Question

(x-1/x) =4 than solve((x)2+(1/(x)2) and((x)4+(1/(x)4)​

Answer 1

[tex]\mathfrak{\huge{Answer:-} } \\x-\frac{1}{x}=4\\Squaring\:on\:both\:the\:sides\\{(x-\frac{1}{x})}^{2}={(4)}^{2}\\{(x)}^{2}+{(\frac{1}{x})}^{2}-2(x)(\frac{1}{x})={(4)}^{2}\\ {x}^{2} + \frac{1}{ {x}^{2} } - 2 = 16 \\ {x}^{2} + \frac{1}{ {x}^{2} } = 16 + 2 \\ \bold{ \large{ \boxed{ {x}^{2} + \frac{1}{ {x}^{2} } = 18 } } } \\Squaring\:on\:both\:the\:sides\\{( {x}^{2} + \frac{1}{ {x}^{2} })}^{2}={(18)}^{2}\\{( {x}^{2} )}^{2}+{(\frac{1}{ {x}^{2} })}^{2} + 2( {x}^{2} )(\frac{1}{ {x}^{2} })={(18)}^{2}\\ {x}^{4} + \frac{1}{ {x}^{4} } + 2 = 324 \\ {x}^{2} + \frac{1}{ {x}^{2} } = 324 - 2 \\ \bold{ \large{ \boxed{ {x}^{4} + \frac{1}{ {x}^{4} } = 322} } } [/tex]

Answer 2

Answer:

$\frac{x - 1}{x} = 4 \\ x - 1 = 4x \\ - 1 = 4x - x \\ - 1 = 3x \\ \large{ \boxed{ \underline{ \underline{ \frac{ - 1}{3} }} = x}} \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } \\ { (\frac{ - 1}{3}) }^{2} + \frac{1}{ { (\frac{ - 1}{3} )}^{2} } \\ = \frac{1}{9} + \frac{1}{ \frac{1}{9} } \\ = \frac{1}{9} + (1 \times \frac{9}{1} ) \\ = \frac{1}{9} + 9 \\ = \frac{1 + 81}{9} \\ = \frac{82}{9} \\ \large{ \boxed{= \underline{ \underline{9 \frac{1}{9} }}}} \\ \\ {x}^{4} + \frac{1}{ {x}^{4} } \\ = { (\frac{ - 1}{3}) }^{4} + \frac{1}{ {( \frac{ - 1}{3} )}^{4} } \\ = \frac{1}{81} + \frac{1}{ \frac{1}{81} } \\ = \frac{1}{81} + (1 \times \frac{81}{1} ) \\ = \frac{1}{81} + 81 \\ = \frac{1 + 6561}{81} \\ = \frac{6562}{81} \\ \large{ \boxed{= \underline{ \underline{ 81 \frac{1}{81} }}}}$