Let \(F\) be a field with \(\ch(F)\neq 2\) and let \(E/F\) be a field extension with \([E:F]=2\). Then \(E/F\) is Galois, and \(E=F(\sqrt{a})\) for some \(a\in F\) that is not a square in \(F\). For \(F=\QQ\), \(a\) can be taken to be a squarefree integer different from \(1\).
Choose \(\alpha\in E\setminus F\); then \(E=F(\alpha)\) and the minimal polynomial of \(\alpha\) over \(F\) is \(x^2+bx+c\) with \(b,c\in F\). Since \(2\) is invertible in \(F\), put \(\beta\coloneqq\alpha+b/2\) and \(a\coloneqq b^2/4-c\in F\). Then \(\beta^2=a\), \(\beta\notin F\), and \(E=F(\beta)=F(\sqrt a)\), so \(a\) is not a square in \(F\). The roots of \(x^2-a\) are \(\pm\beta\in E\), and \(\beta\neq-\beta\) because \(\beta\neq0\) and \(\ch(F)\neq2\). Thus \(E\) is the splitting field of the separable polynomial \(x^2-a\), so \(E/F\) is Galois. For \(F=\QQ\), write \(a=m^2 d/n^2\) with \(m,n\) nonzero integers and \(d\) a squarefree integer; then \(\QQ(\sqrt a)=\QQ(\sqrt d)\), and \(d\neq1\) because \(a\) is not a square.