqrt{5}}{2} \right) \left(x^2 + \frac{3 - \sqrt{5}}{2} \right).$
I have that
$\displaystyle
f(x)=x^4+ax^2+b
$
is irreducible over rationals, and its galois group is
$\displaystyle
D_4
$
And M is the splitting field of $\displaystyle f(x)$
and I have that $\displaystyle
\mathbb Q(\sqrt{d(b^2-4d)})
$
is a quadratic subfield of M.
I think that ZetaX is referring to this thread. In that thread he asks that given that $\displaystyle x^4 + bx^2 + d$ is irreducible over $\displaystyle \mathbb{Q}$ with Galois group $\displaystyle D_4$ then show that the three quadradic extensions (we know there has to be three since $\displaystyle D_4$ has three subgroups of index $\displaystyle 2$) are: $\displaystyle \mathbb{Q}(\sqrt{b^2-4d}),\mathbb{Q}(\sqrt{d}),\mathbb{Q}(\sqrt{d(b^2-4d)})$. He is asking to complete that solution.