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| − | == Problem ==
| + | #REDIRECT[[2002 AMC 12B Problems/Problem 6]] |
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| − | Suppose that <math>a</math> and <math>b</math> are nonzero real numbers, and that the equation <math>x^2+ax+b=0</math> has positive solutions <math>a</math> and <math>b</math>. Then the pair <math>(a,b)</math> is
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| − | <math> \mathrm{(A) \ } (-2,1)\qquad \mathrm{(B) \ } (-1,2)\qquad \mathrm{(C) \ } (1,-2)\qquad \mathrm{(D) \ } (2,-1)\qquad \mathrm{(E) \ } (4,4) </math>
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| − | == Solution ==
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| − | 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==
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| − | {{AMC10 box|year=2002|ab=B|num-b=9|num-a=11}}
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| − | [[Category:Introductory Algebra Problems]]
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