2008 AIME II Problems/Problem 7
Contents
Problem
Let , , and be the three roots of the equation Find .
Video Solution by Punxsutawney Phil
https://www.youtube.com/watch?v=6mYZYh9gJBs
Solution 1
By Vieta's formulas, we have so Substituting this into our problem statement, our desired quantity is Also by Vieta's formulas we have so negating both sides and multiplying through by 3 gives our answer of
Solution 2
By Vieta's formulas, we have , and so the desired answer is . Additionally, using the factorization we have that . By Vieta's again,
Solution 3
Vieta's formulas gives . Since is a root of the polynomial, , and the same can be done with . Therefore, we have yielding the answer .
Also, Newton's Sums yields an answer through the application. http://www.artofproblemsolving.com/Wiki/index.php/Newton's_Sums
Solution 4
Expanding, you get: This looks similar to Substituting: Since , Substituting, we get or, We are trying to find . Substituting:
Solution 5
Write and let . Then Solving for and negating the result yields the answer
Solution 6
Here by Vieta's formulas: --(1)
--(2)
By the factorisation formula: Let , , , (By (1))
So
Solution 7
Let's construct a polynomial with the roots and .
sum of the roots:
pairwise product of the roots:
product of the roots:
thus, the polynomial we get is
as and are roots of this polynomial, we know that (using power reduction)
adding all of the equations up, we see that
Solution 8
We want to find what is which reminds us of Newton sum. So we can see that Notice that so it is just , the desired answer is
~bluesoul
Solution 9
This solution uses Vietas, as with everyone else's solution. Expanding the expression we get
Seeing the cubes, we try to find a and upon doing so, we get
Recall that . Thus, we get
Plugging in we get
~firebolt360
Solution 10
We want to find Let's call this result n.
From vieta's formulas, we find that , , and
Expanding and rearranging gives us
Let
Solving gives us
Therefore, the answer is
Also note that this is the only solution that still would have worked effectively if was nonzero.
See also
2008 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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.