8. Solve the equation $2 \sqrt{x^{2}-121}+11 \sqrt{x^{2}-4}=7 \sqrt{3} x$.
8. Draw a circle with diameter $AB = x (x \geqslant 11)$, and take two points $C, D$ on either side of $B$ such that $BC = 2, BD = 11$. Thus, $AC = \sqrt{x^2 - 4}, AD = \sqrt{x^2 - 121}$. Applying Ptolemy's theorem to quadrilateral $ABCD$, we have $CD \cdot x = 2 \cdot \sqrt{x^2 - 121} + 11 \cdot \sqrt{x^2 - 4}$. Comparing this with the original equation, we get $CD = 7\sqrt{3}$. In $\triangle BCD$, by the cosine rule, we have $\cos \angle CBD = \frac{2^2 + 11^2 - (7\sqrt{3})^2}{2 \cdot 2 \cdot 11} = -\frac{1}{2}$, so $\angle CBD = 120^\circ$. Therefore, $x = AB = \frac{CD}{\sin 120^\circ} = 14$ is the solution.