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>. | 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>. | ||
| − | We note that <math> | + | We note that <math>\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> |
We claim that <math>x_2=\dfrac{1}{\sqrt{2014}}</math> | We claim that <math>x_2=\dfrac{1}{\sqrt{2014}}</math> | ||
| − | by | + | 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> | <math>-x_1-x_2-x_3=-4029\rightarrow x_1+x_2+x_3=4029</math> | ||
| + | |||
| + | == See also == | ||
| + | {{AIME box|year=2014|n=I|num-b=8|num-a=10}} | ||
| + | {{MAA Notice}} | ||
Revision as of 18:43, 14 March 2014
Problem 9
Let 
be the three real roots of the equation 
. Find 
.
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:
See also
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
