I want to show:
Let $F$ be a field with characteristic 2. If $E/F$ is a separable extension with degree 2, then $E= F(\alpha)$ for some $\alpha\in F$ such that $\alpha^2+\alpha\in F$.
My attempt: As $E/F$ is a finite extension, let $\alpha\in E\setminus F$ so that $E = F(\alpha)$. Consider the minimal polynomial $m_\alpha$. Since $m_\alpha$ splits in $E$, $m_\alpha = (x-\alpha)(x-(m\alpha+n))$ where $m,n\in F$. As $m_\alpha\in F[x]$, $x^2-(m\alpha+\alpha+n)x+\alpha(m\alpha+n)\in F[x]$ so in particular, $(m+1)\alpha+n\in F$ so that $m=-1$. Hence, the form should be $(x-\alpha)(x-(\alpha+n))$. Here's where I'm stuck. How can I show $n=1$ is possible? Any comment will be appreciated.