Is it true that any quadratic extension over $F$ has form $F(\sqrt D)$?

Question

where $D \in F$ and $F$ has characteristic $0$.

I showed that since extension is finite and all extensions over a characteristic $0$ field are separable, then by primitive element theorem we have that any quadratic extension is of the form $F(\alpha)$.

But I am not sure where to proceed from here.

Answer 1

Let $f(X)=X^2+aX+b \in F[X]$ be the minimal polynomial of $\alpha$. Use the quadratic formula to see:

$$\alpha= \frac{-a \pm \sqrt{a^2-4b}}{2}.$$

Since $[F(\alpha):F]=2,$ there are no proper intermediate fields. Morevover, $\alpha \notin F \Rightarrow \sqrt{a^2-4b} \notin F$. Therefore, $F(\alpha)=F \left (\sqrt{a^2-4b} \right )$.