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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Updated on: 10-Oct-2022

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