find the value of

$x=9-4 \sqrt{5}$        .......(1)

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

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

$\Rightarrow \frac{1}{x}=\frac{9+4 \sqrt{5}}{9^{2}-(4 \sqrt{5})^{2}}$

$\Rightarrow \frac{1}{x}=\frac{9+4 \sqrt{5}}{81-80}$

$\Rightarrow \frac{1}{x}=9+4 \sqrt{5} \quad \ldots(2)$

Adding (1) and (2), we get

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

$\Rightarrow x+\frac{1}{x}=18$

Squaring on both sides, we get

$\left(x+\frac{1}{x}\right)^{2}=18^{2}$

$\Rightarrow x^{2}+\frac{1}{x^{2}}+2 \times x \times \frac{1}{x}=324$

$\Rightarrow x^{2}+\frac{1}{x^{2}}=324-2=322$

Thus, the value of $x^{2}+\frac{1}{x^{2}}$ is 322 .