Find all real numbers $x$ for which there exists a triangle $A B C$ with circumradius 2 , such that $\angle A B C \geq 90^{\circ}$, and  
$$ x^{4}+a x^{3}+b x^{2}+c x+1=0 $$  
where $a=B C, b=C A, c=A B$.
The answer is $x=-\frac{1}{2}(\sqrt{6} \pm \sqrt{2})$. We prove this the only possible answer. Evidently $x0$ implies 
$$ \{a, c\}=\{\sqrt{6} \pm \sqrt{2}\} $$ 
This gives the $x$ values claimed above; by taking $a, b, c$ as deduced here, we find they work too. 
Remark. Note that by perturbing $\triangle A B C$ slightly, we see a priori that the set of possible $x$ should consist of unions of intervals (possibly trivial). So it makes sense to try inequalities no matter what.