Difference between revisions of "2002 AMC 12B Problems/Problem 6"
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Another method is to use [[Vieta's formulas]]. The sum of the solutions to this polynomial is equal to the opposite of the <math>x</math> coefficient, since the leading coefficient is 1; in other words, <math>a + b = -a</math> and the product of the solutions is equal to the constant term (i.e, <math>a*b = b</math>). Since <math>b</math> is nonzero, it follows that <math>a = 1</math> and therefore (from the first equation), <math>b = -2a = -2</math>. Hence, <math>(a,b) = \boxed{\mathrm{(C)}\ (1,-2)}</math> | Another method is to use [[Vieta's formulas]]. The sum of the solutions to this polynomial is equal to the opposite of the <math>x</math> coefficient, since the leading coefficient is 1; in other words, <math>a + b = -a</math> and the product of the solutions is equal to the constant term (i.e, <math>a*b = b</math>). Since <math>b</math> is nonzero, it follows that <math>a = 1</math> and therefore (from the first equation), <math>b = -2a = -2</math>. Hence, <math>(a,b) = \boxed{\mathrm{(C)}\ (1,-2)}</math> | ||
− | ===Solution 3(Using the Answer Choices)=== | + | ===Solution 3 (Using the Answer Choices)=== |
Note that for roots <math>a</math> and <math>b</math>, <math>ab = b</math>. This implies that <math>a</math> is <math>1</math>, and there is only one answer choice with <math>1</math> in the position for <math>a</math>, hence, <math>(a,b) = \boxed{\mathrm{(C)}\ (1,-2)}</math> | Note that for roots <math>a</math> and <math>b</math>, <math>ab = b</math>. This implies that <math>a</math> is <math>1</math>, and there is only one answer choice with <math>1</math> in the position for <math>a</math>, hence, <math>(a,b) = \boxed{\mathrm{(C)}\ (1,-2)}</math> | ||
Revision as of 18:49, 17 July 2018
- 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.