# Difference between revisions of "2014 AIME I Problems/Problem 9"

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

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+ | We note that <math>x=\dfrac{1}{\sqrt{2014}}</math> is a solution since <math>(\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</math> | ||

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+ | We claim that <math>x_2=\dfrac{1}{\sqrt{2014}}</math> | ||

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+ | by vieta's formula we have that the <math>x^2</math> coefficent is equal to <math>-x_1-x_2-x_3</math> and that the <math>x</math> coeefficent is equal to <math>x_1x_2+x_1x_3+x_2x_3</math> so using the values in the above equation we get: | ||

+ | <math>-x_1-x_2-x_3=-4029\rightarrow x_1+x_2+x_3=4029</math> |

## Revision as of 12:18, 14 March 2014

## Problem 9

## Solution

Let be the three real roots of the equation . Find .

We note that is a solution since

We claim that

by vieta's formula we have that the coefficent is equal to and that the coeefficent is equal to so using the values in the above equation we get: