find the value of


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



$x=9-4 \sqrt{5}$        .......(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


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

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

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

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now