Question

If $\: {x}^{2} + \frac{1}{ {x}^{2} } \: = \: 27$
Then find the value of
$\ \: x - \frac{1}{x}$​

Answer 1

Given :-

• $\sf \: {x}^{2} + \dfrac{1}{ {x}^{2} } \: = \: 27$

To find :-

• $\sf \: x - \dfrac{1}{x}$

Identity used :

• ( a+b )² = a² + b² - 2ab

Solution :-

$\sf : \implies\red {\: \: x - \dfrac{1}{x} }$

$\sf \: : \implies \: \bigg( x - \dfrac{1}{x} { \bigg)}^{2}$

$\sf: \implies \:\bigg( x - \dfrac{1}{x} { \bigg)}^{2} \: = \: {x}^{2} \: + \: \bigg( \dfrac{1}{x} { \bigg)}^{2} \: - \:2 \times \bigg( \dfrac{1}{x} { \bigg)} \bigg( x \bigg)$

$\sf :\implies \: \bigg( x - \dfrac{1}{x} { \bigg)}^{2} \: = \: {x}^{2} \: + \: \bigg( \dfrac{1}{x} { \bigg)}^{2} \: - \:2$

As we are given that:-

$\red { \bigg[ \: \sf \: {x}^{2} \: + \: \bigg( \dfrac{1}{x} { \bigg)}^{2} \: = \:27 \: \: \bigg]}$

$\sf: \implies \: \bigg( x - \dfrac{1}{x} { \bigg)}^{2} \: = \: {x}^{2} \: + \: \bigg( \dfrac{1}{x} { \bigg)}^{2} \: - \:2 \: = \: 27 - 2$

$\sf :\implies \: \: \bigg( x - \dfrac{1}{x} { \bigg)}^{2} = \: 25$

$\sf: \implies \: \bigg( x - \dfrac{1}{x} { \bigg)} \: = \sqrt{25}$

$\sf\red { :\implies \: \bigg( x - \dfrac{1}{x} { \bigg)} \: = \: \pm \: 5}$

Answer 2

Step-by-step explanation:

$\sf \color{red} ANSWER:-$

$\huge \color{red}\mathtt( x - \frac{1}{x}) = ±5$

$\sf \color{red}hope \: its \: helps \: you$