Question
If $\text{x}^2+\frac{1}{\text{x}^2}=66,$ find the value of $\text{x}-\frac{1}{\text{x}}.$
✓
Answer
We have, $\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2+\Big(\frac{1}{\text{x}}\Big)^2-2\times\text{x}\times\frac{1}{\text{x}}$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}-2$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=66-2$
$\big[\therefore\ \text{x}^2+\frac{1}{\text{x}^2}=66\big]$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=64$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=(\pm8)^2$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)=\pm8$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}-2$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=66-2$
$\big[\therefore\ \text{x}^2+\frac{1}{\text{x}^2}=66\big]$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=64$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=(\pm8)^2$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)=\pm8$
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