Let $k \subseteq F$ be a extension, and let $a, b \in k$ be elements which are not squares, but such that $\sqrt{a}, \sqrt{b} \in F$, and are well defined and whatnot. Prove that, if the characteristic of $k$ is not $2$, then $k(\sqrt{a}, \sqrt{b}) = k(\sqrt{a} + \sqrt{b})$. Furthermore, show that the extension has degree 2 or 4 depending on if $ab$ is, or is not, a square in $k$, respectively.
Most of this question I have solved. I have shown the extensions to be equal, and that if $ab$ is a square, then the degree of the extension is $2$. The only part remaining is to show that the degree is $4$ if $ab$ is not a square.
I have made some progress in this direction. I have bounded the degree of the extension by $4$, both abstractly and by finding a monic degree $4$ polynomial with $\sqrt{a} + \sqrt{b}$ as a root. However, none of this is quite enough to show that the degree actually reaches $4$. There's a variety of options I can see, including showing that $\sqrt{ab} \notin k(\sqrt{a})$and showing that the polynomial I found is irreducible, but I can't see how to actually do that. Hints and solutions are greatly appreciated.
Consider $k(\sqrt{ab})\subseteq k(\sqrt{a},\sqrt{b})$. Suppose $ab$ is not a square in $k$. Then either $k(\sqrt{ab})=k(\sqrt{a},\sqrt{b})$ (and the degree is $2$) or the degree is $4$.
Can we have equality? Suppose so. Then $$ \sqrt{a}\in k(\sqrt{ab})=k(\sqrt{a},\sqrt{b}) $$ so there exist $p,q\in k$ such that $$ \sqrt{a}=p+q\sqrt{ab} $$ and so $2pq\sqrt{ab}=a-p^2-abq^2\in k$. This implies either $p=0$ or $q=0$.
Can you finish?