Let $L$ be a field extension of a field $K$ of characteristic $\neq 2$ such that every element of $L\setminus K$ has degree $2$ over $K$, can we show that $[L:K]=2$ by elementary methods, without using the primitive element theorem ?
If $L$ is a separable extension of $K$, it is easy to prove this by using the primitive element theorem. Let $\alpha, \beta \in L\setminus K$. There exists $\gamma \in L$ such that $K(\gamma) = K(\alpha,\beta)$, therefore $[K(\alpha,\beta):K] = 2= [K(\alpha):K]$ and $K(\alpha,\beta)= K(\alpha)$, and $\beta \in K(\alpha)$. Since this is true for all $\beta \in L\setminus K$, $L=K(\alpha)$.
However, we have a counterexample if the characteristic of $K$ equal to $2$ : if $L= \mathbb F_2(X,Y)$ and $K=\mathbb F_2(X^2,Y^2)$, any element $f\in L$ is a root of $T^2-f^2 \in K[T]$, so has degree $1$ or $2$ over $K$, but $[L:K]= 4$.
Take any $\alpha \in L\backslash K$. If $K[\alpha] \not= L$, pick $\beta \in L\backslash K[\alpha]$ so that $[K[\alpha,\beta]:K]=4$ and has a K-basis $\{1,\alpha, \beta, \alpha\beta\}$. Now show that $[K[\alpha + \beta]:K]=4$ by observing that it contains at least three K-linearly independent vectors $\{1, \alpha + \beta, (\alpha\beta + v) \}$ where $v$ lies in span$_K(\alpha,\beta)$ (here you need the fact that 2 is invertible in $K$). This gives a contradiction.