Finding all $\alpha\in\Bbb R$ for which there is some matrix $A\in M_{2005}(\Bbb R)$ s. t. $p_\alpha(A)=0$ 
For which $\alpha\in\Bbb R$ does there exist $A\in M_{2005}(\Bbb R)$ satisfying $5A^2+(6\alpha-2)A+(2\alpha^2-2\alpha+1)I=0$?

My attempt:
First of all, $$\begin{aligned}5A^2+(6\alpha-2)A+(2\alpha^2-2\alpha+1)I&=0\\ \implies 5A^{2005}+(6\alpha-2)A^{2004}+(2\alpha^2-2\alpha+1)A^{2003}&=0\\ \implies (-1)^{2005}A^{2005}-\frac{6\alpha-2}5A^{2004}-\frac{2\alpha^2-2\alpha+1}5A^{2003}&=0\\ \implies\det A&=0\end{aligned}$$
so, $A$ is singular. Let $p_\alpha(A):=5A^2+(6\alpha-2)+(2\alpha^2-2\alpha+1)I$ and let $\mu_A$ denote the minimal polynomial of $A.$ Then $p_\alpha(A)=0\implies\operatorname{deg}(\mu A)\le 2$ so $A$ has at most $2$ distinct eigenvalues, that is $\sigma(A)\subseteq\{0,\lambda\}, \lambda\in\Bbb R\setminus\{0\}.$ ($\lambda$ must be real, otherwise it would've come in pair with $\overline\lambda.$)
Let's first consider the case when $A$ has only one eigenvalue. Since $A$ is necessarily singular, it must be $0.$ Let $v\in\operatorname{Ker}A\subseteq\operatorname{Ker}A^2,v\ne 0.$ Then $5A^2v+(6\alpha-2)Av=0,$ so $(2\alpha^2-2\alpha+1)Iv=0,$ which implies $2\alpha^2-2\alpha+1=\alpha^2+(\alpha-1)^2=0,$ contradiction. Therefore, $A$ must have two distinct eigenvalues. That, together with $\operatorname{deg}(\mu_A)\le2$ implies $\mu_A(\lambda)=\lambda(\lambda-\lambda_0).$ As the irreducible factors are linear, $A$ is diagonalizable, that is, $\exists T\in M_{2005}(\Bbb R)$ invertible s. t. $$D=TAT^{-1}.$$ We can look at $p_\alpha(A)$ in the corresponding basis:
$$\begin{aligned}Tp_\alpha(A)T^{-1}&=5\underbrace{TA^2T^{-1}}_{TAT^{-1}TAT^{-1}}+(6\alpha-2)TAT^{-1}+(2\alpha^2-2\alpha+1)TIT^{-1}\\ &=5D^2+(6\alpha-2)D+(2\alpha^2-2\alpha+1)I\\ &=0,\end{aligned}$$
which yields $$\begin{aligned}5d_{ii}^2+(6\alpha-2)d_{ii}&=-(2\alpha^2-2\alpha+1), i\in\{1,\ldots,2005\}\\ \implies d_{ii}(5d_{ii}+6\alpha-2)&=-(2\alpha^2-2\alpha+1),i\in\{1,\ldots,2005\}\end{aligned}$$ contradiction, since $d_{ii}=0$ for some $i\in \{ 1,\ldots,2005 \},$ while $2\alpha^2-2\alpha+1\ne 0,\forall\alpha\in\Bbb R.$
Therefore, I think there is no such $\alpha\in\Bbb R.$
Is my answer valid?
 A: Hint
Consider the polynomial $p(x)= 5 x^2+(6\alpha-2)x+(2\alpha^2-2\alpha+1)$. Its discriminant is $\Delta = -4( \alpha-2)^2$ which is negative for $\alpha \neq 2$.
Let's first consider the simple case $\alpha = 2$. Then $p(x)=5(x+1)^2$ and $-I$ is a solution of the equation $p(A)=0$.
Now if $\alpha \neq 2$, $\Delta \lt 0$, and $p$ has no real root. You can then prove that it exists $e_1, \dots, e_n$ such that $\{e_1, \dots, e_n, A\cdot e_1, \dots, A \cdot e_n\}$ is a basis of the linear space over which $A$ operate. This family of vectors has an even number of vectors. A contradiction as here the dimension is $2005$.
Hence $\alpha = 2$ is the only case for which $p(A)=0$ has a solution.
A: Fact 1: Minimal polynomial of A divides $5x^2+(6\alpha-2)x+(2\alpha^2-2\alpha+1)=0$.
Fact 2: The characteristic polynomial has odd degree and therefore, has at least one real root. (IVT)
Fact 3: The minimal polynomial must have at least, one real root.
Then, the question is just about finding $\alpha$ such that the polynomial has a real root.
