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This is not an answer to my question but it is just a summary of the relevant results which have been mentioned so far about positive solvability.

So here is a useful result:

Theorem Let $K/F$ be a finite Galois extension and let $M/F$ be a radical extension. Assume that you have an embedding $\iota:K\hookrightarrow M$. Then if $[K:F]$ is odd there exists a root of unity of odd order inside $M$.

A proof of this result may be found for example in Brian Conrad's note: radical tower and roots of unity.

We thus have the following corollary

Corollary Let $f(x)\in\mathbf{Q}[x]$ be a polynomial which has all its roots in $\mathbf{R}$. Let $K$ be the (abstract) splitting field of $f(x)$ and assume that $[K:\mathbf{Q}]$ is odd. Then $K$ cannot be embedded in a radical extension contained in $\mathbf{R}$. In particular, $f(x)$ is not positive solvable.

As an example to illustrate this corollary, let $\zeta_n=e^{2\pi i/n}$. Let $f_n(x)$ be the minimal polynomial of $\zeta_n+\zeta_n^{-1}=2\cos(2\pi/n)$ over $\mathbf{Q}$ and let $K_n$ be its splitting field. Since $[K_n:\mathbf{Q}]=\varphi(n)/2$ we find that if $\varphi(n)/2$ is not a power of $2$ then $f_n(x)$ is not positive solvable.

2 added 10 characters in body

This is not an answer to my question but it is just a summary of the relevant results which have been mentioned so far about positive solvability.

So here is a useful result:

Theorem Let $K/F$ be a finite Galois extension and let $M/F$ be a radical extension. Assume that you have an embedding $\iota:K\hookrightarrow M$. Then if $[K:F]$ is odd there exists a root of unity of odd order inside $M$.

A proof of this result may be found for example in Brian Conrad's note: radical tower and roots of unity.

We thus have the following corollary

Corollary Let $f(x)\in\mathbf{Q}[x]$ be an irreducible a polynomial of odd degree which has all its roots in $\mathbf{R}$. Let $K$ be the (abstract) splitting field of $f(x)$. f(x)$and assume that$[K:\mathbf{Q}]$is odd. Then$K$cannot be embedded in a radical extension contained in$\mathbf{R}$. In particular,$f(x)$is not positive solvable. 1 This is not an answer to my question but it is just a summary of the relevant results which have been mentioned so far about positive solvability. So here is a useful result: Theorem Let$K/F$be a finite Galois extension and let$M/F$be a radical extension. Assume that you have an embedding$\iota:K\hookrightarrow M$. Then if$[K:F]$is odd there exists a root of unity of odd order inside$M$. A proof of this result may be found for example in Brian Conrad's note: radical tower and roots of unity. We thus have the following corollary Corollary Let$f(x)\in\mathbf{Q}[x]$be an irreducible polynomial of odd degree which has all its roots in$\mathbf{R}$. Let$K$be the (abstract) splitting field of$f(x)$. Then$K$cannot be embedded in a radical extension contained in$\mathbf{R}$. In particular,$f(x)\$ is not positive solvable.