Question

If $(x^2+\frac{1}{x^2})=102$, then find the value of $(x+\frac{1}{x})$.

Answer 1

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

${(x+\frac{1}{x})}^2=102+2$

$x+\frac{1}{x}=\sqrt{104}$