Question

If X square + 1 upon x square is equal to 2 then x ki power 8 + 1 upon x square is equal to

Answer 1

Answer:

❑ CORRECT QUESTIONS :-

$\rm \: If \: {x}^{2} + \frac{1}{ {x}^{2} } = 2 \: , \: \: then \: \: the \: value \: \: of \: \\ \\ \rm{x}^{8} + \frac{1}{ { x }^{8} } \: \: is \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

❑ SOLUTION :-

❑ GIVEN :-

$\sf \: {x}^{2} + \frac{1}{ {x}^{2} } = 2$

$\sf \: Now \: \: squaring \: \: both \: \: sides \: \: we \: get -$

$\implies \sf \: \bigg( {x}^{2} + \frac{1}{ {x}^{2} } \bigg)^{2} = 2^{2}$

$\implies \sf \: ( {x}^{2})^{2} \: + \: 2( {x}^{2} ). \: \frac{1}{( {x}^{2}) } + \frac{1}{ ({x}^{2})^{2} } = 4$

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

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

$\sf \: Again \: squaring \: both \: sides -$

$\implies \sf \: \bigg( {x}^{4} + \frac{1}{ {x}^{4} } \bigg)^{2} = 2^{2}$

$\implies \sf \: ( {x}^{4})^{2} \: + \: 2( {x}^{4} ). \: \frac{1}{( {x}^{4}) } + \frac{1}{ ({x}^{4})^{2} } = 4$

$\implies \sf \: {x}^{8} + \frac{1}{ {x}^{8} } + 2 = 4$

$\implies \sf \: {x}^{8} + \frac{1}{ {x}^{8} } = 2$

$\: \: \: \: \: \: \: \boxed{ \sf \: {x}^{8} + \frac{1}{ {x}^{8} } = 2}$