If \(\frac{7(x^2+1)-17x}{7x}\) = 2 then, \( { \left(\frac{x+1}{\sqrt {x }}\right) }^{2} \) = ?

6.42

7(x² + 1) - 17x = 14x

7(x² + 1) = 14x + 17x

7(x² + 1) = 31 x

x² + 1 = \(\frac{31}{7}\)x

x + \(\frac{1}{x} = \frac{31}{7}\)

[If x + \(\frac{1}{x}\) = a, then \(\sqrt {x } + \frac{1}{\sqrt {x }} \) = \(\sqrt {a + 2}\)]

Therefore,

\(\sqrt {x } + \frac{1}{\sqrt {x }} \) =\(\sqrt { \frac{31}{7}+2}\) = \(\sqrt { \frac{45}{7}}\)

\( { \left(\sqrt {x } + \frac{1}{\sqrt {x }}\right) }^{2} \) = \(\frac{45}{7}\)

\( { \left(\sqrt {x } + \frac{1}{\sqrt {x }}\right) }^{2} \) = 6.42

⇒ \( { \left(\frac{x+1}{\sqrt {x }}\right) }^{2} \) = 6.42