Question

For $n=\frac{1}{3+\sqrt{8}}$, if $n^2$ is $17-6\sqrt{m}$ then $m$ is:

• A. $7$
• B. $8$
• C. $4$
• D. $5$

Answer 1

The correct option is B $8$

By rationalising the denominator, we get

$n=\frac{1}{3+\sqrt{8}}\times \frac{3-\sqrt{8}}{3-\sqrt{8}}$

$n=\frac{3-\sqrt{8}}{9-8}$

$n=\frac{3-\sqrt{8}}{1}$

$n^2=(3-\sqrt{8})(3-\sqrt{8})$

$=9+8-6\sqrt{8}$

$=17-6\sqrt{8}$

Hence, $m=8$.