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If $ x-\frac{1}{x}=-1 $, find the value of $ x^{2}+\frac{1}{x^{2}} $.
Given:
\( x-\frac{1}{x}=-1 \)
To do:
We have to find the value of \( x^{2}+\frac{1}{x^{2}} \).
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,
$x-\frac{1}{x}=-1$
Squaring both sides, we get,
$(x-\frac{1}{x})^{2}=(-1)^{2}$
$\Rightarrow x^{2}+\frac{1}{x^{2}}-2 \times x \times \frac{1}{x}=1$
$\Rightarrow x^{2}+\frac{1}{x^{2}}=1+2=3$
The value of \( x^{2}+\frac{1}{x^{2}} \) is 3.
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