Question

If (x - 1/ x ) =2 find the value of i) (x + 1/x ) ii) ( x2 + 1x/2 ) iii) ( x4 + 1x/4 )​

Answer 1

Answer:

$Given \: \: \: \: \: \: \: x - \frac{1}{x} = 2 \\ {(x - \frac{1}{x}) }^{2} = {x}^{2} + \frac{1}{ {x}^{2} } - 2 \\ = > {2}^{2} = {x}^{2} + \frac{1}{ {x}^{2} } - 2 \\ = > 4 = {x }^{2} + \frac{1}{{x}^{2} } - 2 \\ = > {x}^{2} + \frac{1}{ {x}^{2} } = 4 + 2 \\ = > {x}^{2} + \frac{1}{ {x}^{2} } = 6 \\ so , \\ (x + \frac{1}{x} )^{2} = {x}^{2} + \frac{1}{ {x}^{2} } + 2 \\ = > {(x + \frac{1}{x}) }^{2} = 6 + 2 \\ = > {(x + \frac{1}{x} )}^{2} = 8 \\ = > x + \frac{1}{x} = \sqrt{8}$

${( {x}^{2} + \frac{1}{ {x}^{2} }) }^{2} = {x}^{4} + \frac{1}{ {x}^{4} } + 2 \\ = > {(6)}^{2} = {x}^{4} + \frac{1}{ {x}^{4} } + 2 \\ = > 36 = {x}^{4} + \frac{1}{ {x}^{4} } + 2 \\ = > {x}^{4} + \frac{1}{ {x}^{4} } = 36 - 2 \\ = > {x}^{4} + \frac{1}{ {x}^{4} } = 34$

x + 1/x = √8

x^2 + 1/x^2 = 6

x^4 + 1/x^4 = 34