Stuck with solving inequality to find the product of highest and lowest integer solutions The inequality in question is
$$\sqrt{(5+2\sqrt{6})^{2x}}+\sqrt{(5-2\sqrt{6})^{2x}}\leq98$$
This time our job is to find the product of highest and lowest integer solutions.

My attempt
$$\sqrt{(5+2\sqrt{6})^{2x}}+\sqrt{(5-2\sqrt{6})^{2x}}\leq98$$
$$\sqrt{((5+2\sqrt{6})^{x})^2}+\sqrt{((5-2\sqrt{6})^{x})^2}\leq98$$
$$(5+2\sqrt{6})^x+(5-2\sqrt{6})^x\leq98$$
What next?

The solution is $-4$.

As $(5+2\sqrt6)(5-2\sqrt6)=1$
if $(5+2\sqrt6)^x=a>0$
$$a+\dfrac1a\le98\iff a^2-98a+1\le0$$
$a^2-98a+1=0\implies a=\dfrac{98\pm40\sqrt6}2=49\pm20\sqrt6=(5\pm2\sqrt6)^2$
$\implies(5+2\sqrt6)^{-2}=(5-2\sqrt6)^2\le a\le(5+2\sqrt6)^2 $
$\implies-2\le x\le2$
