2002 AMC 12B Problems/Problem 6
- The following problem is from both the 2002 AMC 12B #6 and 2002 AMC 10B #10, so both problems redirect to this page.
Contents
[hide]Problem
Suppose that and are nonzero real numbers, and that the equation has solutions and . Then the pair is
Solution
Solution 1
Since , it follows by comparing coefficients that and that . Since is nonzero, , and . Thus .
Solution 2
Another method is to use Vieta's formulas. The sum of the solutions to this polynomial is equal to the opposite of the coefficient, since the leading coefficient is 1; in other words, and the product of the solutions is equal to the constant term (i.e, ). Since is nonzero, it follows that and therefore (from the first equation), . Hence,
Solution 3(Using the Answer Choices)
Note that for roots and , . This implies that is , and there is only one answer choice with in the position for , hence,
See also
2002 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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 10 Problems and Solutions |
2002 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 5 |
Followed by Problem 7 |
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.