Question
If $\text{x}^2+\frac{1}{\text{x}^2}=79,$ find the value of $\text{x}+\frac{1}{\text{x}}.$
✓
Answer
In the given problem, we have to find $\Big(\text{x}+\frac{1}{\text{x}}\Big)$
Given $\text{x}^2+\frac{1}{\text{x}^2}=79$ Adding and subtracting $2$ on left hand side, $\text{x}^2+\frac{1}{\text{x}^2}+2-2=79$
$\Big(\text{x}^2+\frac{1}{\text{x}^2}+2\times\text{x}\times\frac{1}{\text{x}}\Big)-2=79$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2-2=79$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=79+2$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=81$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\sqrt{81}$
$\text{x}+\frac{1}{\text{x}}=\sqrt{9\times9}$
$\text{x}+\frac{1}{\text{x}}=\pm9$
Hence the value of $\Big(\text{x}+\frac{1}{\text{x}}\Big)$ is $\pm9$
Given $\text{x}^2+\frac{1}{\text{x}^2}=79$ Adding and subtracting $2$ on left hand side, $\text{x}^2+\frac{1}{\text{x}^2}+2-2=79$
$\Big(\text{x}^2+\frac{1}{\text{x}^2}+2\times\text{x}\times\frac{1}{\text{x}}\Big)-2=79$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2-2=79$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=79+2$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=81$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\sqrt{81}$
$\text{x}+\frac{1}{\text{x}}=\sqrt{9\times9}$
$\text{x}+\frac{1}{\text{x}}=\pm9$
Hence the value of $\Big(\text{x}+\frac{1}{\text{x}}\Big)$ is $\pm9$
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