2014 AIME I Problems/Problem 9
Contents
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 and we're done.
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
Solution 5
Let The original equation simplifies to Here we clearly see that is a root. Dividing from the sum we find that From simple bounding we see that is the middle root. Therefore
Solution 6
occurs multiple times, so let k = .
The equation becomes . Since we want to relate k and x, we should solve for one of them. We can't solve for x, since that would require the cubic formula, so we solve for k, and express it in terms of a quadratic, and apply the quadratic formula.
We get the roots are:
, and .
In the first case, .
In the second case, . The solutions are . The sum of these 2 solutions is , and is the middle solution, and thus,
Solution 7 (Estimation and Intuitive Function Analysis)
We will estimate the roots of the polynomial. (This strategy normally doesn't work on AIME #9, but playing around with a function is often good strategy for getting an intuition for the problem. In this problem, estimation happens to be a valid solution path. It isn't a proof, but given the constraint that the answer is an integer, we can be certain that our answer is correct.)
Let . We start by estimating , , and (A natural first step for function analysis.):
We conclude by Intermediate Value Theorem (or just common sense), that there is a root on and another root on .
We know that and that . We conclude that the third root is on . Therefore, , , and .
We will estimate . For , the constant term of is negligible. We simplify and get . Solving for (We can divide by because we know ), we get . We can intuitively bound between and .
We will now estimate and . and are close to . As a result, the term is negligible. We simplify and get . Solving for , we get . We can intuitively bound between and . Similarly, we can intuitively bound between and .
We calculate that the minimum possible value of is and the maximum possible value is . The only integer that falls is this range is .
~numerophile
Solution 8
Let . We have , so or . As such, the solutions are at and . Note that , so . ( by vietas.)
-zhoujef000
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.