## 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 $\inline 5 - 2\sqrt{6}$

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

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

## Thursday, July 15, 2010

1. Solve the following equation for
$\fn_jvn \small \left ( 5+2\sqrt{6} \right )^{x^2-3}+\left ( 5-2\sqrt{6} \right )^{x^2-3}=10$

Hi All,
In this forum I would like to post math questions of at the senior school level. The solutions would be discussed subsequently. Many of these questions have been asked in engineering entrance exams in India.

Lok