<s>[INST] Is something wrong with this precalculus question? If $x^{2}-4x+6=0$, then what can be the value of $1-\frac{4}{3x}+\frac{2}{x^{2}}$?
My answer is $1-\frac{4}{3x}+\frac{2}{x^{2}}=\frac{3x^{2}-4x+6}{3x^{2}}=\frac{3x^{2}-x^{2}}{3x^{2}}=\frac{2}{3}$ for $x\neq 0$. But according to the book answer is 2. What is the point i miss?
 [/INST] $3x^2-4x+6=2x^2+(x^2-4x+6)=2x^2$
You're right. Note that $x$ is complex and not real, but apart from this, the value of the other expression is $2/3$.
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