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Algebra Difficulty 4.9 AIME Prove it United States

Quadratic polynomials $P(x)$ and $Q(x)$ have leading coefficients of $2$ and $-2$ , respectively. The graphs of both polynomials pass through the two points $(16, 54)$ and $(20, 53)$ . Find $P(0) + Q(0)$ .

Solution

Because the leading coefficients of $P(x)$ and $Q(x)$ are negatives of each other, the polynomial $R(x) = P(x) + Q(x)$ is linear. Furthermore, $R(16) = 54 + 54 = 108$ and $R(20) = 53 + 53 = 106$ . It follows that $R(x) = 116 - 0.5x$ , so $P(0) + Q(0) = R(0) = 116$ .

Note that
$P(x) = 2x^2 - \frac{289}{4}x + 698 \quad \text{and} \quad Q(x) = -2x^2 + \frac{287}{4}x - 582.$

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