# If x^2 + 1/x^2 = 27. Then find x-1/x ?

## If ${x}^{2} + \frac{1}{x} ^ 2 = 27$. Then find $x - \frac{1}{x}$

Jan 27, 2018

$x - \frac{1}{x} = \pm 5$

#### Explanation:

just expand, rearrange n substitute,!
${\left(x - \frac{1}{x}\right)}^{2}$

$= {x}^{2} - 2 \cdot x \cdot \frac{1}{x} + {\left(\frac{1}{x}\right)}^{2}$

$= {x}^{2} - 2 + {\left(\frac{1}{x}\right)}^{2}$

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

since we have the value of ${x}^{2} + \frac{1}{x} ^ 2$ as $27$,

$= 27 - 2$

$= 25$

$\implies {\left(x - \frac{1}{x}\right)}^{2} = 25$

$\implies x - \frac{1}{x} = \pm 5$

-Sahar

Jan 27, 2018

$x - \frac{1}{x} = \pm 5$

#### Explanation:

$\text{note that } {\left(x - \frac{1}{x}\right)}^{2} = {x}^{2} - 2 + \frac{1}{x} ^ 2$

$\Rightarrow {x}^{2} - 2 + \frac{1}{x} ^ 2 = 27 - 2 = 25$

$\Rightarrow {\left(x - \frac{1}{x}\right)}^{2} = 25$

$\textcolor{b l u e}{\text{take the square root of both sides}}$

$\Rightarrow x - \frac{1}{x} = \pm 5 \leftarrow \textcolor{b l u e}{\text{note plus or minus}}$