Question

$x-\frac{1}{x}=8 \: then \: find \: (x^{2}+\frac{1}{x^{2}}) \: and \: (x^{4}+\frac{1}{x^{4}})$
pls solve
urgent​

Answer 1

Question :

$\sf \small \: x-\frac{1}{x}=8 \: then \: find \: (x^{2}+\frac{1}{x^{2}}) \: and \: (x^{4}+\frac{1}{x^{4}})$

To find :

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

Solution :

$\sf \small \: i) \: ( {x}^{2} + \frac{1}{ {x}^{2} } )$

$\sf \small \rightarrow(x - \frac{1}{x} )^{2} = {x}^{2} + \frac{1}{ {x}^{2} } - 2$

$\sf \small \rightarrow {(8)}^{2} = {x}^{2} + \frac{1}{ {x}^{2} } - 2$

$\sf \small \rightarrow {x}^{2} + \frac{1}{ {x}^{2} } = 64 + 2$

$\boxed{ \sf \small \rightarrow {x}^{2} + \frac{1}{ {x}^{2} } = 66}$

So now we know the value of x² + 1/x² i.e. 66

$\sf \small \: ii) ( {x}^{4} + \frac{1}{ {x}^{4} } )$

$\sf \small \rightarrow( {x}^{2} + \frac{1}{ {x}^{2} } ) ^{2} = {(66)}^{2}$

$\sf \small \rightarrow {x}^{4} + \frac{1}{ {x}^{4} } - 2 = {(66)}^{2}$

$\sf \small \rightarrow {x}^{4} + \frac{1}{ {x}^{4} } = 4356 + 2$

$\boxed{ \sf \small \rightarrow {x}^{4} + \frac{1}{ {x}^{4} } = 4358 }$

Final Answer :

• The we got the value of x² + 1/x² is 66

• The value of x⁴ + 1/x⁴ is 4358