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

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== Solution ==
 
== Solution ==
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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:
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<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 $x_1< x_2 < x_3$ 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 $x=\dfrac{1}{\sqrt{2014}}$ 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 $x_2=\dfrac{1}{\sqrt{2014}}$

by vieta's formula we have that the $x^2$ coefficent is equal to $-x_1-x_2-x_3$ and that the $x$ coeefficent is equal to $x_1x_2+x_1x_3+x_2x_3$ so using the values in the above equation we get: $-x_1-x_2-x_3=-4029\rightarrow x_1+x_2+x_3=4029$

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