Difference between revisions of "2010 AMC 12B Problems/Problem 23"
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The minimum value of <math> P(x)</math> is <math> - b</math>, and the minimum value of <math> Q(x)</math> is <math> - d</math>. Thus, our answer is <math> - (b + d) = - 100</math>, or answer <math> \boxed{\textbf{(A)}}</math>. | The minimum value of <math> P(x)</math> is <math> - b</math>, and the minimum value of <math> Q(x)</math> is <math> - d</math>. Thus, our answer is <math> - (b + d) = - 100</math>, or answer <math> \boxed{\textbf{(A)}}</math>. | ||
− | + | == Bashy Solution== | |
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==See Also== | ==See Also== | ||
{{AMC12 box|ab=B|year=2010|num-a=24|num-b=22}} | {{AMC12 box|ab=B|year=2010|num-a=24|num-b=22}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 11:39, 10 May 2020
Contents
[hide]Problem 23
Monic quadratic polynomial and have the property that has zeros at and , and has zeros at and . What is the sum of the minimum values of and ?
Solution
. Notice that has roots , so that the roots of are the roots of . For each individual equation, the sum of the roots will be (symmetry or Vieta's). Thus, we have , or . Doing something similar for gives us . We now have . Since is monic, the roots of are "farther" from the axis of symmetry than the roots of . Thus, we have , or . Adding these gives us , or . Plugging this into , we get . The minimum value of is , and the minimum value of is . Thus, our answer is , or answer .
Bashy Solution
See Also
2010 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 |
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