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Assume $K/F$ is quadratic extension i.e., $[K:F]=2$ and $F$ is of characteristic 2. Show that $K=F(\alpha)$ where $\alpha$ is a root of some polynomial of form $x^2+a$ or $x^2+x+a$ in $F[x]$. ( F is not necessarily the field of 2 elements)

I know if I can prove that $x^2+a$ and $x^2+x+a$ are irreducible in $F[x]$ then the result follows but I am not able to show that these polynomials are irreducible.

Thanks for the help.

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    As you see from the (+1) answer from user218931 you are approaching the problem a bit from the wrong end. The polynomials you listed are not necessarily irreducible. The upshot is that we can choose the generatiing element $\alpha$ in such a way that its minimal polynomial over $F$ is of one of those two forms. – 2017-02-07
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    $[K:F] = n$ really means that $K$ is a $F$-vector space of dimension $n$. So for any $a \in K$, there is a non-trivial linear relation on $\{1,a,a^2,\ldots,a^n\}$ of the form $\sum_{m=0}^d c_m a^m = 0, c_m \in K, c_d \ne 0, d \le n$. – 2017-02-07

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Take any $x\in K$ with $x\notin F$. Because of $[K:F] = 2$ we have $K = F(x)$ in any case. Let $X^2+aX+b\in F[X]$ be the minimal polynomial of $x$. If $a=0$, then $x$ is a root of $X^2+b$, so we are done.

Assume now that $a\neq 0$. Let $y:= \frac{x}{a}$. Then $y$ is a root of $X^2+X+\frac{b}{a^2}\in F[X]$. Since $y = \frac{x}{a} \notin F$, we must have $K = F(y)$ and $y$ is as desired.