Question

If x=9-4√5, find the value of x²+1/x²

Answer 1

$x = 9 - 4 \sqrt{5} \\ \\ \frac{1}{x} = \frac{1}{9 - 4 \sqrt{5} } \\ \\ \frac{1}{x} = \frac{1}{9 - 4 \sqrt{5} } \times \frac{9 + 4 \sqrt{5} }{9 + 4 \sqrt{5} } \\ \\ \frac{1}{x} = \frac{9 + 4 \sqrt{5} }{ {(9)}^{2} - {(4 \sqrt{5} )}^{2} } \\ \\ \frac{1}{x} = \frac{9 + 4 \sqrt{5} }{81 - 80} \\ \\ \frac{1}{x} = 9 + 4 \sqrt{5}$

Now,

${x}^{2} + \frac{1}{ { x}^{2} } = {(9 - 4 \sqrt{5}) }^{2} + {(9 + 4 \sqrt{5} )}^{2} \\ \\ = 81 + 80 - 72 \sqrt{5} + 81 + 80 + 72 \sqrt{5} \\ \\ = 81 + 80 + 81 + 80 \\ \\ = 161 + 161 \\ \\ = 322$

Answer 2

Answer:

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

Step-by-step explanation:

Given : If x=9-4√5

To find : The value of $x^2+\frac{1}{x^2}$

Solution :

First we find, $\frac{1}{x}$

Substitute $x = 9 -4 \sqrt{5}$

$\frac{1}{x} =\frac{1}{9 - 4 \sqrt{5}}\\ \frac{1}{x}=\frac{1}{9- 4\sqrt{5}}\times\frac{9+4\sqrt{5}}{9+4\sqrt{5}}\\\\ \frac{1}{x}=\frac{9+ 4\sqrt{5}}{{(9)}^{2}-{(4\sqrt{5} )}^{2}} \\\\ \frac{1}{x} = \frac{9+ 4\sqrt{5}}{81 - 80}\\\\ \frac{1}{x}= 9 + 4\sqrt{5}$

Now, substitute in $x^2+\frac{1}{x^2}$

${x}^{2}+\frac{1}{ { x}^{2}}={(9 - 4\sqrt{5})}^{2}+{(9 + 4\sqrt{5} )}^{2}\\ \\= 81 + 80 - 72 \sqrt{5}+81 + 80 + 72\sqrt{5}\\= 81 + 80 + 81 + 80\\= 161 +161\\= 322$

Therefore, $x^2+\frac{1}{x^2}=322$