= Solution
{c}
If $M^2=Q^2$ with $M>0$, then
$$
\boxed{F=(1-M/r)^2,\qquad r_+=r_-=M.}
$$
The <curvature singularity> remains at $r=0$. At $r=M$ the static chart has a <coordinate singularity>, removed by the <Eddington-Finkelstein coordinates> above. The <Kretschmann scalar> there equals $8/M^4$, which is finite. The coincident horizons form a <degenerate Killing horizon>, and its future outer branch is the <event horizon> of the <extremal black hole>. A double zero of $F$ does not remove the horizon.
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