Difference between revisions of "2014 AIME I Problems/Problem 9"
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We claim that <math>x_2=\dfrac{1}{\sqrt{2014}}</math> | We claim that <math>x_2=\dfrac{1}{\sqrt{2014}}</math> | ||
+ | [please clarify] | ||
− | + | 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> coefficent is equal to <math>x_1x_2+x_1x_3+x_2x_3</math>. 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> | ||
Revision as of 18:55, 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 [please clarify]
By Vieta's formula we have that the coefficent is equal to and that the coefficent is equal to . 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 |
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