Question

If $(x^2+\frac{1}{25x^2})=9\frac{2}{5}$, find the value of $(x+\frac{1}{5x})$

Answer 1

$(x^2+\frac{1}{25x^2})=9\frac{2}{5}=\frac{47}{5}$
$\because {(x+\frac{1}{5x})}^2={\left(x\right)}^2+{\left(\frac{1}{5x}\right)}^2-2.x.\frac{1}{5x}$
$x^2+\frac{1}{25x^2}-\frac{2}{5}$
$=\frac{47}{5}-\frac{2}{5}$
$=\frac{47-2}{5}=\frac{45}{5}=9$
$\Rightarrow x-\frac{1}{5x}=\pm 3$