Semi-simple quadratic extension over $\mathbb{Q}$ An old qual problem

Let $K$ be a semi-simple quadratic extension over $\mathbb{Q}$ and
  consider the regular representation $\rho: K \to M_2(\mathbb{Q})$.
  Compute the index of $\rho(K^\times)$ in the normalizer of
  $\rho(K^\times)$ in $GL_2(\mathbb{Q})$, and justify your answer.

I'm actually having trouble figuring out what they mean. What's a "semi-simple quadratic extension of $\mathbb{Q}$? Do they mean a field extension? Those would all be separable and so fit the definition of a semisimple algebra (I think). So that would be redundant.
edit: Suppose they just mean a semisimple $Q$-algebra of dimension 2. The  algebra is either $\mathbb{Q} \times \mathbb{Q}$ or $\mathbb{Q}(\sqrt{d})$. In the former case I claim the index is two. In the latter case, I'm not so sure. What do we know about the index of something in its normalizer? Or about the normalizer of matrices like
$$\begin{pmatrix} a& db \\ b& a \end{pmatrix}?$$
 A: Claim: Any two-dimensional semisimple $\mathbb{Q}$ algebra is either $\mathbb{Q} \times \mathbb{Q}$ or a degree two field extension of $\mathbb{Q}$. The only other possible thing in a Wedderburn decomposition would be a two-dimensional division algebra over $\mathbb{Q}$. The center of such an algebra would be a $\mathbb{Q}$-vector space, so necessarily either the whole thing, or just $\mathbb{Q}$. The center being $\mathbb{Q}$ isn't possible, because a central division algebra over a field is always of square dimension. $\square$
Claim: When the algebra is $\mathbb{Q} \times \mathbb{Q}$, the index of $\rho(K^\times)$ in its normalizer is two. For then $\rho(K^\times)$ are the invertible diagonal matrices. The normalizer of the invertible diagonal matrices in $GL_n(F)$ (perhaps I should say in characteristic not 2?) are the generalized permutation matrices, which have structure $\texttt{diag} \rtimes S_n$. Therefore the index of $\rho(K^\times)$ in its normalizer is $|S_2|=2$.
Claim: In case the algebra is $\mathbb{Q}(\sqrt{d})$, the normalizer is two. 
Lemma: The subgroup $\rho(K^\times)$ is self-centralizing.
Lemma: The subgroup $\rho(K^\times)$ is not self-normalizing. For think about automorphisms of $K/\mathbb{Q}$: the one sending $\sqrt{d} \mapsto -\sqrt{d}$. I should be able to get that from conjugation, since I can change basis from $1, \sqrt{d}$ to $1, -\sqrt{d}$. In particular, conjugation by 
$$\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$$ 
is nontrivial.
Now the normalizer mod the centralizer is a subgroup of $\operatorname{Gal}(K/\mathbb{Q})$, so $|N/C|$ is at most two, and we just showed it's more than one. $\square$
