Question

if x = 3 + √8, then find the value of
x²+1/x²

solve it ...​

Answer 1

Answer:

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Answer 2

Given:

$x = 3 + \sqrt{8}$

To find:

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

Solution:

$x = 3 + \sqrt{8} \\ \\ \frac{1}{x} = \frac{1}{3 + \sqrt{8} } \\ \\ rationalizing \: the \: denominator \\ \\ = > \frac{1}{3 + \sqrt{8}} \frac{ \times \: (3 - \sqrt{8}) }{ \times \: (3 - \sqrt{8}) } \\ \\ using \: identity \: ( {a }^{2} - {b}^{2}) = (a + b)(a - b) \\ = > \frac{3 - \sqrt{8} }{ {3}^{2} - ( \sqrt{8}) {}^{2} } \\ \\ = > \frac{3 - \sqrt{8} }{9 - 8} \\ \\ = > \frac{1}{x} = 3 - \sqrt{8}$

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

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

$= > {x}^{2} + \frac{1}{ {x}^{2} } = (x + \frac{1}{x} ) {}^{2} - 2 \: \: - - (1)$

$The \: value \: of \: \\ {x} + \frac{1}{x} \\ \\ sustituting \: values \\ = 3 + \sqrt{8} + (3 - \sqrt{8} ) \: \\ \\ = > x + \frac{1}{x} = 6 \: \: \: \:</u><u>----</u><u> (2)$

We have,

${x}^{2} + \frac{1}{ {x}^{2} } = ( {x} + \frac{1}{x} ) {}^{2} - 2 \: \:$

$sustituting \: values \: from \: \: 2 \: in \: \: 1$

We get,

${x}^{2} + \frac{1}{ {x}^{2} } = ( {6})^{2} - 2 \\ \\ = > 36 - 2 \\ \\ = > 34$

Hence, 34 is the answer.

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