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 ?
 A: I think your working is perfect. If the book is correct then
$$1-\frac4{3x}+\frac2{x^2}=2\implies 3x^2-4x+6=6x^2\implies3x^2+4x-6=0$$
whereas according to what is written there it must be $\;3x^2+4x-6=4x^2\;$, so I'd say the book's answer is wrong...or you forgot a coefficient $\;3\;$ for the quadratic term.
A: Divide both sides by $x^2$ to get  $$1-\dfrac4x+\dfrac6{x^2}=0\iff\dfrac4x-\dfrac6{x^2}=1$$
$$1-\dfrac4{3x}+\dfrac2{x^2}=1-\dfrac13\left(\dfrac4x-\dfrac6{x^2}\right)=?$$
A: I think that the book is wrong.
$$1-\frac{4}{3x}+\frac{2}{x^2}=\frac{3x^2}{3x^2}\cdot \frac{1-\frac{4}{3x}+\frac{2}{x^2}}{1}=\frac{3x^2-4x+6}{3x^2}=\frac{2x^2+(x^2-4x+6)}{3x^2}$$
$$=\frac{2x^2+0}{3x^2}=\frac{2}{3},x\ne 0$$
A: 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
