$$\sqrt{(5+2\sqrt6)^{2x}}+\sqrt{(5-2\sqrt6)^{2x}}\le98$$
I noticed that $5+2\sqrt6=(\sqrt2+\sqrt3)^2$ but that hardly helps.
After cancelling the square root and the power of two, I tried multiplying the whole thing with $(5+2\sqrt6)^{x}$ and replacing $(5+2\sqrt6)^{x}=t$ would get a quadratic equation that wouldn't help me without some serious approximation which I don't trust myself enough to do.
Is there an easier way to go around this?
Firstly note that $$(5+2\sqrt{6})(5-2\sqrt{6}) = 25-24=1$$ so $(5+2\sqrt{6})^{-1}= (5-2\sqrt{6})$. Putting $t=(5+2\sqrt{6})^x$ you get the inequality $$t + \frac{1}{t} \leq 98$$ which has solution $$(5+2\sqrt{6})^{-2} = 49 - 20 \sqrt{6} \le t \le 49 + 20 \sqrt{6} = (5+2\sqrt{6})^2$$ which means exactly $$-2 \le x \le 2$$