Question

If x^+1/x^=66 then find x-1/x​

Answer 1

Given -  x^{2} +  \frac{1}{ x^{2} } = 66

We have the identity,

x^{2} +  \frac{1}{ x^{2} } =   (x - \frac{1}{x})^{2} + 2

(Derived from the original identity,  

(a - b)^{2} =  a^{2} +  b^{2}  - 2ab )

By putting values,

(x - \frac{1}{x})^{2} + 2 = 66

(x- \frac{1}{x}) ^{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:

$\bf \underline{Solution-}$

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

$\bf \underline{To find-}$

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

$\bf \underline{Solution-}$

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

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

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

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