Question

If $(5+2\sqrt{6}{)}^{x^2-3}+(5-2\sqrt{6}{)}^{x^2-3}=10$, find x.

Answer 1

$\to$ As $5+2\sqrt{6}=\sqrt{24}+\sqrt{25}$ and

$(\sqrt{25}+\sqrt{24}{)}^{\lambda }(\sqrt{25}-\sqrt{24}{)}^{\lambda }=1$

let $(\sqrt{25}+\sqrt{24}{)}^{x^2-3}=\alpha$

$\Rightarrow$ Expression : $\alpha +\frac{1}{\alpha }=10$

$\Rightarrow {\alpha }^2-10\alpha +1=0$

$\Rightarrow \alpha =\frac{10\pm \sqrt{100-4}}{2}=5\pm \sqrt{24}$

$\Rightarrow (\sqrt{25}+\sqrt{24}{)}^{x^2-3}=(\sqrt{25}\pm \sqrt{24})$

$\Rightarrow x^2-3=\pm 1\Rightarrow x^2=4,2\Rightarrow x=\pm \sqrt{2},2$