I want to show that
$[L:K]=2$ and $char(K)\neq 2$ $\Rightarrow$ $L=K(\sqrt{a})$ for an $a \in K$.
I know that, since $[L:K]=[K(a):K]=[a:K]$ for all $a\in L/K$, the minimal polynomial of $a$ has to have degree 2. Hence I have as minimal polynomial $X^2-a=0 \Rightarrow L=K(\sqrt{a})$.
Why I need to have that the $char(K)\neq 2$?
I have also seen this forumpage:
Let $[L:K]=2$ and char $K\neq 2$ show that $L/K$ is a simple 2-radical extension.
But I really don't understand the solution
Take $a\in L\setminus K$. Then consider the minimal polynomial $m_{a}(x)$ of $a$ over $K$ . Then $1\leq\deg(m_{a})\leq 2$ but $\deg(m_{a})\neq 1$ as $a\notin K$.
So $m_{a}(x)=x^{2}+bx+c$ say for some $b,c\in K$.
Then notice that $2^{-1}(-b+\sqrt{b^{2}-4c})$ and $2^{-1}(-b-\sqrt{b^{2}-4c})$ both are roots of the polynomial. Here by $\sqrt{b^{2}-4c}$ I mean any root of the polynomial $x^{2}-(b^{2}-4c)$ in $K[x]$ . But notice that the polynomial $x^{2}-(b^{2}-4c)$ cannot have a root in $K$ as otherwise $a\in K$ . So $x^{2}-(b^{2}-4c)$ is the minimal polynomial of $\sqrt{b^{2}-4c}$ over $K[x]$ and hence $[K(\sqrt{b^{2}-4c}):K]=2$ . But $K(\sqrt{b^{2}-4c})\subseteq L$ so we must have $L=K(\sqrt{b^{2}-4c})$
If you want a more elaborate explanation of the above result then you can look at Dummit and Foote pg $522$.
We used that $\text{char}(K)\neq 2$ when we took $2^{-1}$ .
But say you are working over $char(K)=2$. Say $K=\Bbb{F}_{2}=\{0,1\}$ and $L=\Bbb{F}_{4}$
Say $L=K(\sqrt{a})$ for some $a\in K$
Then $a=0$ or $a=1$. But this means that $\sqrt{a}$ is either a root of $x^{2}-0$ or $x^{2}-1$. Which means that $\sqrt{a}=0$ or $\sqrt{a}=1$. And hence $L\neq K(\sqrt{a})=K$.