Question

If x = 3+2V2, find the value of Vx - 1 √x
please help ​

Answer 1

$x = 3 + 2 \sqrt{2} \\ \\ \frac{1}{x} = \frac{1}{3 + 2 \sqrt{2} } \\ \frac{1}{x} = \frac{1}{3 + 2 \sqrt{2} } \times \frac{3 - 2 \sqrt{2} }{3 - 2 \sqrt{2} } \\ \frac{1}{x} = \frac{3 - 2 \sqrt{2} }{9 - 8} \\ \frac{1}{x} = 3 - 2 \sqrt{2} \\ \\ x + \frac{1}{x} = 6 \\ x + \frac{1}{x} - 2 = 4 \\ ( { \sqrt{x} - \frac{1}{ \sqrt{x} } })^{2} = 4 \\ \sqrt{x} - \frac{1}{ \sqrt{x} } = ±2$