My answer don't match with the answer of the book Let $x^{2}-4x+6=0$. What can be the result of $1-\frac{4}{3x}+\frac{2}{x^{2}}$ ?
A) $-\frac{2}{3}~~$ B) $-\frac{1}{3}~~$ C) $\frac{1}{3}~~$ D) $\frac{2}{5}~~$ E) $2$
My answer is: $1-\frac{4}{3x}+\frac{2}{x^{2}}=1-\frac{1}{x^{2}}(\frac{4x}{3}-2)=1-\frac{1}{x^{2}}(\frac{4x-6}{3})=1-\frac{1}{x^{2}}(\frac{x^{2}}{3})=\frac{2}{3}$.
But according to the book the right answer is option E, i.e, $2$.  
Is the book right or me ?

By solving the quadratic equation,$x = 2+\sqrt 2 *i$ or  $x = 2-\sqrt 2 *i$ where $i$ is a unit of image number. 


*

*When $x = 2+\sqrt 2 *i$,  $1-\frac{4}{3x}+\frac{2}{x^{2}}$ = $1-\frac{4}{3*(2+\sqrt 2 *i)}+\frac{2}{(2+\sqrt 2 *i)^{2}} = 1-\frac{4}{9}+\frac{2*\sqrt 2}{9}*i + \frac{1}{9} - \frac{2*\sqrt 2}{9}*i = \frac{2}{3}$ w

*When $x = 2-\sqrt 2 *i$, $1-\frac{4}{3x}+\frac{2}{x^{2}} = \frac{2}{3}$


reference:


*

*Solving quadratic equation: http://mathworld.wolfram.com/QuadraticEquation.html

*Complex number :http://mathworld.wolfram.com/ComplexNumber.html
