Show: For every $\alpha \in \mathbb{Q}$, the polynomial $X^2+\alpha$ has endless different roots in $\mathbb{Q}^{2\times2}$. 
Show that for every $\alpha \in \mathbb{Q}$, the polynomial $X^2+\alpha$ has endless different roots in $\mathbb{Q}^{2\times2}$.

What I did:
Let $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \mathbb{Q}^{2\times2}, f=X^2+\alpha \in \mathbb{Q}[X].$
Consider $$f(A) = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \cdot \begin{pmatrix} a & b \\ c & d \end{pmatrix}+\begin{pmatrix} \alpha & 0 \\ 0 & \alpha \end{pmatrix} = \\ \begin{pmatrix} a^2+bc & ab+bd \\ ac+cd & bc+d^2 \end{pmatrix} + \begin{pmatrix} \alpha & 0 \\ 0 & \alpha \end{pmatrix} = \\\begin{pmatrix} a^2+bc + \alpha & ab+bd \\ ac+cd & bc+d^2 + \alpha \end{pmatrix}.$$
How do I get to the claim from here on? 
Thanks in advance!

Pick some $X$ that maps some vector $v$ to another (independent) vector $w$ and that $w$ maps to $-\alpha v$. 


*

*What does that imply for $X^2$?

*How many different $X$ can you obtain this way? (Say, let $v=e_1$, then how many choices for $w$ are possible?)

