Question
If $\text{x}-\frac{1}{\text{x}}=-1$ find the value of $\text{x}^2+\frac{1}{\text{x}^2}.$
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Answer
We have, $\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2+\big(\frac{1}{\text{x}}\big)^2-2\times\text{x}\times\frac{1}{\text{x}}$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}-2$
$\Rightarrow(-1)^2=\text{x}^2+\frac{1}{\text{x}^2}-2$
$\big[\because\text{x}-\frac{1}{\text{x}}=-1\big]$
$\Rightarrow2+1=\text{x}^2+\frac{1}{\text{x}^2}$
$\therefore\ \text{x}^2+\frac{1}{\text{x}^2}=3.$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}-2$
$\Rightarrow(-1)^2=\text{x}^2+\frac{1}{\text{x}^2}-2$
$\big[\because\text{x}-\frac{1}{\text{x}}=-1\big]$
$\Rightarrow2+1=\text{x}^2+\frac{1}{\text{x}^2}$
$\therefore\ \text{x}^2+\frac{1}{\text{x}^2}=3.$
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