Let $K/F$ be a finite field extension such that $F$ contains a primitive root of unity of order the degree of $K/F$. A theorem states that if the characteristic of $F$ is $0$ and $Gal(K/F)$ is a solvable group, then $K/F$ is a radical extension (not necessarily simple).
Question: what happens if we don't assume that the characteristic of $F$ is $0$? i.e if $K/F$ is a finite field extension, with $char(F)=p$, such that $F$ contains a primitive root of unity of order $[K:F]$ (which cannot be prime though). If $Gal(K/F)$ is solvable, does it follow that $K/F$ is radical?