$x^2+\frac{1}{x^2}=34$, then find the value of $x+\frac{1}{x}$.


Given: $x^2+\frac{1}{x^2}=34$.

To do: To find the value of $x+\frac{1}{x}$.

Solution:

$x^2+\frac{1}{x^2}=34$

On adding $( 2)$ to both sides we have,

$\Rightarrow x^2+\frac{1}{x}²+2=36$

$\Rightarrow ( x)^2+( \frac{1}{x^2})+2.x.\frac{1}{x}=( \pm6)^2$

$\Rightarrow ( x+\frac{1}{x})^2=(\pm6)^2$

$\Rightarrow x+\frac{1}{x}=\pm6$

Thus, $x+\frac{1}{x}=\pm6$

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Updated on: 10-Oct-2022

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