1995 AIME Problems/Problem 5
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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
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
.
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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