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!

By definition,
$\alpha \in \Bbb Q \Longrightarrow \exists p, q \in \Bbb Z, \; \alpha = \dfrac{p}{q} = pq^{-1}; \tag 1$
set
$Y = \begin{bmatrix} 0 & -p \\ q^{-1} & 0 \end{bmatrix} ; \tag 2$
then
$Y^2 = \begin{bmatrix} 0 & -p \\ q^{-1} & 0 \end{bmatrix} \begin{bmatrix} 0 & -p \\ q^{-1} & 0 \end{bmatrix}$
$= \begin{bmatrix} -pq^{-1} & 0 \\ 0 & -pq^{-1} \end{bmatrix} = \begin{bmatrix} -\alpha & 0 \\ 0 & -\alpha \end{bmatrix} = -\alpha I, \tag 3$
that is,
$Y^2 + \alpha I = 0; \tag 4$
thus $Y$ satisfies
$X^2 + \alpha \in \Bbb Q[X]; \tag 5$
note further for
$S \in \Bbb Q^{2 \times 2}, \tag 6$
with $S$ invertible, that
$(S^{-1}YS)(S^{-1}YS) = S^{-1}Y(SS^{-1})YS = S^{-1}YIYS$
$= S^{-1}Y^2S = S^{-1}(-\alpha I)S = -\alpha S^{-1}IS = -\alpha I, \tag 7$
or
$(S^{-1}YS)^2 + \alpha I = 0; \tag 8$
thus $S^{-1}YS$ also satisfies (5).
To see that there are an infinite number of such matrices, let
$S = \begin{bmatrix} s & 0 \\ 0 & 1 \end{bmatrix}, \;0 \ne s \in \Bbb  Q; \tag 9$
we have
$S^{-1}YS =  \begin{bmatrix} s^{-1} & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 0 & -p \\ q^{-1} & 0 \end{bmatrix} \begin{bmatrix} s & 0 \\ 0 & 1  \end{bmatrix}$
$= \begin{bmatrix} s^{-1} & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 0 & -p \\ q^{-1}s & 0 \end{bmatrix} = \begin{bmatrix} 0 & -ps^{-1} \\ q^{-1}s & 0 \end{bmatrix}, \tag{10}$
which are distinct for distinct values of $s$.
By further adjusting the vaues of $p$ and $q$ even more such matrices may be obtained; for example, every
$\forall p, q \in \Bbb Z, \; \alpha = pq^{-1} \tag{11}$
yields such a matrix $S^{-1}YS$.
