Question

If x minus 1 by x is equal to 5 then find x to the power of 2 minus 1 by x to the power of 2 and x to the power of 4 minus 1 by x to the power of 4

Answer 1

Given that:

x - \frac{1}{x} = 7 \: \: \: \: \: \: \: \: ...(1)

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To find:

{x}^{2} + { (\frac{1}{x} )}^{2} \: and \: {x}^{4} + ( { \frac{1}{x} )}^{4}

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Solution:

({x - \frac{1}{x} )}^{2} = {x}^{2} +( { \frac{1}{x}) }^{2} - 2x \times \frac{1}{x} \\ \\ = > due \: to \: (1) \\ \\ = > {7}^{2} = {x}^{2} +( { \frac{1}{x}) }^{2} - 2 \\ \\ = > 49 + 2 = {x}^{2} +( { \frac{1}{x}) }^{2} \\ \\ = > 51 = {x}^{2} +( { \frac{1}{x}) }^{2}

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{ ( {x}^{2} +( { \frac{1}{x}) }^{2} )}^{2} = {x}^{4} + {( \frac{1}{x} )}^{4} + 2 {x}^{2} \times ( { \frac{1}{x} })^{2} \\ \\ = > due \: to \: (2) \\ \\ = > {51}^{2} = {x}^{4} + {( \frac{1}{x} )}^{4} + 2 \\ \\ = > 2601 - 2 = {x}^{4} + {( \frac{1}{x} )}^{4} \\ \\ = > 2599 = {x}^{4} + {( \frac{1}{x} )}^{4}

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