Question

If $m=\frac{1}{3-\sqrt{8}}$ and $n=\frac{1}{3+\sqrt{8}}$.

If $m^2$ is equal to $17+6\sqrt{n}$, then value of $n$ is?

Answer 1

Given, $m=\frac{1}{3-\sqrt{8}}$ and $n=\frac{1}{3+\sqrt{8}}$.

We rationalize the denominator,
Therefore, $m=\frac{1}{3-\sqrt{8}}\times \frac{3+\sqrt{8}}{3+\sqrt{8}}$
$\Rightarrow m=\frac{3+\sqrt{8}}{9-8}$
$\Rightarrow m=\frac{3+\sqrt{8}}{1}$.
Therefore, $m^2=(3+\sqrt{8})(3+\sqrt{8})$
$=9+8+6\sqrt{8}$
$=17+6\sqrt{8}$.

Comparing this with $17+6\sqrt{n}$, we get,

the value of $n=8$.