Now, $$\kappa^2-4(2-\lambda)\leq0$$ or
$$\kappa^2+4\lambda\leq8.$$
0
Bumbble Comm
On
$x^2+1 > 0$ so you can multiply the equation by the denominator. Then you get a new quadratic equation $q(x)\geq 0$, i.e. a parabola above the x-axis. Investigate when this is the case.
hint: $D \le 0$ is all you need !