Field having an archimedean ordering and a non archimedean ordering

Let $F$ be a formally real field, so that $F$ has at least one ordering (compatible with the operations of $F$).

Question. is it possible for $F$ to have an archimedean ordering $\leq_1$ and a non-archimedean ordering $\leq_2$ ?

I know that the answer is NO in several cases:

• formally real number fields (because of the description of orderings of finite extensions of $\mathbb{Q}$)

• real closed fields (since such fields have a unique ordering).

• formally real pythagorean fields (pythagorean: every sum of two squares is a square), but the arguments I have are extremely complicated and indirect... To be quick, I think I can show that if $F$ is pythagorean and admits an archimedean ordering , $SO_3(F)$ is a simple group, while if $F$ is pythagorean and admits a non archimedean ordering, I can show that $SO_3(F)$ is not a simple group.

I am pretty sure the answer is NO in general, but I cannot come up with simple arguments ( I am probably missing something obvious...)

If needed, I am mainly interested to have a simple proof in the case of formally real pythagorean fields.

EDIT. Apparently, there are fields $F$ for which the answer is YES (see GEdgar's comment).

The real question would then be:

Question (V2). Let $F$ be a formally real pythagorean field. Find a direct proof of the fact that $F$ cannot have an archimedean ordering $\leq_1$ and a non-archimedean ordering $\leq_2$.

Take $\mathbb Q$ together with one transcendental $𝑥$. Then the countable field $\mathbb Q(x)$ can be ordered:
(a) so that $x=\pi$,
but it can also be ordered
(b) so that $𝑥 > r$ for all $r \in \mathbb Q$.