2004 AMC 12B Problems/Problem 23
Contents
Problem
The polynomial has integer coefficients and three distinct positive zeros. Exactly one of these is an integer, and it is the sum of the other two. How many values of are possible?
Solution 1
Let the roots be , and let . Then
and by matching coefficients, . Then our polynomial looks like and we need the number of possible products .
Since and , it follows that , with the endpoints not achievable because the roots must be distinct. Because cannot be an integer, there are possible values of .
Solution 2
Letting the roots be , , and , where , we see that by Vieta's Formula's, , and so . Therefore, is a factor of . Letting gives that because . Letting and noting that for some , we see that is the sum of the roots of , and , and so . Now, we have that has roots and , and we wish to find the number of possible values of . By the quadratic formula, we see that are the two values of noninteger positive real numbers and , neither of which is equal to . This information gives us that , and so since is evidently not a square, we have possible values of .
See also
2004 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.