Minimal polynomial of extension of degree 2 over a finite field with characteristic 2 I'm struggling to solve the following question.

Let $F$ be a finite field with characteristic 2 and $L/F$ be a finite extension with $[L:F]=2$.  Prove that there exists $\alpha\in L$ such that $L = F(\alpha) $ and the minimal polynomial of $\alpha$ , $\text {Irr}_F \alpha(x) = x^2-x-a$ for some $a\in F$

My attempt:
Let $\alpha\in L\text \F$, then $F\subset F(\alpha) \subset L$ is a tower, so we have $2 = [L:F] = [L:F(\alpha)][F(\alpha):F]$.
Because $\alpha\notin F$, $[F(\alpha):F]=2$ and $[L:F(\alpha)]=1$. 
So we get $L=F(\alpha)$.
Let $p_\alpha(x) = \text{Irr}_F\alpha(x)$.
Note that deg $(p_\alpha(x)) = [F:F(\alpha)] = 2$ . So we can assume that $p_\alpha(x) = x^2+bx+a \in F[x]$, here $a\in F, b\in F.$
Then, I want to show that there exists $\alpha \in L\text\F$ such that $b=1$.
First, I'm trying to find  $\alpha$  such that b $\neq 0$:
Observe that if  $\alpha^2\in F$, then the minimal polynomial $p(x)$ will be $x^2-\alpha^2$.
So I want to find $\alpha$ such that $\alpha^2\notin F$, if such $\alpha$ exists, then $b\neq 0$.
But I have no ideal about how to prove the existence of this $\alpha$.
And here comes another problem: If we can find $\alpha\in L\text\\F$ satisfy the condition above, how can we say that $b=1$ ?
So I would like to know that my thoughts are correct or not, and want to solve this question in elementary method (without Galois Theory).
Thanks for any help!

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 $L$.
