Bifurcation points of differential equation (example) Assume the differential equation:
$$
x'=\lambda^2-8a\lambda x+2x^2, \quad a\in \mathbb{R}.
$$
The critical points are the solutions to the equation:
$$
x'=0 \iff 2x^2-8a\lambda x +\lambda^2=0\tag{1}
$$
which admits solutions:
$$
x=\lambda \cdot \frac{8a \pm \sqrt{64a^2-8}}{2}
$$
Now, having in mind that by bifurcation point, we mean a point where change of the number of equilibrium points occur, it seems to me that the number of $(1)$'s solutions depends only on $a$ and not on $\lambda$. Is this the case or did I get something wrong?
 A: It appears to me our OP LoneBone is correct; I can find no errors in his/her logic.
A: To study bifurcation points for 
$$
x'=\lambda^2-8a\lambda x+2x^2, \quad a\in \mathbb{R}, 
$$
we see that $x'=0$ implies that 
$$
x = \frac{\lambda}{2}\left(4a \pm \sqrt{2(8a^2-1)}\right). 
$$
$\textbf{Case 1}$. If $8a^2 < 1$, then the expression in the radical is a negative (real) number. So there are no bifurcation points, i.e., there doesn't exist an $x\in \mathbb{R}$ that will give $x'=0$. 
$\textbf{Case 2}$. If $8a^2 = 1 $, then $x^*=2a\lambda$, which is a fixed point (it is a double root). If you plug in values of $x< 2a\lambda$ into $x'$, then $x'>0$. You also see that if you plug in values of $x > 2a\lambda$, then $x'>0$. So if you perturb the fixed point $x^*$ a little to the left, it appears to be an attractor. But if you perturb $x^*$ a little to the right, it appears to be a repeller. So $x^*$ is neither an attractor nor a repeller; it is a semistable point. 

$\textbf{Case 3}$.  If $8a^2 > 1$, then there are two subcases to consider. 
$\color{green}{\textbf{Subcase 1}}$. If $\lambda\not=0$, then there are two fixed points: 
$$
x_1^* = \frac{\lambda}{2}\left(4a - \sqrt{2(8a^2-1)}\right), 
\quad 
x_2^* = \frac{\lambda}{2}\left(4a + \sqrt{2(8a^2-1)}\right). 
$$
We can use a similar analysis as in $\textbf{Case 2}$, or we can plot the differential equation on a planar graph to see that $x_1^*$ is an attracting (stable) fixed point, while $x_2^*$ is a repelling (unstable) fixed point. 

$\color{green}{\textbf{Subcase 2}}$. If $\lambda=0$, then $x^*=0$ is a (double) fixed point, which is analogous to $\textbf{Case 2}$.
