2010 AMC 12B Problems/Problem 23
Contents
Problem
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 1
. 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 .
Solution 2 (Bash)
Let and .
Then is , which simplifies to:
We can find by simply doing and to get:
The sum of the zeros of is . From Vieta, the sum is . Therefore, .
The sum of the zeros of is . From Vieta, the sum is . Therefore, .
Plugging in, we get:
Let's tackle the coefficients, which is the sum of the six double-products possible. Since gives the sum of these six double products of the roots of , we have:
Similarly with , we get:
Thus, our polynomials are and .
The minimum value of happens at , and is .
The minimum value of happens at , and is .
The sum of these minimums is . -srisainandan6
Solution 3 (Mild Bash)
Let and . Notice that the roots of are and the roots of are Then we get:
The two possible equations are then and . The solutions are . From Vieta's we know that the total sum so the roots are paired and . Let and .
We can similarly get that and , and . Add the first two equations to get This means .
Once more, we can similarly obtain Therefore .
Now we can find the minimums to be and Summing, the answer is
~Leonard_my_dude~
Solution 4
Let , .
Notice how the coefficient for has to be the same for the two quadratics that are multiplied to create , and .
, , , , ,
, , ,
,
.
See Also
2010 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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