Question

please solve this question​

Answer 1

Answer:

$x = 5 - 2 \sqrt{6} \\ \frac{1}{x} = \frac{1}{x} = \frac{1}{5 - 2 \sqrt{6} \times \frac{5 + 2 \sqrt{6} }{5 + 2 \sqrt{6} } = 5 + 2 \sqrt{6} \\ {x}^{2} + { \frac{1}{x} }^{2} = {5 - 2 \sqrt{6} }^{2} + {5 + 2 \sqrt{6} }^{2} \\ = (25 + {2 \sqrt{6} }^{2} - 20 \sqrt{6} + {2 \sqrt{6} }^{2} ) + (25 + 20 \sqrt{6} + {2 \sqrt{6} }^{2} \\ = 50 + 48 \\ = 98 \\ answer$

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Answer 2

Answer:

98

Step-by-step explanation:

Given:

$x = 5 - 2 \sqrt{2}$

To find:

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

Putting the value of x

${5 - 2 \sqrt{6} }^{2} + \frac{1}{? \\ {(5 - 2 \sqrt{6} }^{2} }$

$25 + 24 - 2(5 \times 2 \sqrt{6} ) + \frac{1}{25 + 24 - 2(5 \times \sqrt{6} }$

$25 + 24 - 2(5 \times 2 \sqrt{6} + \frac{1}{49 - 2(5 \times 2 \sqrt{6} }$

Rationalising the second term

$25 + 24 - 2(5 \times 2 \sqrt{6} + \frac{1}{49 - 20 \sqrt{6} } \times \frac{49 + 20 \sqrt{6} }{49 + 20 \sqrt{6}$

$25 + 24 - 20 \sqrt{6} + \frac{49 + 20 \sqrt{6} }{ {49}^{2} - {20 \sqrt{6} }^{2} }$

$49 - 20 \sqrt{6} + \frac{49 + 20 \sqrt{6} }{1} \\ elimimate \: the \: opposites \\ 49 + 49 = 98$