Question

If $a=3+2\sqrt{2}$, find $\frac{a^2+1}{a^2}$.

Answer 1

Given,

$a=3+2\sqrt{2}$.

Then,
$a^2={(3+2\sqrt{2})}^2$ $=9+8+12\sqrt{2}$
$a^2=17+12\sqrt{2}$.

$\frac{a^2+1}{a^2}$$=1+\frac{1}{a^2}$

$=1+\frac{1}{17+12\sqrt{2}}$
$=1+\frac{17-12\sqrt{2}}{(17+12\sqrt{2})(17-12\sqrt{2})}$
$=1+\frac{17-12\sqrt{2}}{{\left(17\right)}^2-{\left(12\sqrt{2}\right)}^2}$
$=1+\frac{17-12\sqrt{2}}{289-288}$
$=1+17-12\sqrt{2}$
$=18-12\sqrt{2}$
$=6(3-2\sqrt{2})$.