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$x^2+\frac{1}{x^2}=34$, 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, $x+\frac{1}{x}=\pm6$
Updated on: 10-Oct-2022
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