## Wednesday, December 29, 2010

To solve this question, we observe that,

$\left ( 5+2\sqrt{6} \right )\left ( 5-2\sqrt{6} \right )=25-24 = 1$

So we can write,
$5-2\sqrt{6}=\frac{1}{5+2\sqrt{6}}$

The given equation can now be written as:

$\left ( 5+2\sqrt{6} \right )^{x^{2}-3}+\frac{1}{\left ( 5-2\sqrt{6} \right )^{x^{2}-3}}=10$
Now let's put
$\left ( 5+2\sqrt{6} \right )^{x^{2}-3}=y$

The equation then becomes
$y+\frac{1}{y}=10$

or
$y^{2}-10y+1=0$

Solving this we get,

$y=\frac{10\pm \sqrt{100-4}}{2}=\frac{10\pm 4\sqrt{6}}{2}=5 \pm 2\sqrt{6}$

But,

$y=\left ( 5+2\sqrt{6} \right )^{x^{2}-3}$

Therefore,
$\inline \left ( 5+2\sqrt{6} \right )^{x^{2}-3} = 5 + 2\sqrt{6}$ or $\inline 5 - 2\sqrt{6}$

$\inline \therefore x^{2}-3= +1 \mbox{ or} -1$

$\inline \Rightarrow x^{2}=4 \mbox{ or } x^{2}=2$