Question: Suppose $K$ is a field of characteristic $\neq 2$ and $L$ is a quadratic extension of $K$. Prove that $L$ is Galois over $K$ and there is an $a \in K$ such that $L=K(\sqrt{a})$
First part ($L$ over $K$ is Galois)
I found the solution given here to be helpful: Extensions of degree two are Galois Extensions. Let $L/K$ be field extension of degree $2$. $L $ \ $K \implies p(t)=min_K(\alpha, t)$ has degree 2 and splits over $L$
The only part I was unsure about is why does $p'(t) \neq 0$ imply separability?
Second part (there is an $a$ such that $L=K(\sqrt{a})$
I thought that if char$K \neq 2$ then any any cyclic quadratic extension of $K$ is of the form $L=K(\sqrt{a})$ so I though all elements in our example would have this property?
First part
You have to first prove that a monic polynomial $f$ over a field $F$ has repeated roots over a splitting field $K$ iff $\gcd(f,f')$ is not a unit, where $f'$ is the formal derivative. This involves looking at the formal derivative of the factorized form of $f$ over $K$. The simplest way is via the product rule for the formal derivative, which you can prove before that.
After that, prove that if $f$ is irreducible, then $\gcd(f,f')$ is a unit iff $f' \ne 0$. This follows by expanding definitions, because $f'$ has strictly lower degree than $f$.
Second part
$K,L$ are given, and the question hence asks for you to find some $a \in K$ such that $L = K(\sqrt{a})$. Clearly not all $a \in K$ work, since $K(\sqrt{1}) = K \ne L$. $\def\less{\smallsetminus}$
However, it is true that $L = K(b)$ for any $b \in L \less K$, since the only possible finite extension of $K$ within $L$ is $L$ itself.