Eq. (\[2eq29\]) has at least one positive root $\bar{r}$ if $Q>0$. The black hole horizon $r_+$ is just the largest positive root of Eq. (\[2eq29\]).

On the other hand, if $m$ is even, one then has $s=m/2$, and $$\bar{r}^{m}\widehat{\alpha}_{0}\pm\bar{r}^{m-2}\widehat{\alpha}_{1}+\cdots
+(\pm1)^{s-1}\bar{r}^2\widehat{\alpha}_{s-1}+(\pm1)^s\widehat{\alpha}_{s}=Q\,
.$$ From the theory of polynomial, the above equation has at least one negative root and one positive root if $\widehat{\alpha}_{0}\left[(\pm1)^s\widehat{\alpha}_{s}-Q\right]<0$. Recall $\widehat{\alpha}_{0}>0$, and this condition can be always satisfied if $Q>\widehat{\alpha}_{s}$. Namely, black hole horizon exists in this case.

In summary, black hole horizon always exists provided $Q-\widehat{\alpha}_{s}>0$, even when the parameter $M$ vanishes.

Examples of exact solutions
---------------------------

To be more explicit, in this subsection, we give some simple examples of exact solutions given in (\[Fequ\]).

### The case of $m=3$, $p=2$, $D\geq 5$

In this case, Eqs. (\[lineq\]) give $$\label{c0c2r0D} c_0=\frac{1}{2}(D^2-3D-6)r_0^{-2}\, ,\quad
c_2=\frac{1}{2(D-3)(D-4)}r_0^2\, .$$ We then have $$\widehat{\alpha}_0=\frac{2(2D-9)}{3(D-3)}r_0^{-2}\, ,\quad
\widehat{\alpha}_1=\frac{2}{D-3}\, .$$ Equation (\[Fequ\]) for $F$ becomes $$\widehat{\alpha}_0+\widehat{\alpha}_1
F=\frac{M}{r^2}+\frac{Q}{r^3}\, ,$$ which has the solution $$f=\frac{1}{\widehat{\alpha}_1}\left(-M-\frac{Q}{r}+\widehat{\alpha}_0
r^2\right)\, .$$ Here we have used the fact that the $(m-2)$-dimensional constant curvature space is always Ricci flat for $m=3$, i.e., $\delta=0$. Since $\widehat{\alpha}_0$ is always positive, this solution is just a BTZ black hole deformed by the additional charge $Q$. This kind 