InterviewSolution
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InterviewSolution
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If x = 7+4√3 and xy = 1, then = \(\frac{1}{x^2}\) + \(\frac{1}{y^2}\)A. 64B. 134C. 194D. \(\frac{1}{49}\) |
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Answer» Given, x = 7 + 4√3 , xy = 1 Y = \(\frac{1}{x}\) = \(\frac{1}{7}\)+ 4√3 = 7 - 4√3 Y2 = \(\frac{1}{x^2}\) = 49 + 48 - 56√3 = 97 - 56√3 Similarly, x = \(\frac{1}{y}\) = x2 = \(\frac{1}{y^2}\) = ( 7 + 4√3)2 = 49 + 48 + 56√3 = 97+ 56√3 So, \(\frac{1}{x^2}\) + \(\frac{1}{y^2}\) = 97 + 56√3 + 97 – 56√3 = 194 |
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