Q: Extension of degree 2 with characteristic not 2 I am told that $K \leq L$ is a field extension of degree 2 and that $K$ has characteristic not 2. I am then asked to show that $\exists \: \alpha \in L$ such that $L = K(\alpha), \; \alpha^2\in K $
So far I have that:
$\exists \; \alpha_1, \alpha_2 \in L$ such that $L = K( \alpha_1, \alpha_2)$
$\Rightarrow |L:K| = |L:K(\alpha_1)| |K(\alpha):K| $ so $|K(\alpha):K| = 1$ or $2$
$|K(\alpha_1):K| = 2 \Rightarrow L = K(\alpha_1)$ and $deg(f_{\alpha_1}(t)) = 2  $ where $f_{\alpha_1}(t)$ is the minimal polynomial of $\alpha_1$ over $K$ 
So $f_{\alpha_1}(t) = t^2 + at + b$ for some $a,b \in K$
Also, $|K(\alpha_1):K|=1 \Rightarrow K(\alpha_1) = K$ and $L = K(\alpha_2)$ which means $degf_{\alpha_2}(t) = 2$ where $f_{\alpha_2}(t)$ is the minimal polynomial of $\alpha_2$ over $K$. 
So $f_{\alpha_2}(t) = t^2 + at + b$ for some $a,b \in K$
Given that we consider the same polynomial each time, relabel to consider: 
$f_{\alpha}(t) = t^2 + at+ b $ for some $a,b \in K$
We have already demonstrated that $L=K(\alpha)$ for some $\alpha \in L$ so now we much show that $\alpha^2 \in K$
$\alpha^2 \in K \Leftrightarrow a = 0_K$
I now have no idea how to show that this coefficient of $f_{\alpha}(t)$ is necessarily $0$. 
I know that $f_{\alpha}(t)$ has two roots in $L$ and that because $K$ and thus $L$ has characteristic not 2, they cannot be the same root, but I don't know how to then conclude that they must sum to $0$
Any help would be greatly appreciated. 

A: Let $w \in L{\setminus}K$.

Since $K(w) \subseteq L$, we have $[K(w):K] \le 2$.

Since $w \notin K$, we have $[K(w):K] > 1$.

It follows that  $[K(w):K] = 2$, hence $K(w)=L$.

Let $p \in K[x]$ be the minimal polynomial for $w\;$over $K$.

Then $p(x) = x^2 + ax + b$, for some $a,b \in K$.
\begin{align*}
\text{Then}\;\;&w^2+aw+b = 0\\[4pt]
\implies\;&w^2+aw = -b\\[4pt]
\implies\;&4w^2+4aw = -4b\\[4pt]
\implies\;&4w^2+4aw + a^2 = a^2-4b\\[4pt]
\implies\;&(2w+a)^2 = a^2-4b\\[4pt]
\end{align*}
Letting $u = 2w+a$, then since$\;\text{char}(K) \ne 2$, we get $K(u) = K(w)$, hence


*

*$L = K(u)$.$\\[6pt]$

*$u \notin K$.$\\[6pt]$

*$u^2 \in K$.


so the element $u\;$satisfies the required conditions.
