Question

if x^2+1/x^2=18, then the value of x-1/x will be-​

Answer 1

$\begin{gathered}\begin{gathered}\bf \: Given \:  - \begin{cases} &\sf{ {x}^{2} + \dfrac{1}{ {x}^{2} } = 18 } \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: To \: find \:  - \begin{cases} &\sf{x - \dfrac{1}{x} }  \end{cases}\end{gathered}\end{gathered}$

$\large\underline\purple{\bold{Solution :- }}$

Given that

$\rm :\implies\: {x}^{2} + \dfrac{1}{ {x}^{2} } = 18$

On Subtracting 2 on both sides, we get

$\rm :\implies\: {x}^{2} + \dfrac{1}{ {x}^{2} } - 2 = 18 - 2$

$\rm :\implies\: {x}^{2} + \dfrac{1}{ {x}^{2} } - 2 \times 1= 16$

$\rm :\implies\: {x}^{2} + \dfrac{1}{ {x}^{2} } - 2 \times x \times \dfrac{1}{x} = 16$

$\rm :\implies\: {(x - \dfrac{1}{x}) }^{2} = 16$

$\rm :\implies\:x - \dfrac{1}{x} = \: \pm \: 4$