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fighting_cax
- Senior | Next Rank: 100 Posts
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1. x^2 = 2x means that x^2 - 2x = 0, or that x(x-2) = 0. So x can be either 2 or 0, therefore we cannot establish if x > 0. So 1 is insufficient.
2. x^3 = 3x is equivalent to x^3 - 3x = x(x^2 - 3) = x[x - sqrt(3)][x+ sqrt(3)] = 0 (obtained using the (x-y)(x+y) = x^2 - y^2 formula). This means that x can be either 0, sqrt(3) or -sqrt(3). Again, we cannot establish if x is or not greater than 0.
Both statements work together for the answer: there is only one number that is a common solution for both 1 and 2 and that is 0. Since x = 0 = 0 (huh...), it is obvious that x is not greater than 0.
