Question

Solve $(5-2\sqrt{6}{)}^x+(5+2\sqrt{6}{)}^x=98$

Answer 1

$(5-2\sqrt{6}{)}^x+(5+2\sqrt{6}{)}^x=98$

Let $A=5-2\sqrt{6}$ then $5+2\sqrt{6}=\frac{1}{A}$

Therefore $A^x+\frac{1}{A^x}=98$

$\Rightarrow A^{2x}-98A^x+1=0$ ……….$\left(1\right)$

Let $A^x=z$, then equation (i) changes to

$z^2-98z+1=0$

$\therefore z=\frac{98\pm \sqrt{9600-4}}{2}=\frac{98\pm \sqrt{9600}}{2}$

$=\frac{98\pm 10\sqrt{96}}{2}=\frac{98\pm 40\sqrt{6}}{2}$

$=49\pm 20\sqrt{6}$

Now $A^x=(5-2\sqrt{6}{)}^x=(49-20\sqrt{6})$

$\Rightarrow x=2$.