Question

If $m=\frac{1}{3-\sqrt{8}}$ and $n=\frac{1}{3+\sqrt{8}}$, find if $n^2$ is equal to $17-6\sqrt{m}$, then value of m is?

Answer 1

We rationalize the denominator,
$n=\frac{1}{3+\sqrt{8}}\times \frac{3-\sqrt{8}}{3-\sqrt{8}}$
$n=\frac{3-\sqrt{8}}{9-8}$
$n=\frac{3-\sqrt{8}}{1}$.
$\therefore n^2=(3-\sqrt{8})(3-\sqrt{8})$
$=9+8-6\sqrt{8}$
$=17-12\sqrt{2}$

$=17-6\sqrt{8}$.

Thus the value of $m$ from the above question is $8$.