Difference between revisions of "2002 AMC 10B Problems/Problem 10"
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== Solution == | == Solution == | ||
| − | {{ | + | From [[Vieta's Formulas]], <math>ab=b</math> and <math>a+b=-a</math>. Since <math>b\ne 0</math>, we have <math>a=1</math>, and hence <math>b=-2</math>. Our answer is <math>\boxed{(1,-2)\Rightarrow\text{(C)}}</math>. |
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| + | ==See Also== | ||
| + | {{AMC10 box|year=2002|ab=B|num-b=9|num-a=11}} | ||
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
Revision as of 14:11, 27 December 2008
Problem
Suppose that 
and 
are nonzero real numbers, and that the equation 
has positive solutions and . Then the pair 
is
Solution
From Vieta's Formulas, 
and 
. Since 
, we have 
, and hence 
. Our answer is 
.
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 | ||
