Problem: Find all real numbers $x$ for which
there exists a triangle $ABC$ with circumradius $2$,
such that $\angle ABC \ge 90\dg$, and
\[ x^4 + ax^3 + bx^2 + cx + 1 = 0 \]
where $a = BC$, $b = CA$, $c = AB$.

Solution: The answer is $x = -\half (\sqrt6 \pm \sqrt 2)$.

We prove this the only possible answer.
Evidently $x < 0$.
Now, note that
\[ a^2+c^2 \le b^2 \le 4b \]
since $b \le 4$ (the diameter of its circumcircle).
Then,
\begin{align*}
  0 &= x^4 + ax^3 + bx^2 + cx + 1 \\
  &= x^2 \left[ \left( x + \half a \right)^2
    + \left( \frac1x + \half c \right)^2
    + \left( b - \frac{a^2+c^2}{4} \right) \right] \\
  &\ge 0+0+0 = 0.
\end{align*}
In order for equality to hold,
we must have $x = -\half a$, $1/x = -\half c$,
and $a^2+c^2 = b^2 = 4b$.
This gives us $b = 4$, $ac = 4$, $a^2+c^2=16$.
Solving for $a,c > 0$ implies
\[ \{a,c\} = \left\{ \sqrt6 \pm \sqrt 2 \right\}.  \]
This gives the $x$ values claimed above;
by taking $a$, $b$, $c$ as deduced here,
we find they work too.

\begin{remark*}
  Note that by perturbing $\triangle ABC$ slightly,
  we see \emph{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.
\end{remark*}