If the value of 2x-\(\frac{1}{x}\)=\(\sqrt {281}\)

find the value of \(\frac{(2x+\frac{1}{x})^2+60}{3}\)

\(\frac{349}{3}\)

If 2x-\(\frac{1}{x}\)=\(\sqrt {281}\)

than 2x+\(\frac{1}{x}\) = \(\sqrt{(\sqrt{281})^2 + 4×2x×\frac{1}{x}}\)

2x+\(\frac{1}{x}\) = \(\sqrt{281+8}\)

2x+\(\frac{1}{x}\) = 17 ⇒ Put in find out

⇒ \(\frac{(2x+\frac{1}{x})^2+60}{3}\) = \(\frac{({17})^{2}+60}{3}\)

= \(\frac{349}{3}\)