2014 AIME I Problems/Problem 9

Problem 9

Let $x_1<x_2<x_3$ be the three real roots of the equation $\sqrt{2014}x^3-4029x^2+2=0$ . Find $x_2(x_1+x_3)$ .

Solution

Substituting $n$ for $2014$ , we get $\sqrt{n}x^3 - (1+2n)x^2 + 2 = \sqrt{n}x^3 - x^2 - 2nx^2 + 2 = x^2(\sqrt{n}x - 1) - 2(nx^2 - 1) = 0$ . Noting that $nx^2 - 1$ factors as a difference of squares to $(\sqrt{n}x - 1)(\sqrt{n}x+1)$ , we can factor the left side as $(\sqrt{n}x - 1)(x^2 - 2(\sqrt{n}x+1))$ . This means that $\frac{1}{\sqrt{n}}$ is a root, and the other two roots are the roots of $x^2 - 2\sqrt{n}x - 2$ . Note that the constant term of the quadratic is negative, so one of the two roots is positive and the other is negative. In addition, by Vieta's Formulas, the roots sum to $2\sqrt{n}$ , so the positive root must be greater than $2\sqrt{n}$ in order to produce this sum when added to a negative value. Since $0 < \frac{1}{\sqrt{2014}} < 2\sqrt{2014}$ is clearly true, $x_2 = \frac{1}{\sqrt{2014}}$ and $x_1 + x_3 = 2\sqrt{2014}$ . Multiplying these values together, we find that $x_2(x_1+x_3) = \boxed{002}$ .

2014 AIME I (Problems • Answer Key • Resources)
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2014 AIME I Problems/Problem 9

Contents

Problem 9

Solution

Solution 2

See also