Question

if x is equal to 2 + under root 5 prove that is X square + 1 upon x square is equal to 18​

Answer 1

Answer:

$x = 2 + \sqrt{5} \\ \\ \frac{1}{x} = \frac{1}{2 + \sqrt{5} } \times \frac{2 - \sqrt{5} }{2 - \sqrt{5} } \\ \\ \frac{1}{x} = \frac{2 - \sqrt{5} }{(2) {}^{2} - ( \sqrt{5} ) {}^{2} } \\ \\ \frac{1}{x} = \frac{2 - \sqrt{5} }{4 - 5} \\ \\ \frac{1}{x} = - 2 + \sqrt{5} \\ \\ x + \frac{1}{x} = 2 + \sqrt{5} - 2 + \sqrt{5} \\ \\ x + \frac{1}{x} = 2 \sqrt{5} \\ \\ (x + \frac{1}{x}){}^{2} = (2 \sqrt{5} ) {}^{2} \\ \\ x {}^{2} + \frac{1}{x {}^{2} } + 2 = 20 \\ \\ x {}^{2} + \frac{1}{x {}^{2} } = 20 - 2 \\ \\ x {}^{2} + \frac{1}{x {}^{2} } = 18 \\ \\ \quad \quad \huge{ \bf proved.}$

Answer 2

Answer:

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