I am solving the following function for stability
$\dot x=x(x^2-1-\kappa)$
The linearisation of the function at rest position $x_1= \sqrt{1+\kappa}$ gives us the form $\frac{df}{dx}(x_1)= 3\,\sqrt{1+\kappa}^2-1-\kappa$.
Now the question is:
Will this solution be real or would it also include imaginary numbers for the case of $\kappa>-1$. Should I square the roots and then get only real values? or keep as it is?
If $\kappa > -1$ then $f(x)= x(x^2-(1+ \kappa))$ has three distinct zeros $0, \pm \sqrt{1+\kappa}$ and you can check that $f'(0) <0$ and $f'(\pm \sqrt{1+\kappa} )>0$ hence $0$ is stable and the other two are not.
If $\kappa <-1$ then $f$ has one zero at $x=0$ and $f'(0) >0$ so is unstable.
If $\kappa = -1$ then $f$ has one zero at $x=0$ but $f'(0) = 0$ so the linearisation is not conclusive. (It is not hard to show that $x=0$ is unstable, but it does not follow from lineaarisation.)
Under the assumption that you are dealing with a real system, you only care about the real equilibrium points, that is, the real zeros of $f$.