1995 AIME Problems/Problem 5
Contents
Problem
For certain real values of and the equation has four non-real roots. The product of two of these roots is and the sum of the other two roots is where Find
Solution 1
Since the coefficients of the polynomial are real, it follows that the non-real roots must come in complex conjugate pairs. Let the first two roots be . Since is not real, are not conjugates, so the other pair of roots must be the conjugates of . Let be the conjugate of , and be the conjugate of . Then, By Vieta's formulas, we have that .
Solution 2
Let's assume that the 2 roots multiplied together are p+qi, and r+si, and the two roots added together are the conjugates of the previous roots. Using Vieta, we get
We are now stuck. We can't simplify further. But, we look back to the problem and see that the two roots that are multiplied together give a product of , and the two roots that are added give . This gets three equations necessary for solving the problem. So, alright. Let's use the first equation to get that , and substitute that in. Now, the equation becomes:
We wish that we can turn the 2pr into 2qs. Fortunately, we can do that. By using the second equation, we can manipulate it to be , and substitute that in.
We can square both sides of the third equation, and get We substitute that in and we get
- AlexLikeMath
- Corrections by VSPuzzler
- small correction be Marshall_Huang
See also
1995 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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