Question
If $\text{x}+\frac{1}{\text{x}}=11$ find the value of $\text{x}^2+\frac{1}{\text{x}^2}.$
✓
Answer
We have $\text{x}+\frac{1}{\text{x}}=11$ Now, $\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$
$\Rightarrow(11)^2=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\Rightarrow121=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=119.$
$\Rightarrow(11)^2=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\Rightarrow121=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=119.$
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