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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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