Question

if x=9+4root5 , find the value of root x- 1/root x

Answer 1

$x = 9 - 4 \sqrt{5} \\ \frac{1}{x } = \frac{1}{9 - 4 \sqrt{5} } \\ = \frac{1}{9 - 4 \sqrt{5} } \times \frac{9 + 4 \sqrt{5} }{9 + 4 \sqrt{5} } \\ = \frac{9 + 4 \sqrt{5} }{(9) {}^{2} -( 4 \sqrt{5}) {}^{2} } = \frac{9 + 4 \sqrt{5} }{81 - 80 } \\ \frac{1}{x} = 9 + 4 \sqrt{5} \\ (\sqrt{x} + \frac{1}{ \sqrt{x} } ) {}^{2} = ( \sqrt{x} ) {}^{2} + ( \frac{1}{ \sqrt{x} }) {}^{2} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: + 2 \times \sqrt{x} \times \frac{1}{ \sqrt{x} } \\ (\sqrt{x} + \frac{1}{ \sqrt{x} } ) {}^{2} = x + \frac{1}{x} + 2 \\ = 9 - 4 \sqrt{5} + 9 + 4 \sqrt{5} \\ = 18 \\ so \:( \sqrt{x} + \frac{1}{ \sqrt{x} } ) {}^{2} = 18 \\ therefore \sqrt{x} + \frac{1}{ \sqrt{x} } = \sqrt{18}$