If $ x^{2}+\frac{1}{x^{2}}=66 $, find the value of $ x-\frac{1}{x} $.

Given:

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

To do:

We have to find the value of \( x-\frac{1}{x} \).

Solution:

We know that,

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

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

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

Therefore,

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

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

$\Rightarrow (x-\frac{1}{x})^{2}=64$

$\Rightarrow x-\frac{1}{x}=\sqrt{64}$

$\Rightarrow x-\frac{1}{x}=\pm 8$

Hence, $x-\frac{1}{x}=\pm 8$.

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Updated on: 10-Oct-2022

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