Question

If
$x = 3 - \sqrt[2]{2}$
,find
${x}^{2} + 1 \div {x}^{2}$
​

Answer 1

Answer:

$\frac{193 - 132\sqrt{2} }{11-6\sqrt{2} }$

Step-by-step explanation:

$x = 3 - \sqrt{2}$

To find, $x^{2} + \frac{1}{x^{2} }$

$x^{2} = (3-\sqrt{2})^{2}\\x^{2} = 9 + 2 - 6\sqrt{2} \\x^{2} = 11 - 6\sqrt{2}$

Now,

$\frac{1}{x^{2} } = \frac{1}{11 - 6\sqrt{2} }$

So, $x^{2} + \frac{1}{x^{2} }$ = $11-6\sqrt{2} + \frac{1}{11-6\sqrt{2} }$

= $\frac{(11-6\sqrt{2})^{2}+1 }{11-6\sqrt{2} }$

= $\frac{121 + 72 - 132\sqrt{2} }{11- 6\sqrt{2} }$

= $\frac{193 - 132\sqrt{2} }{11-6\sqrt{2} }$

Hope this helped!

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