Assume $F$ is a field of characteristic $\neq 2$. Let $(V,q)$ be a quadratic space such that $\rm dim~ q\geq 3$. When $q$ is irreducible it is known that

there exist a purely transcendental field extension $K/F$ such that $[F(q):K]=2$. Here $F(q)$ is a function field of a scheme $X_q:=\rm Proj (S(V^*)/q)$ associated to $q$, where $S(V)$ is a symmetric algebra.

I want to show above fact, but I am not able to do it. I was trying to find some isotropic sub-form of codimension 2 of $q$, because in that case this sub-form will have function field which will be purely transcendental over $F$. Also note that the form $q$ will be irreducible since $\rm dim ~q\geq 3$.

Can somebody suggest proof or a reference to the proof.

Thanks!