I'm given a discrete dynamical system in which $f$ is given by $$f(x,y)=(1+y-ax^2,bx). a,b \in \mathbb{R}, b\neq 1$$
I'm asked to find the equilibria which I have done below:
$$f(x,y)=(x,y) $$ $$\Rightarrow 1+y-ax^2=x \text{ and } bx=y$$ $$ \Rightarrow ax^2-(b-1)x-1=0$$
$$\Rightarrow x=\frac{(b-1) \pm \sqrt{(b-1)^2+4a}}{2a}$$
Next I'm told that when $a=\frac{-(b-1)^2}{4}$ there is only one equilibrium. I'm asked what can be said about it's stability.
I have found that the equilibrium is $x=\frac{b-1}{2a}.$
To determine its stability I have found $f'(x,y)=\begin{pmatrix} -2ax & 1 \\ b & 0 \end{pmatrix}$
At the equilibrium position this is $\begin{pmatrix} 1-b & 1 \\ b & 0\end{pmatrix} $
Then $\det(A-\lambda I)=\begin{vmatrix} 1-b-\lambda && 1 \\ b && -\lambda \end{vmatrix} \Rightarrow \lambda=\frac{1-b \pm \sqrt{(1-b)^2+4b}}{2a}=\frac{-2\big(1-b \pm\sqrt{(1-b)^2+4b} \big)}{(b-1)^2} $
What do I do next?