<s>[INST] Can this determinant expression ever equal $0$? My issue is that when calculating the eigenvalues of a matrix, there are cases where the eigenvalue is $0$, and so I was trying to confirm this by calculating the determinant and checking when it is equal to $0$. 
The setup is: $a^2 - 4bc \neq 0$, $bc \neq 0$, $(a,b,c) \in \mathbb R^3$, and $n \geq 1$ is an integer 
and I have this expression: 
\begin{equation}
\dfrac1{\sqrt{a^2-4bc}} \left( \left( \dfrac{a + \sqrt{a^2-4bc}}2\right)^{n+1} - \left( \dfrac{a - \sqrt{a^2-4bc}}2\right)^{n+1}\right) = 0
\end{equation}
Which I simplified to 
\begin{equation}
  \left( \dfrac{a + \sqrt{a^2-4bc}}2\right)^{n+1} = \left( \dfrac{a - \sqrt{a^2-4bc}}2\right)^{n+1} \implies \sqrt{a^2-4bc} = -\sqrt{a^2-4bc}
\end{equation}
From here, I am unsure how to get a solution, since $a^2 - 4bc \neq 0$, and for real numbers, $x$, I think the only case for $+x = -x$ is when $x = 0$, which is not true here. So since the square root can be complex, I am unsure if there are cases where this can be true. 
The eigenvalue expression is 
\begin{equation}
\lambda_k = a + 2b\sqrt{\frac{c}{b}}\text{cos}(\frac{k\pi}{n+1}), \qquad k \in \{1, \dots, n\}
\end{equation}
I am not experienced with complex numbers, but using the eigenvalue equations, it seems like the case of $b = 1 \text{, } c = 1, a= -2\cdot\text{cos}(\frac{1\pi}{n+1})$ (for any $n$, but I used $n = 5$), should lead to a $0$ determinant/singular matrix, but this results (for $n=5$) in $a^2 - 4bc = -1$, and so it'd require $\sqrt{-1} = -\sqrt{-1}$, but I can't see how that'd be true.
Thanks in advance for the help! (Also if you want to see the matrix I am working with, this question has the determinant expression: How to compute the determinant of a tridiagonal matrix with constant diagonals?)
 [/INST] From $$\left(\frac{a+\sqrt{a^2-4bc}}2\right)^{n+1}=\left(\frac{a-\sqrt{a^2-4bc}}2\right)^{n+1} $$
we get, by dividing by the non-zero(!) rught hand side
$$ \left(\frac{a+\sqrt{a^2-4bc}}{a-\sqrt{a^2-4bc}}\right)^{n+1}=1,$$ 
or after making the denominator rational,
$$ \left(\frac{(a+\sqrt{a^2-4bc})^2}{4bc}\right)^{n+1}=1,$$
so the condition is rather
$$ \tag1\frac{(a+\sqrt{a^2-4bc})^2}{4bc}=e^{\frac{2k\pi i}{n+1}},\quad 0\le k\le n.$$
Now distinguish cases depending on the value of $bc$.
If $bc<0$, the left hand side in $(1)$ is real and negative,  and we arrive at
$$ \left|a+\sqrt{a^2-4bc}\right|=2\sqrt{-bc},\quad n\text{ odd}, k=\frac{n+1}2.$$
If $0<bc<\frac14a^2$, the left hand side in $(1)$ is real and positive, and we arrive at
$$ \left|a+\sqrt{a^2-4bc}\right|=2\sqrt{bc},\quad n\text{ arbitrary}, k=0.$$
Finally, if $bc>\frac14a^2$, we really encounter complex solutions, namely for
$$ a=\sqrt{4bc}\cos \phi, \quad(n+1)\phi\equiv 0\pmod{2\pi}.$$ 
</s>