\(3\sqrt x - \frac1{4\sqrt x} = \sqrt 7\)
Squaring on both sides, we get
\(9x + \frac1{16x} - 2(3\sqrt x) \left(\frac1{4\sqrt x}\right) = 7\) \((\because (a-b)^2 = a^2 + b^2 - 2ab)\)
⇒ \(9x + \frac1{16x} = 7 + \frac32 = \frac{17}2\)
Again squaring on both sides
\(81x^2 + \frac1{256x^2}+ 2(9x)\left(\frac1{16x}\right)= \frac{17}2\)
⇒ \(81x^2 + \frac1{256x^2} = \frac{17}2- \frac98= \frac{68-9}{8}= \frac{59}{8}\)
Hence, the value of \(81x^2 + \frac1{256x^2} = \frac{59}{8}\).
