Question

If x2 + 1/x2 = 66, find the value of 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

Hope it helped! 

Answer 2

Given :

$x^2 + 1/x^2 = 66$

To find :

$x -1/x$

Step-by-step explanation:

using,

$a^2 + b^2 -2ab = (a-b)^2$

we get ;

(x - 1/x)^2 =  $x^2 + 1/x^2 -2$

               = 66 - 2 = 64

$x-1/x = \sqrt{64} = 8$

Hence, the answer is 8 .

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