Question

$x ^{2} - \frac{1}{x ^{2} } = 2find \sqrt{x} + \sqrt \frac{1}{x}$
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Answer 1

Answer:

${x} - \frac{1}{x} = \sqrt{ {x}^{2} + \frac{1}{ {x}^{2} } - 2} \\ \\x + \frac{1}{x} = \sqrt{ {x}^{2} + \frac{1}{ {x}^{2} } + 2 } \\ \\ = > multiply \: \: them - \\ \\ {x}^{2} - \frac{1}{ {x}^{2} } = \sqrt{ {( {x}^{2} + \frac{1}{ {x}^{2} } )}^{2} - 4} \\ \\ = > 2 = \sqrt{ {( {x}^{2} + \frac{1}{ {x}^{2} } )}^{2} - 4} \\ \\ = > 8 = {( {x}^{2} + \frac{1}{ {x}^{2} } )}^{2} \\ \\ = > {x}^{2} + \frac{1}{ {x}^{2} } = 2 \sqrt{2} \\ \\ but - \\ \\ = > {(x + \frac{1}{x}) }^{2} = {x}^{2} + \frac{1}{ {x}^{2} } + 2 \\ \\ = > {(x + \frac{1}{x} })^{2} = 2 \sqrt{2} + 2 \\ \\ = > x + \frac{1}{x} = \sqrt{2 \sqrt{2} + 2 } \\ \\ = > {( \sqrt{x} + \frac{1}{ \sqrt{x} } ) }^{2} = x + \frac{1}{x} + 2 \\ \\ = > { (\sqrt{x} + \frac{1}{ \sqrt{x} }) }^{2} = \sqrt{2 \sqrt{2} + 2} + 2 \\ \\ = > \sqrt{x} + \frac{1}{ \sqrt{x} } = \sqrt{ \sqrt{2 \sqrt{2} + 2 } + 2}$

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