2014 AIME I Problems/Problem 9

Revision as of 19:12, 14 March 2014 by NeilOnnsu (talk | contribs) (Solution)

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

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

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 $x_2(x_1+x_3) = \boxed{002}$.

See also

2014 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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