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>. | ||
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| + | ==Solution 2== | ||
== See also == | == See also == | ||
{{AIME box|year=2014|n=I|num-b=8|num-a=10}} | {{AIME box|year=2014|n=I|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 12:55, 15 March 2014
Contents
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 
.
Solution 2
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. 
