Question

if x2 +1/x2=18 find the value of √x -1/√x

Answer 1

$(x - \frac{1}{x} {)}^{2} = {x}^{2} + \frac{1}{ {x}^{2} } - 2 \\ \\ put \: the \: value \\ \\ (x - \frac{1}{ {x}^{} } ) {}^{2} = 18 - 2 \\ \\ (x - \frac{1}{x} {)}^{2} = 16 \\ \\ (x - \frac{1}{x} ) = \sqrt{16} \\ \\ (x - \frac{1}{x} ) = 4 \\ \\ now \: taking \: both \: side \: sqaure \: root \\ \\ \sqrt{x - \frac{1}{x} } = \sqrt{4} \\ \\ \sqrt{x} - \frac{1}{ \sqrt{x} } = 2$
I hope its helps