2008 AIME II Problems/Problem 7
Let , , and be the three roots of the equation Find .
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
By Vieta's formulas, we have , and so the desired answer is . Additionally, using the factorization we have that . By Vieta's again,
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
Expanding, you get: This looks similar to Substituting: Since , Substituting, we get or, We are trying to find . Substituting:
Write and let . Then Solving for and negating the result yields the answer
Here by Vieta's formulas: --(1)
By the factorisation formula: Let , , , (By (1))
Let us 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
|2008 AIME II (Problems • Answer Key • Resources)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15|
|All AIME Problems and Solutions|