Difference between revisions of "2014 AIME I Problems/Problem 9"
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Substituting <math>n</math> for <math>2014</math>, we get <math>\sqrt{n}x^3 - (1+2n)x^2 + 2 = \sqrt{n}x^3 - x^2 - 2nx^2 + 2 = x^2(\sqrt{n}x - 1) - 2(nx^2 - 1) = 0</math>. Noting that <math>nx^2 - 1</math> factors as a difference of squares to <math>(\sqrt{n}x - 1)(\sqrt{n}x+1)</math>, we can factor the left side as <math>(\sqrt{n}x - 1)(x^2 - 2(\sqrt{n}x+1))</math>. This means that <math>\frac{1}{\sqrt{n}}</math> is a root, and the other two roots are the roots of <math>x^2 - 2\sqrt{n}x - 2</math>. Note that the constant term of the quadratic is negative, so one of the two roots is positive and the other is negative. In addition, by Vieta's Formulas, the roots sum to <math>2\sqrt{n}</math>, so the positive root must be greater than <math>2\sqrt{n}</math> in order to produce this sum when added to a negative value. Since <math>0 < \frac{1}{\sqrt{2014}} < 2\sqrt{2014}</math> is clearly true, <math>x_2 = \frac{1}{\sqrt{2014}}</math> and <math>x_1 + x_3 = 2\sqrt{2014}</math>. Multiplying these values together, we find that <math>x_2(x_1+x_3) = \boxed{002}</math>. | Substituting <math>n</math> for <math>2014</math>, we get <math>\sqrt{n}x^3 - (1+2n)x^2 + 2 = \sqrt{n}x^3 - x^2 - 2nx^2 + 2 = x^2(\sqrt{n}x - 1) - 2(nx^2 - 1) = 0</math>. Noting that <math>nx^2 - 1</math> factors as a difference of squares to <math>(\sqrt{n}x - 1)(\sqrt{n}x+1)</math>, we can factor the left side as <math>(\sqrt{n}x - 1)(x^2 - 2(\sqrt{n}x+1))</math>. This means that <math>\frac{1}{\sqrt{n}}</math> is a root, and the other two roots are the roots of <math>x^2 - 2\sqrt{n}x - 2</math>. Note that the constant term of the quadratic is negative, so one of the two roots is positive and the other is negative. In addition, by Vieta's Formulas, the roots sum to <math>2\sqrt{n}</math>, so the positive root must be greater than <math>2\sqrt{n}</math> in order to produce this sum when added to a negative value. Since <math>0 < \frac{1}{\sqrt{2014}} < 2\sqrt{2014}</math> is clearly true, <math>x_2 = \frac{1}{\sqrt{2014}}</math> and <math>x_1 + x_3 = 2\sqrt{2014}</math>. Multiplying these values together, we find that <math>x_2(x_1+x_3) = \boxed{002}</math>. |
Revision as of 19:15, 14 March 2014
Problem 9
Let be the three real roots of the equation . Find .
Solution
Substituting for , we get . Noting that factors as a difference of squares to , we can factor the left side as . This means that is a root, and the other two roots are the roots of . Note that the constant term of the quadratic is negative, so one of the two roots is positive and the other is negative. In addition, by Vieta's Formulas, the roots sum to , so the positive root must be greater than in order to produce this sum when added to a negative value. Since is clearly true, and . Multiplying these values together, we find that .
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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