Question

if x = 17+12√2 then find the value of √x - 1 / √x​

Answer 1

$x = 17 + 12 \sqrt{2} \\ \frac{1}{x} = \frac{1}{17 + 12 \sqrt{2} } \\ \frac{1}{x} = \frac{1}{17 + 12 \sqrt{2} } \times \frac{17 - 12 \sqrt{2} }{17 - 12 \sqrt{2} } \\ \frac{1}{x} = \frac{17 - 12 \sqrt{2} }{289 - 288} \\ \frac{1}{x} = 17 - 12 \sqrt{2} \\ \\ x + \frac{1}{x} = 17 + 12 \sqrt{2} + 17 - 12 \sqrt{2} \\ x + \frac{1}{x} = 34 \\ x + \frac{1}{x} - 2 = 32 \\ ( { \sqrt{x} })^{2} + ( { \frac{1}{ \sqrt{x} } })^{2} - 2 \: \sqrt{x } \: \frac{1}{ \sqrt{x} } = 32 \\ ( { \sqrt{x} - \frac{1}{ \sqrt{x} } })^{2} = 32 \\ \sqrt{x} - \frac{1}{ \sqrt{x} } = \sqrt{32} = 4 \sqrt{2}$