Characteristic fields and extensions

Question

$K$ is a field of characteristic $0$. Let $L$ be an extension of $K$ with$ [L : K] = 2.$ How do you prove that there exists an $u ∈ L$ such that $L = K(u)$ and $ u ^2 ∈ K$.

Answer 1

Take $v\in L\setminus K$. Since $[L:K]=2$, $\{1,v,v^2\}$ is linearly dependent over $K$. That is, there are $a,b,c\in K$ such that $av^2+bv+c=0$ and that at least one of them is not $0$. Then $a\neq0$. Otherwise $bv=-c\in K$ and it would follow that $b=0$ or that $v\in K$, and none of them can be true. But then\begin{align}av^2+bv+c=0&\iff a\left(v+\frac b{2a}\right)^2-\frac{b^2}{4a}+c=0\\&\iff\left(v+\frac b{2a}\right)^2=\frac{b^2-4ac}{4a^2}(\in K)\end{align}and therefore$$L=K\left(v+\frac b{2a}\right).$$