 Solution

Choose $\alpha \in K\setminus F$. Since $[K:F]=2$, the linearly independent elements $1,\alpha$ form a basis of $K$ over $F$. Thus there are $a,b\in F$ with ${\alpha }^2=a+b\alpha$. The assumption on the characteristic of a field allows division by two. SetCompleting the square givesAlso $\delta \notin F$, since otherwise $\alpha =\delta +b/2$ would lie in $F$. Because $\alpha \in F(\delta )$ and $\delta \in F(\alpha )$, these generated subfields agree. The basis representation shows $F(\alpha )=K$, henceThe square-root generator is obtained by translating an arbitrary generator to remove the linear term of its quadratic relation.