Question

a =
$6 - \sqrt{35}$
${a}^{2} \frac{1}{ {a}^{2} ?}$
​

Answer 1

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$a = 6 - \sqrt{25} \\ take \: a \: as \: x \\ \\ \frac{1}{x} = \frac{1}{6 - \sqrt{35} } \times \frac{6 + \sqrt{35} }{6 + \sqrt{35} } \\ \\ = \frac{6 + \sqrt{35} }{(6) {}^{2} } - ( \sqrt{35)} {}^{2} \\ \\ = 6 + \sqrt{35} \\ x + \frac{1}{x} = 6 + \sqrt{35} + 6 - \sqrt{35} \\ \\ = 6 + 6 = 12 \\ \\ (x + \frac{1}{x} {})^{2} = 12 {}^{2} \\ = 144 \\ \\ (x + \frac{1}{x} ) {}^{2} = x {}^{2} + \frac{1}{x {}^{2} } + 2 \\ \\ = x {}^{2} + \frac{1}{x {}^{2} } + 2 = 144 \\ \\ x {}^{2} + \frac{1}{x {}^{2} } = 144 - 2 = 142$

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