When is the given matrix not diagonizable? 
Given the following matrix $$A = \begin{bmatrix} 0 & 1\\ bk+b & -bk\end{bmatrix}$$ where $b \neq 0$ and $k \neq 0$, find when the given matrix is not diagonizable.

My initial thought would be when the matrix doesn't have $2$ linearly independent eigenvectors. 
In this case it would be when the eigenvalue has a multiplicity of 2.
I.e. $\lambda_1=\lambda_2 $.
I find the eigenvalues:
$
\lambda_1=\frac{-bk+\sqrt{(bk)^2-4\cdot(-bk-b)}}{2}\\
\lambda_2=\frac{-bk-\sqrt{(bk)^2-4\cdot(-bk-b)}}{2}
$
I then let them equal each other:
$
(1)\qquad\frac{-bk+\sqrt{(bk)^2-4\cdot(-bk-b)}}{2}=\frac{-bk-\sqrt{(bk)^2-4\cdot(-bk-b)}}{2}\\
\quad\\
\quad\\
(2)\qquad\sqrt{(bk)^2-4\cdot(-bk-b)}=-\sqrt{(bk)^2-4\cdot(-bk-b)}
$
Here my initial thought would be that the only case this is true if the term under the square root is equal zero.
$
b^2k^2+4bk+4b=0\\
b(bk^2+4k+4)=0
$
Since $b\neq0$ I need the term inside the parantheses to be equal 0.
$
bk^2+4k+4=0
$
I'm a bit lost from here.
I know the solution is 
$
-bk-k=(\frac{-bk}{2})^2
$.
The solution was provided by someone else here on Stack Exchange, he used the determinant and the trace of the matrix to come to that conclusion, which I can't seem to replicate.

$$b\neq 0, k\neq 0, A=\begin{pmatrix} 
0 & 1\\
bk+b & -bk
\end{pmatrix}\Rightarrow \det (A-xI)=\begin{vmatrix}
-x & 1\\
bk+b & -bk-x
\end{vmatrix}=0\Rightarrow$$
$$\Rightarrow x^2+bkx-(bk+b)=0\Rightarrow x=\dfrac{-bk\pm \sqrt{b^2k^2+4bk+4b}}{2}$$
$$x_1=x_2\Rightarrow b^2k^2+4bk+4b=0\Rightarrow b^2k^2=-4(bk+b)\Rightarrow \left( \dfrac{bk}{2}\right)^2=-bk-b$$
Let us now look at the eigenvectors.
$$x=-\dfrac{bk}{2}\Rightarrow A-xI=\begin{pmatrix}
\frac{bk}{2} & 1\\
-\left( \frac{bk}{2}\right)^2 & -\frac{bk}{2}
\end{pmatrix}$$
$$\begin{pmatrix}
\frac{bk}{2} & 1\\
-\left( \frac{bk}{2}\right)^2 & -\frac{bk}{2}
\end{pmatrix}\begin{pmatrix}
x\\
y
\end{pmatrix}=\begin{pmatrix}
0\\
0
\end{pmatrix}\Rightarrow \left\{ \begin{array}{lcc}
             \dfrac{bk}{2}x+y=0 \\
             \\ -\left( \dfrac{bk}{2}\right)^2x-\dfrac{bk}{2}y=0
             \end{array}
   \right.\Rightarrow bkx+2y=0\Rightarrow$$
$$E_{-\frac{bk}{2}}=\langle (-2,bk)\rangle \Rightarrow \dim \left( E_{-\frac{bk}{2}}\right) =1$$
Not diagonalizable.
