Degree 2 Field extensions 
Are all degree $2$ field extensions Galois?   

I know that this is true over the rationals. But is it true in general?

It is true over any field of characteristic different from $2$. 
If $F$ has characteristic different from $2$, and $K$ is of degree $2$ over $F$, then let $\alpha\in K$, $\alpha\notin F$. Then $K=F(\alpha)$, since $2=[K:F]=[K:F(\alpha)][F(\alpha):F]$, and $[F(\alpha):F]\gt 1$. Let $p(x)$ be the minimal polynomial of $\alpha$ over $K$. Then $p(x) = x^2 + rx+t$ for some $r,t\in F$, and $\alpha = \frac{-r+\sqrt{r^2-4t}}{2}$ or $\alpha=\frac{-r-\sqrt{r^2+4t}}{2}$ (since the characteristic is not $2$). Moreover, the polynomial is irreducible and separable, and $\sqrt{r^2-4t}\notin F$. So $K = F(\sqrt{r^2-4t})$ and $K$ is a splitting field of an irreducible separable polynomial (namely, $x^2 - (r^2-4t)$), hence is Galois over $F$. 
If the characteristic is $2$, then the result is true for perfect fields, but not in general, as the examples by Zev Chonoles and Giovanni De Gaetano show. 
