Question about Galois Theory. Extension of a field of odd characteristic. 
Let $F$ be a field of characteristic $\ne 2$, and let $K$ be an extension of $F$ with $[K:F]=2$. Show that $K = F(\sqrt{a})$ for $a \in F$; that is, $K = F(\alpha)$ with $\alpha^{2}=a$. Moreover, show that $K$ is Galois over $F$.

My doubt is in the first part.
Take $\alpha \in K\setminus F$. Then $\lbrace 1, \alpha, \alpha^{2} \rbrace$ are LD over $F$, so $\alpha^{2} + p\alpha + q = 0$ with $p,q \in F$ ($1$ and $\alpha$ are LI over $F$, since $\alpha \not\in F$). Completing the squares, we have
$$\left(\alpha + \frac{p}{2}\right)^{2} = \frac{p^{2}}{4} - q$$
because char$(F) \neq 2$. Let $\displaystyle a = \frac{p^{2}}{4} - q$, so $\displaystyle \sqrt{a}=\alpha + \frac{p}{2}$ and $\sqrt{a} \not\in F$.
Here is my doubt: I know $K/F$ is finite, then $K$ is finitely generated, but why $\sqrt{a} \not\in F$ ensure that $K=F(\sqrt{a})$?

The quickest way to see this that I know is to note that $\sqrt a \notin F$ implies $1$ and $\sqrt a$ are linearly independent over $F$, since for $\alpha, \beta \in F$, not both zero,
$\alpha + \beta \sqrt a = 0 \Longrightarrow \sqrt a = -\dfrac{\alpha}{\beta} \in F \Rightarrow \Leftarrow \sqrt a \notin F; \tag 1$
therefore $F(\sqrt a)$ is a two-dimensional subspace of $K$; since $[K:F] = 2$, we are done.
