1996 AIME Problems/Problem 5
Problem
Suppose that the roots of are , , and , and that the roots of are , , and . Find .
Video Solution
https://youtu.be/3dfbWzOfJAI?t=2785
~ pi_is_3.14
Solution 1
By Vieta's formulas on the polynomial , we have , , and . Then
This is just the definition for .
Alternatively, we can expand the expression to get
Solution 2
Each term in the expansion of has a total degree of 3. Another way to get terms with degree 3 is to multiply out . Expanding both of these expressions and comparing them shows that:
A way to realize :
(Add an extra )
.
The value of is the negation of this, which is
~Extremelysupercooldude
Solution 3
We have that for roots In the second cubic function the roots are
By Vieta's formulae, we see that As we know that the sum of the roots of the first polynomial, is by applying Vieta's again.
Using this fact, we can rewrite as
Seeing this, we can find the value of the product of the roots by applying this to the first equation. This can be done by setting so, from this, we see that we should plug in for in .
After simplifying, we get that the polynomial is Given that the product of the roots of this equation is equivalent to our desired value for , we can apply Vieta's formulae for a third time to find that
~MathWhiz35
See also
1996 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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.