Minimal polynomial of extension of degree 2 over a finite field with characteristic 2

You cannot have $b=0$.

If $|K|=2^n$ then $a=a^{2^n}$ for every $a\in K$. If $a=0$ this is trivial. Otherwise $a$ belongs to the group $(K^*,\cdot )$ of order $2^n-1$ so, by Lagrange's theorem, $a^{2^n-1}=1$. We multiply this relation by $a$ and we get $a^{2^n}=a$.

Suppose now that $b=0$ so $p_\alpha (x)=x^2+a$. Since $a=a^{2^n}=(a^{2^{n-1}})^2$ and we are in characteristic $2$, we get $p_\alpha (x)=x^2+(a^{2^{n-1}})^2=(x+a^{2^{n-1}})^2$. But this is impossible, since $p_\alpha$ is irreducible.

Hence $b\neq 0$. When we divide the relation $\alpha^2+b\alpha +a=0$ by $b^2$ we get $(\alpha/b)^2+\alpha/b+a/b^2=0$ so $p_{\alpha/b}(x)=x^2+x+a/b^2=x^2-x-a/b^2$, which is what you want.

BTW you have a more general result. If $F$ is finite of characteristic $p$ and $[L:F]=p$ then $L=F(\alpha )$ for some $\alpha$ whose minimal polynomial over $F$ is of the form $x^p-x-a$ for some $a\in F$. But for this you need some Galois theory in positive characteristic. Look for Artin-Schreier extensions.

Consider the polynomials $p_a(x)=x^2-x-a$ for $a\in F$.

Observe that for $a\neq b$, both in $F$, the roots of $p_a$ and $p_b$ are disjoint. This is because a common root $r$ implies $r^2-r-a=0=r^2-r-b$, from where $a=b$ follows.

Also, for a given $a\in F$, the two roots, $r_1$ and $r_2$, of $p_a$ satisfy $r_1+r_2=1$. If they there equal, then $0=2r_1=1$.

If $p_a$ were reducible for all $a\in F$, then the total number of the roots of these polynomials would would be $2|F|$ different elements of $F$.

Therefore, at least one of those polynomials $p_a$ is irreducible over $F$.

Then $F[x]/(p_a)$ is a finite field of characteristic $2$ with the same cardinality as $L$.