8. [20] Find the positive real number(s) $x$ such that $\frac{1}{2}\left(3 x^{2}-1\right)=\left(x^{2}-50 x-10\right)\left(x^{2}+25 x+5\right)$.
Answer: $\mathbf{2 5 + 2 \sqrt { 1 5 9 }}$. Write $a=x^{2}-50 x-10$ and $b=x^{2}+25 x+5$; the given becomes $\frac{a+2 b-1}{2}=a b$, so $0=2 a b-a-2 b+1=(a-1)(2 b-1)$. Then $a-1=x^{2}-50 x-11=0$ or $2 b-1=2 x^{2}+50 x+9=0$. The former has a positive root, $x=25+2 \sqrt{159}$, while the latter cannot, for obvious reasons.