 Solution

At a horizon, use $Q^2=2M{r}_0^2-r_0^4$ in $V^{\prime }(r)=4M/r^3-4Q^2/r^5$ to obtain $V^{\prime }(r_0)=4(r_0^2-M)/r_0^3$. A double root therefore requires $r_0^2=M$, and the horizon equation then gives $Q^2=M^2$. Conversely at $|Q|=M$,Hence $\boxed{|Q|=M\text{ gives a degenerate horizon at }{r}_h=\sqrt{M}}$. It is a degenerate Killing horizon of an extremal black hole. For distinct charged roots, the normalized signed surface gravities are ${\kappa }_{\pm }=\pm 2\sqrt{M^2-Q^2}/r_{\pm }^3$, so neither is degenerate. The neutral positive-radius horizon is also simple.