If $ x^{2}+\frac{1}{x^{2}}=62, $ find the value of

1. $ x+\frac{1}{x} $
2. $ x-\frac{1}{x} $

Given:

\( x^{2}+\frac{1}{x^{2}}=62 \)
To do:

We have to find the value 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,

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

$x^{2}+\frac{1}{x^{2}}+2\times x \times \frac{1}{x}=64$

$(x+\frac{1}{x})^2=(8)^2$

This implies,

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

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

Subtracting 2 on both sides, we get,

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

$x^{2}+\frac{1}{x^{2}}-2\times x \times \frac{1}{x}=60$

$(x-\frac{1}{x})^2=(\sqrt{15\times4})^2$

$(x-\frac{1}{x})^2=(2\sqrt{15})^2$

This implies,

$x-\frac{1}{x}=2\sqrt{15}$.

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Updated on: 10-Oct-2022

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