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Answer 1

Given


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

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


Solution :-


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

Add 2x²(1/x²) on both sides.

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

Now, x⁴ + 1/x⁴ + 2x²(1/x)² = (x² + 1/x²)²

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

(since, x² × 1/x² = 1)

=>
$( {x}^{2} + \frac{1}{ {x}^{2} } ) {}^{2} = 4 \\ \\ = > {x}^{2} + \frac{1}{ {x}^{2} } = \sqrt{4} \\ \\ = > {x}^{2} + \frac{1}{ {x}^{2} } = 2$

Repeat the steps again

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

=>
$(x + \frac{1}{x} ) {}^{2} = 2 + 2 \\ \\ = > (x + \frac{1}{x} ) {}^{2} = 4 \\ \\ = > x + \frac{1}{x} = \sqrt{4} \\ \\ = > x + \frac{1}{x} = 2$


Answer :- 2