= Solution
At a horizon, use $Q^2=2Mr_0^2-r_0^4$ in $V'(r)=4M/r^3-4Q^2/r^5$ to obtain $V'(r_0)=4(r_0^2-M)/r_0^3$. A double root therefore requires $r_0^2=M$, and the horizon equation then gives $Q^2=M^2$. Conversely at $|Q|=M$,
$$
V(r)=\left(1-\frac M{r^2}\right)^2,\qquad V'(\sqrt M)=0,\qquad V''(\sqrt M)=\frac8M\ne0.
$$
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 $\kappa_\pm=\pm2\sqrt{M^2-Q^2}/r_\pm^3$, so neither is degenerate. The neutral positive-radius horizon is also simple.
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