If $ x^{2}+\frac{1}{x^{2}}=62 $, find the value of
(a) $ x+\frac{1}{x} $
(b) $ x-\frac{1}{x} $

Given :

The given expression is \( x^{2}+\frac{1}{x^{2}}=62 \).

To do :

We have to find the values of

(a) \( x+\frac{1}{x} \)
(b) \( x-\frac{1}{x} \)

Solution :  

a) $x^{2}+\frac{1}{x^{2}}=62$

Adding 2 on both sides, we get,

$x^2+\frac{1}{x^2}+2=62+2$

$(x)^2+(\frac{1}{x})^2+2.x.\frac{1}{x}=64$

$(x+\frac{1}{x})^2=8^2$                      $[(a+b)^2=a^2+b^2+2ab]$

$\Rightarrow x+\frac{1}{x}=8$

b) $x^{2}+\frac{1}{x^{2}}=62$

Subtracting 2 from both sides, we get,

$x^2+\frac{1}{x^2}-2=62-2$

$(x)^2+(\frac{1}{x})^2-2.x.\frac{1}{x}=60$

$(x-\frac{1}{x})^2=(\sqrt{60})^2$                      $[(a-b)^2=a^2+b^2-2ab]$

$\Rightarrow x-\frac{1}{x}=\sqrt{60}$

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Updated on: 10-Oct-2022

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