x-2-1-x-2-27-x-1-x-

Question Number 93170 by ar247 last updated on 11/May/20

x^{2} + \frac{1}{x^{2}} = 27

\sqrt{x} + \frac{1}{\sqrt{x}} = \dots .

Commented by mr W last updated on 11/May/20

{(x + \frac{1}{x})}^{2} = x^{2} + \frac{1}{x^{2}} + 2 = 27 + 2 = 29

x + \frac{1}{x} = \pm \sqrt{29}

\sqrt{x} + \frac{1}{\sqrt{x}} = t > 0

t^{2} = x + \frac{1}{x} + 2 = \sqrt{29} + 2

\Rightarrow t = \sqrt{\sqrt{29} + 2}

Commented by i jagooll last updated on 11/May/20

let \sqrt{x} + \frac{1}{\sqrt{x}} = t

x + \frac{1}{x} = t^{2} - 2

x^{2} + \frac{1}{x^{2}} + 2 = {(t^{2} - 2)}^{2}

29 = {(t^{2} - 2)}^{2}

t^{2} = 2 + \sqrt{29}

t = \sqrt{2 + \sqrt{29}}