Question about quadratic field extension of $\mathbb{Q}$ I saw a question about quadratic extension today:

Let $K \subset \mathbb{C}$ be a quadratic field extension of $\mathbb{Q}$. Show that there exists $D \in \mathbb{Q}$ such that $K = \{a+b\sqrt{D}: a,b \in \mathbb{Q}\}$ as a subset of $\mathbb{C}$.

Now I'm puzzled by the way the question is asked. Isn't it precisely the definition of a quadratic extension of $\mathbb{Q}$? What's there to show?
 A: I'll provide a general proof for a field $F$ of characteristic not $2$ and it's degree $2$ extension $K$.
Take any $\alpha \in K\setminus F$
Then by definition the elements $1,\alpha,\alpha^{2}$ are linearly dependent over $F$. (as dimension of $K$ as a vector space over $F$ is $2$.)
Then they satisfy a relation $c+b\alpha +a\alpha^{2}=0\,\,,a,b,c\in F$ . Observe that $a\neq 0$ as otherwise $\alpha\in F$.
So we can assume that that $a=1$ as we can just divide by $a$ and get the coefficient of $\alpha^{2}=1$.
So the minimal polynomial of $\alpha$ is $x^{2}+bx +c $ . This polynomial is irreducible as $\alpha\notin F$ and hence you cannot factor into linear factors over $F$.
Thus we can say by the quadratic formula(Which is valid as $char F\neq 2$) that :-
the roots are $\displaystyle \frac{-b\pm\sqrt{b^{2}-4c}}{2}$.
Where the $\sqrt{b^{2}-4c}$ is just a symbol for a root of the irreducible polynomial $x^{2}-(b^{2}-4c)$. (Note that this polynomial is indeed irreducible as otherwise if one of the roots lie in $F$ then $\alpha$ must lie in $F$ ).
Now define $F(\sqrt{b^{2}-4c})\cong \frac{F[x]}{(x^{2}-(b^{2}-4c))}$.
Now $\alpha\in F(\sqrt{b^{2}-4c})$ and also $\sqrt{b^{2}-4c}=\mp (2\alpha+b)$ . Thus in any case $F(\sqrt{b^{2}-4c})=F(\alpha)$.
Note :- $F(\alpha)\cong \frac{F[x]}{(x^{2}+bx+c)}$
Now we have $F(\alpha)\subset K$ but also $[F(\alpha):F]=2=[K:F]$. Hence $K=F(\alpha)=F(\sqrt{b^{2}-4c})=\{a_{1}+a_{2}\sqrt{b^{2}-4c}:a_{1},a_{2}\in F\}$.
Thus any quadratic extension of a field with characteristic not $2$ can be written as $F(\sqrt{D})$ with $D\in F$.
