Let $x^{4}+ax^{2}+b\in K[x]$ (with $\mathrm{char}K\neq 2$) be irreducible with Galois group $G$.
I have shown that (a) if $b$ is a square in $K$, then $G$ is the Klein 4-group, (b) if $b$ is not a square in $K$ and $b(a^{2}-4b)$ is a square in $K$, then $G\cong\mathbb{Z}_{4}$.
I need to show that in the remaining case, i.e. if neither $b$ nor $b(a^{2}-4b)$ is a square in $K$, then $G\cong D_{4}$.
The roots of the polynomial are some $\pm\alpha,\pm\beta$. Moreover, $a=-\alpha^{2}-\beta^{2}$ and $b=\alpha^{2}\beta^{2}$. The roots of the resolvant cubic are $a,2\alpha\beta,-2\alpha\beta$. From my work in part (b), I know that in the last case we must have either $G\cong\mathbb{Z}_{4}$ or $G\cong D_{4}$. But I can't give an argument to conclude that it is $D_{4}$.
Update: I was thinking of arguing in the following way (basically showing the converse of part (b)) but I'm not sure whether it is right. Suppose $G\cong\mathbb{Z}_{4}$. Then $G$ is generated by the automorphism $\sigma:\alpha\mapsto-\beta,\beta\mapsto\alpha$ (since it has order 4). Now $\sigma$ fixes $\alpha\beta(\alpha^{2}-\beta^{2})$ so $G$ fixes it and it is an element in $K$. But its square is $b(a^{2}-4b)$, giving a contradiction.
