Difference between revisions of "2010 AMC 12B Problems/Problem 23"
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We now have <math> P(x) = (x + 54)^2 - b, Q(x) = (x + 19)^2 - d</math>. Since <math> Q</math> is monic, the roots of <math> Q(x) = a + \sqrt {b}</math> are "farther" from the axis of symmetry than the roots of <math> Q(x) = a - \sqrt {b}</math>. Thus, we have <math> Q( - 23) = - 54 + \sqrt {b}, Q( -21) =- 54 - \sqrt {b}</math>, or <math> 16 - d = - 54 + \sqrt {b}, 4 - d = - 54 - \sqrt {b}</math>. Adding these gives us <math> 20 - 2d = - 108</math>, or <math> d = 64</math>. Plugging this into <math> 16 - d = - 54 + \sqrt {b}</math>, we get <math> b = 36</math>. | We now have <math> P(x) = (x + 54)^2 - b, Q(x) = (x + 19)^2 - d</math>. Since <math> Q</math> is monic, the roots of <math> Q(x) = a + \sqrt {b}</math> are "farther" from the axis of symmetry than the roots of <math> Q(x) = a - \sqrt {b}</math>. Thus, we have <math> Q( - 23) = - 54 + \sqrt {b}, Q( -21) =- 54 - \sqrt {b}</math>, or <math> 16 - d = - 54 + \sqrt {b}, 4 - d = - 54 - \sqrt {b}</math>. Adding these gives us <math> 20 - 2d = - 108</math>, or <math> d = 64</math>. Plugging this into <math> 16 - d = - 54 + \sqrt {b}</math>, we get <math> b = 36</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>. | 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>. | ||
| + | |||
| + | Alternate solution at: | ||
| + | http://artofproblemsolving.com/community/c4h1256144_2010_amc_12b | ||
==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 15:05, 14 June 2016
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 
.
Alternate solution at: http://artofproblemsolving.com/community/c4h1256144_2010_amc_12b
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
