<s>[INST] Calculating $x$ from $x^3 = 2x^2 - x$, getting wrong result I am working with a text book problem, trying to find the solutions to $x^3 = 2x^2 - x$, but I keep ending up with a different result than the book gives. I'm thinking the problem lie in getting the wrong square root towards the end, seeing as the text book gives different results for $x_1$ and $x_2$, but I can't find any mistakes.
$$x^3 = 2x^2 - x$$
$$\iff x^2 = 2x - 1$$
$$\iff x^2 - 2x + 1 = 0$$
$$\iff (x - \frac{2}{2})^2 = (\frac{2}{2})^2 - 1$$
$$\iff (x - 1)^2 = 0$$
$$\iff x - 1 = \pm \sqrt{0}$$
$$\iff x = 1 + \pm \sqrt{0}$$
$$\iff x = 1 + \pm 0$$
$$x_1 = 1 + 0 = 1$$
$$x_2 = 1 - 0 = 1$$
 [/INST] You can not just divide by $x$, since $x$ could be $0$.
You have $x^3=2x^2-x$, which is equivalent to:
$$0=x^3-2x^2+x=x(x^2-2x+1)=x(x-1)^2$$
Hence $x=0$ or $x=1$.
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