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If \(x-\frac{1}{x}\) =3, find the values of \(x^2+\frac{1}{x^2}\) and \(x^4+\frac{1}{x^4}\). |
Answer» (i) Given that, x - \(\frac{1}{x}\) = 3 Squaring both sides, we get (x - \(\frac{1}{x}\))2 = (3)2 x2 - 2 × x × \(\frac{1}{x}\) + (\(\frac{1}{x}\))2 = 9 x2 - 2 + \(\frac{1}{x\times x}\) = 9 x2 + \(\frac{1}{x\times x}\) = 11 (ii) Squaring both sides, we get (x2 + \(\frac{1}{x\times x}\))2 = (11)2 (x2)2 + 2 × x2 × \(\frac{1}{x\times x}\) + (\(\frac{1}{x\times x}\))2 = 121 x4 + 2 + \(\frac{1}{x\times x\times x\times x}\) = 121 x4 + \(\frac{1}{x\times x\times x\times x}\) = 119 |
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