If negative discriminant then cannot have a cyclic order 4 Galois group 
Let $f(x)$ be an irreducible polynomial of degree 4 in $\mathbb{Q}[x]$ with discriminant $D$. Prove that if $D < 0$ then $K$ cannot be cyclic of degree 4 over $\mathbb{Q}$.

I can see that if $D$ is negative, then $K$ has to have a complex subfield, namely $\mathbb{Q}(\sqrt{D})$. Then complex conjugation has to be an order 2 element of the Galois group. I want to get a contradiction from this, but I don't know how to finish this argument.
EDIT: Further ideas I'm thinking of is manipulating $D = \prod_{i < j} (\alpha_i - \alpha_j)^2$, in some way to get $\sqrt{-D}$ and then we could divide to get $i$ somehow? 
 A: To expand on Mathmo123's comment, one may show the following:
Proposition. Let $K$ be a field of characteristic different from $2$. If $L/K$ is cyclic of order $4$, then the discriminant $D$ of $L/K$ is a sum of two squares. 
In particular, if $K\subset\mathbb{R}$, $D$ is positive.
Sketch of proof.  Let $L/K$ be a cyclic extension of degree $4$. We know that $D$ is not a square, and that $\sqrt{D}\in L$. Hence $K(\sqrt{D})$ is the unique quadratic subfield of $L$ by Galois correspondence. 
Now $L/K(\sqrt{D})$ is quadratic, hence $L=K(\sqrt{u+v\sqrt{D}})$ for some $u,v\in K$.
Set $\alpha= \sqrt{u+v\sqrt{D}}$, $\beta=\sqrt{u-v\sqrt{D}}$. 
Then $\mu_{\alpha,K}=(X^2-\alpha^2)(X^2-\beta^2)=X^4-2uX^2+u^2-Dv^2$ (it is a monic polynomial with coefficients in $K$ of degree $4$, and $\alpha$ is a root of this polynomial). 
Hence the automorphisms are $Id_L, \alpha\mapsto -\alpha, \alpha\mapsto \beta,\alpha\mapsto -\beta$.
One may show that $\alpha\beta\notin K$ (otherwise, computation shows that the Galois group is the Klein group). But $(\alpha\beta)^2=u^2-Dv^2\in K$.
Hence $K(\alpha\beta)$ is a quadratic subfield of $L$, so $K(\alpha\beta)=K(\sqrt{D})$, and Kummer theory says that $\alpha^2\beta^2=\lambda^2 D=u^2-Dv^2$. Thus, $D=\dfrac{u^2}{\lambda^2+ v^2}=\Bigl(\dfrac{u\lambda}{\lambda^2+v^2}\Bigr)^2+\Bigl(\dfrac{uv}{\lambda^2+v^2}\Bigr)^2$ is a sum of two squares.
Side remark. The converse is true. If $D\in K$ is a sum of two squares, but not a square, then $D$ is the discriminant of a cyclic extension of degree $4$.
Indeed, if $D=u^2+v^2$, then $L=K(\sqrt{q(D+u\sqrt{D})})$ is cyclic for all $q\in K^\times$.
