Question

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Answer 1

Required answer :-

Question :

•$If \: (x + \frac{1}{x}) \: = \: \sqrt{7} \: then \: find \: the \: values \: of \: {x}^{2} + \frac{1}{x {}^{2} } \: and \: x {}^{4} + \frac{1}{x {}^{4} }$

Solution :

Given,

•$(x + \frac{1}{x}) \: = \: \sqrt{7}$

To find,

• $values \: of \: {x}^{2} + \frac{1}{x {}^{2} } \: and \: x {}^{4} + \frac{1}{x {}^{4} }$

Formula used :

•. (a+b)² = a² + b² + 2ab

Step by step explaination : -

According the question we have to square both the terms -

$x {}^{2} + \frac{1}{x {}^{2} } \: + \: 2x \: \times \: \frac{1}{x} = 7 \:$

$x {}^{2} + \frac{1}{x {}^{2} } + 2 \: = \: 7 \: \\ x {}^{2} + \frac{1}{x {}^{2} } = 5$

$Hence, \: value \: of \: {x}^{2} + \frac{1}{x {}^{2} } \: is \: 5$

Now again we have to square both for calculating the value of second term.

$(x {}^{2} + \frac{1}{x {}^{2} } ) {}^{2} = 5$

$x {}^{4} + \frac{1}{x {}^{4} } + 2x {}^{2} \times \frac{1}{x {}^{2} } = 25$

$x {}^{4} + \frac{1}{x {}^{4} } + 2 = 25$

$x {}^{4} + \frac{1}{x {}^{4} } = 23$

$Hence, \: value \: of \: x {}^{4} + \frac{1}{x {}^{4} } \: is \: 23$

Answer 2

answer is

value of x⁴ +¼ is 23