Question

if x ^2 + 1/ x square equal to 66 then find the value of x minus 1 by X

Answer 1

${x}^{2} + \frac{1}{ {x}^{2} } = 66 \\ {x}^{2} + \frac{1}{ {x}^{2} } - 2 = 66 - 2 \\ ( {x - \frac{1}{x}) }^{2} = 64 \\ (x - \frac{1}{x} ) = \sqrt{64} \\ (x - \frac{1}{x} ) = 8$

Answer 2

Step-by-step explanation:

$\bf \underline{Solution-}$

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

$\bf \underline{To find-}$

$\sf{the \: value \: of :\: x - \frac{1}{x} = \: ?}$

$\bf \underline{Solution-}$

$\sf {\bigg(x - \frac{1}{x} \bigg) ^{2} = {x}^{2} + \frac{1}{ {x}^{2} } - 2 \: \: \: \: \: \: [ \because \: (a - b {)}^{2} = {a}^{2} + {b}^{2} - 2ab ]} \\ \\ = 66 - 2 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\rm{ [ \because {x}^{2} + \frac{1}{ {x}^{2}} = 66 \:(Given) ] }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:= 64$

$\sf{ \therefore \: \: \: \: \: \: x - \frac{1}{x} = \sqrt{64} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = ± \: 8$

$\bf\underline{Hence,the \: value \: of : \: x - \frac{1}{x} \: is \: ± \: 8.}$