<s>[INST] 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 $-4$.
 [/INST] We have
$$
(a+b)^2+(a-b)^2= 2a^2+2b^2 = 98
$$
with $a = 5$ and $b = 2\sqrt 6$
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