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If x = 7+4√3 and xy = 1, then = \(\frac{1}{x^2}\) + \(\frac{1}{y^2}\)

A. 64

B. 134

C. 194

D. \(\frac{1}{49}\)

1 Answer

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Best 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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