Difference between revisions of "2014 AIME I Problems/Problem 9"
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==Solution 4== | ==Solution 4== | ||
By Vieta's, we are seeking to find <math>x_2(x_1+x_3)=x_1x_2+x_2x_3=-x_1x_3=\frac{2}{\sqrt{2014}x_2}</math>. Substitute <math>n=-x_1x_3</math> and <math>x_2=\frac{2}{\sqrt{2014}n}</math>. Substituting this back into the original equation, we have <math>\frac{4}{1007n^3}-\frac{8058}{1007n^2}+2=0</math>, so <math>2n^3-\frac{8058}{1007}n+\frac{4}{1007}=2n^3-\frac{8058n-4}{1007}=0</math>. Hence, <math>8058n-4\equiv 2n-4 \equiv 0 \pmod{1007}</math>, and <math>n\equiv 2\pmod{1007}</math>. But since <math>n\le 999</math> because it is our desired answer, the only possible value for <math>n</math> is <math>\boxed{002}</math> | By Vieta's, we are seeking to find <math>x_2(x_1+x_3)=x_1x_2+x_2x_3=-x_1x_3=\frac{2}{\sqrt{2014}x_2}</math>. Substitute <math>n=-x_1x_3</math> and <math>x_2=\frac{2}{\sqrt{2014}n}</math>. Substituting this back into the original equation, we have <math>\frac{4}{1007n^3}-\frac{8058}{1007n^2}+2=0</math>, so <math>2n^3-\frac{8058}{1007}n+\frac{4}{1007}=2n^3-\frac{8058n-4}{1007}=0</math>. Hence, <math>8058n-4\equiv 2n-4 \equiv 0 \pmod{1007}</math>, and <math>n\equiv 2\pmod{1007}</math>. But since <math>n\le 999</math> because it is our desired answer, the only possible value for <math>n</math> is <math>\boxed{002}</math> | ||
− | + | BEST PROOOFFFF | |
Stormersyle & mathleticguyyy | Stormersyle & mathleticguyyy | ||
Revision as of 14:58, 12 June 2020
Problem 9
Let be the three real roots of the equation . Find .
Solution 1
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
From Vieta's formulae, we know that and Thus, we know that .
Now consider the polynomial with roots and . Expanding the polynomial , we get the polynomial Substituting the values obtained from Vieta's formulae, we find that this polynomial is . We know is a root of this polynomial, so we set it equal to 0 and simplify the resulting expression to .
Given the problem conditions, we know there must be at least 1 integer solution, and that it can't be very large (because the term quickly gets much larger/smaller than the other 2). Trying out some numbers, we quickly find that is a solution. Factoring it out, we get that . Since the other quadratic factor clearly does not have any integer solutions and since the AIME has only positive integer answers, we know that this must be the answer they are looking for. Thus, , so .
Solution 3
Observing the equation, we notice that the coefficient for the middle term is equal to . Also notice that the coefficient for the term is . Therefore, if the original expression was to be factored into a linear binomial and a quadratic trinomial, the term of the binomial would have a coefficient of . Similarly, the term of the trinomial would also have a coefficient of . The factored form of the expression would look something like the following: where are all positive integers (because the term of the original expression is negative, and the constant term is positive), and .
Multiplying this expression out gives . Equating this with the original expression gives . The only positive integer solutions of this expression is or . If then setting yields and therefore which clearly isn't equal to as the constant term. Therefore, and the factored form of the expression is: . Therefore, one of the three roots of the original expression is . Using the quadratic formula yields the other two roots as and . Arranging the roots in ascending order (in the order ), . Therefore, .
Solution 4
By Vieta's, we are seeking to find . Substitute and . Substituting this back into the original equation, we have , so . Hence, , and . But since because it is our desired answer, the only possible value for is BEST PROOOFFFF Stormersyle & mathleticguyyy
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.