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

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

We note that $\dfrac{1}{\sqrt{2014}}$ is a solution since $(\dfrac{1}{\sqrt{2014}})^3*\sqrt{2014}-4029(\dfrac{1}{\sqrt{2014}})^2+2=\dfrac{1}{2014}-\dfrac{4029}{2014}+2=\dfrac{1-4029+4028}{2014} = 0$

We claim that $x_2=\dfrac{1}{\sqrt{2014}}$ [please clarify]

By Vieta's formula we have that the $x^2$ coefficent is equal to $-x_1-x_2-x_3$ and that the $x$ coefficent is equal to $x_1x_2+x_1x_3+x_2x_3$ . Using the values in the above equation we get: $-x_1-x_2-x_3=-4029\rightarrow x_1+x_2+x_3=4029$

2014 AIME I (Problems • Answer Key • Resources)
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

Page

Toolbox

Search

2014 AIME I Problems/Problem 9

Problem 9

Solution

See also