Question:
If $x^{2}+1 / x^{2}=79$, find the value of $x+1 / x$
Solution:
We have,
$(x+1 / x)^{2}=x^{2}+(1 / x)^{2}+2 * x * 1 / x$
$\Rightarrow(x+1 / x)^{2}=x^{2}+1 / x^{2}+2$
$\Rightarrow(x+1 / x)^{2}=79+2 \quad\left[\therefore x^{2}+1 / x^{2}=79\right]$
$\Rightarrow(x+1 / x)^{2}=81$
$\Rightarrow(x+1 / x)^{2}=(\pm 9)^{2}$
$\Rightarrow x+1 / x=\pm 9$
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