Here is an **observation**, I don't know yet if it has consequences :    
The unitary highest weight representations of the Virasoro algebra are completely given by the pair $(c,h)$. The number $c$ is called the **central charge**.  
If $0 \le c < 1$, then the FQS criterion says that $c = c_{m} = 1 - \frac{6}{m(m+1)}$ with $m = 2,3,4,5,...$  
Now, forget for a moment that $0 \le c < 1$ and just consider the equation : $c=c(x) = 1 - \frac{6}{x(x+1)}$    
Then : $(c-1)x(x+1)+6=0$  
And so : $(c-1)x^{2} + (c-1)x + 6  = 0$    
But: $\Delta = (c-1)^{2}-24(c-1) = (c-1)(c-25)$   

> **Conclusion** : $\Delta = 0$ iff $c=1$ or  $25$    

$\rightarrow$ I guess it's coherent with the answer of Scott on the negative highest weight representations.   

**N=1 super-Virasoro case** (Neveu-Schwarz and Ramond algebras) :    
In this case, the discrete series central charge is given by : $c = c_{m} = \frac{3}{2}(1 - \frac{8}{m(m+2)})$  
Let the equation $c=c(x) = \frac{3}{2}(1 - \frac{8}{x(x+2)})$  
Then : $(\frac{2}{3}c-1)x^{2} + 2(\frac{2}{3}c-1)x + 8 = 0$  
And: $\Delta = 4(\frac{2}{3}c-1)^{2}-32(\frac{2}{3}c-1) = 4(\frac{2}{3}c-1)(\frac{2}{3}c-9)$   

> **Conclusion** : $\Delta=0$ iff  $c=\frac{3}{2}$ or $\frac{27}{2}$