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\( 3 \sqrt{x}-\frac{1}{4 \sqrt{x}}=\sqrt{7} \), find the value of\( 81 x^{2}+\frac{1}{256 x^{2}} \) |
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Answer» \(3\sqrt x - \frac1{4\sqrt x} = \sqrt 7\) 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 \(81x^2 + \frac1{256x^2} = \frac{59}{8}\). |
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