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Annual Harvard-MIT November Tournament

Algebra Difficulty 4.4 AIME Prove it United States

Problem:
Find the minimum of $x^{2}-2x$ over all real numbers $x$ .

Solution

Solution:
Write $x^{2}-2x = x^{2}-2x+1-1 = (x-1)^{2}-1$ . Since $(x-1)^{2} \geq 0$ , it is clear that the minimum is $-1$ .

Alternate method: The graph of $y = x^{2}-2x$ is a parabola that opens up. Therefore, the minimum occurs at its vertex, which is at $\frac{-b}{2a} = \frac{-(-2)}{2} = 1$ . But $1^{2}-2 \cdot 1 = -1$ , so the minimum is $-1$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.