Question

if x minus one by x is equal to 7 then find the value of x square + 1 by x square and x to the power 4 + 1 by x to the power 4

Answer 1

I hope this helps you

Answer 2

$\textbf{ANSWER:}$

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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That's your answer