Question

If $x=9-4\sqrt{5}$, find the value of $x^2+\frac{1}{x^2}$

Answer 1

Given: $x=9-4\sqrt{5}\dots \dots \left(i\right)$

Taking reciprocal, we have

$\Rightarrow \frac{1}{x}=\frac{1}{9-4\sqrt{5}}$

$\Rightarrow \frac{1}{x}=\frac{[1\times (9+4\sqrt{5}\left)\right]}{\left[\right(9-4\sqrt{5})\times (9+4\sqrt{5}\left)\right]}$ [On rationalising]

$\Rightarrow \frac{1}{x}=\frac{9+4\sqrt{5}}{(81-80)}$ [$\because (a+b)(a-b)=a^2-b^2$]

$\Rightarrow \frac{1}{x}=9+4\sqrt{5}\dots \dots \left(ii\right)$

We have to find the value of

$x^2+\frac{1}{x^2}$

$=(9-4\sqrt{5}{)}^2+(9+4\sqrt{5}{)}^2$ [Using equation $\left(i\right)$ and $\left(ii\right)$]

$=[9^2+(4\sqrt{5}{)}^2-2\times 9\times 4\sqrt{5}]+[9^2+(4\sqrt{5}{)}^2+2\times 9\times 4\sqrt{5})$ [$\because (a\pm b{)}^2=a^2+b^2-\pm 2ab$ ]

$=[81+80-72\sqrt{5}]+[81+80+72\sqrt{5}]$

$=161-72\sqrt{5}+161+72\sqrt{5}=322$