I have a problem but i don't know what method i should use to solve it.

If x^3 – 2x^2 + 3x = 6 prove x=2

Apologies for the poor format of the equation.

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- Aug 26th 2009, 07:06 AMMaths Stigproblem
I have a problem but i don't know what method i should use to solve it.

If x^3 – 2x^2 + 3x = 6 prove x=2

Apologies for the poor format of the equation. - Aug 26th 2009, 07:32 AMLiwuinan
subtract 6 from both sides and factorize the left side:

$\displaystyle x^3-2x^2+3x-6 = x^2(x-2) + 3(x-2) = (x^2+3)(x-2)$

Assuming x should be a real number, this can equal to zero if and only if at least one of the brackets is zero and this happens if and only if x=2 (first bracket is never zero). In complex numbers there would be also other solutions, not only x=2. - Aug 26th 2009, 07:34 AMDefunkt
We will solve this using the fact that $\displaystyle x^3 -2x^2 +3x - 6 = (x-2)(x^2+3)$

If we assume $\displaystyle x \in \mathbb{R}$ (where $\displaystyle \mathbb{R}$ are the real numbers), then the solution for equation would be either the solution for $\displaystyle x-2=0$ or $\displaystyle x^2 + 3 = 0$ . Obviously, $\displaystyle x=2$ is a solution, and there are no other solutions for $\displaystyle x^2 + 3 = 0$ in $\displaystyle \mathbb{R}$, thus $\displaystyle x=2$

If you want to solve this equation over $\displaystyle \mathbb{C}$ (the complex numbers), then $\displaystyle \sqrt{3}i$ , $\displaystyle -\sqrt{3}i$ are also solutions, but I don't think this is the case.