Question

if
$\frac{1}{x } = 3 - \sqrt{2}$
then find the value of
$x - \frac{1}{x}$
and
$x + \frac{1}{x}$
​

Answer 1

Step-by-step explanation:

$if \: \frac{1}{x } = 3 - \sqrt{2} \\ then \: x = \frac{1}{3 - \sqrt{2} }$

$to \: find \: x - \frac{1}{x} \\ by \: appling \: formula \: \\ \: (a - b {)}^{2} = {a}^{2} - 2ab \: + {b}^{2} \\ where \: a \: = x \: and \: b = \frac{1}{x}$

$(x - \frac{1}{x} {)}^{2} = {x}^{2} - 2xy + { \frac{1}{x} }^{2} \\ (x - \frac{1}{x} {)}^{2} = (\frac{1}{3 - \sqrt{2} } {)}^{2} - 2( \frac{1}{3 - \sqrt{2} } )(3 - \sqrt{2} ) + (3 - \sqrt{2} {)}^{2}$