Let $F$ be a field of characteristic $0$. Let $K = F(b,c)$ the field of rational functions in two variables over $F$. Let f = $X^4 + bX^2 +c \in K[X]$
Show that two of the roots of f are algebraically independent over $F$.
Clearly the two roots that we want are
$\alpha = \sqrt{\frac{b}{2} + \frac{\sqrt{b^2-4c}}{2}}$ , $\beta = \sqrt{\frac{b}{2} - \frac{\sqrt{b^2-4c}}{2}}$
I can intuitively see that no polynomial over $F$ in two variables can have $(\alpha,\beta)$ as a root but I'm not sure how to prove it.
I have tried assuming that such a polynomial exists and attempted to reach a contradiction, but that didn't work out.
Alternatively, I tried considering the field $L = F(\alpha,\beta)$ which I showed to be the splitting field of f over $K$. Then clearly $L : F(\alpha,\beta)$ is an algebraic extension (since they are equal) and so I believe {$\alpha,\beta$} is a transcendental spanning set of $L$ over $F$. If $L$ is isomorphic to a field of rational functions over $F$ then we are done since then {$\alpha,\beta$} is algebraically independent but I don't know how to show that $L$ is isomorphic to said field. (I'm not entirely sure this works).
Any help would be appreciated!