The biggest and smallest integer solution of $\sqrt{(5+2\sqrt6)^{2x}}+\sqrt{(5-2\sqrt6)^{2x}}\le98$ are? $$\sqrt{(5+2\sqrt6)^{2x}}+\sqrt{(5-2\sqrt6)^{2x}}\le98$$
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
$$-2 \le x \le 2$$
