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:
