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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.

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 divisible by an odd prime $p$, there exists a non-trivial 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 divisible by an odd prime $p$. 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. We know that $[K_n:\mathbf{Q}]=\varphi(n)/2$. In particular, if $\varphi(n)/2$ is not a power of $2$ then $f_n(x)$ is not positive solvable.

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.

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.

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.

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.