# Question related to separable extension.

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.

Since $$\alpha$$ is separable, $$n\neq 0$$. Now set $$\beta=\dfrac{\alpha}{n}$$. Then $$E=F(\alpha)=F(\beta)$$, and $$\beta^2+\beta=\dfrac{\alpha^2+n\alpha}{n^2}=\dfrac{\alpha(\alpha+n)}{n^2}\in F$$.