# Galois group of an irreducible biquadratic polynomial

We have studied that the Galois group of the splitting field of an irreducible biquadratic polynomial like $x^4+ax^2+b$ in $Q[x]$ over $Q$ can be $Z_{2}XZ_{2}$,$Z_{4}$ or $D_4$. We have proven this by taking into account tha values of $b$ and $\Delta=a^2-4b$, and the Galois group changed based on the fact if they were squares or not (actually I think we checked if $b$ was a square in $Q(\sqrt(\Delta))$ if I remember well).
However, I am having a hard time because I don't understand how we have been able to prove this, and what are the cases in which I could say only by looking at the polynomial what is its Galois group. Thanks for the help.

Let be $q(x)=x^4+bx^2+c\in k[x]$ irreducible polynomial over $k$ and has two distict roots $\alpha, \beta$ and $G=Gal(k(\alpha, \beta),k)$ then
$1)$ if $\sqrt c \in k$ then $G\cong\mathbb Z_2 \times \mathbb Z_2$
$2)$ if $\sqrt c \notin k$ and $\sqrt{c(b^2-4c)} \in k$ then $G\cong\mathbb Z_4$ and $q_1(x)=x^2+bx+c$ irreducible over $k(\sqrt c)$
$3)$ if $\sqrt c \notin k$ and $\sqrt{c(b^2-4c} \notin k$ then $G\cong D_4$ and $q_1(x)=x^2+bx+c$ irreducible over $k(\sqrt c)$