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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.

$$x^3 = 2x^2 - x$$ $$\iff x^2 = 2x - 1$$ $$\iff x^2 - 2x + 1 = 0$$ $$\iff (x - \frac{2}{2})^2 = (\frac{2}{2})^2 - 1$$ $$\iff (x - 1)^2 = 0$$ $$\iff x - 1 = \pm \sqrt{0}$$ $$\iff x = 1 + \pm \sqrt{0}$$ $$\iff x = 1 + \pm 0$$ $$x_1 = 1 + 0 = 1$$ $$x_2 = 1 - 0 = 1$$

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up vote 8 down vote accepted

Note that by dividing by $x$, you are implicitly assuming that $x$ is not $0$. It turns out that this is a solution. You can check this directly.

You are right that $1$ is a double root of the equation. If you move everything to one side and factor, you find that


which confirms our answer.

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Thanks! It never struck me that $x$ could be $0$ and that I discounted that possibility so easily. – Quispiam Aug 18 '12 at 11:41
Did you perhaps mean $+x$ instead of $+1$? (It's too few characters for me to edit.) – pimvdb Aug 18 '12 at 20:05
@pimvdb Yes, thank you. – Potato Aug 18 '12 at 22:33

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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Your solution $x=1$ perfectly well solves $x^3 = 2x^2 - x$, as you can easily check yourself (did you try?). But so does $x=0$, a possibility that you lost along the way (can you see where?). If your book gives any other solutions, you can check that they don't solve the equation. Checking solutions is much easier than finding them, you should always do it for the solutions found.

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I did actually try $x=1$ in the original equation, but were confused as I was supposed to find two separate for $x$. I didn't even stop to think it could be $0$. Thanks! – Quispiam Aug 18 '12 at 11:33

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