2014 AIME I Problem 9

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Concepts:polynomialfactoringVieta’s Formulas

Difficulty rating: 2560

9.

Let x1<x2<x3x_1 \lt x_2 \lt x_3 be the three real roots of the equation 2014x34029x2+2=0.\sqrt{2014}\,x^3 - 4029x^2 + 2 = 0. Find x2(x1+x3).x_2(x_1 + x_3).

Solution:

Write a=2014,a = \sqrt{2014}, so the equation is ax3(2a2+1)x2+2=0.ax^3 - (2a^2 + 1)x^2 + 2 = 0. It factors as (ax1)(x22ax2)=0,\left(ax - 1\right)\left(x^2 - 2ax - 2\right) = 0, as expanding confirms. So one root is 1a,\frac{1}{a}, and the other two are a±a2+2,a \pm \sqrt{a^2 + 2}, with product 2-2 and sum 2a.2a.

Since aa2+2<0<1a<a+a2+2,a - \sqrt{a^2+2} \lt 0 \lt \frac{1}{a} \lt a + \sqrt{a^2+2}, the middle root is x2=1a,x_2 = \frac{1}{a}, and x1+x3=2a.x_1 + x_3 = 2a. Therefore x2(x1+x3)=1a2a=2.x_2(x_1 + x_3) = \frac{1}{a} \cdot 2a = 2.

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