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1995 AIME Problems/Problem 5

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Problem

For certain real values of a, b, c, and d_{}, the equation x^4+ax^3+bx^2+cx+d=0 has four non-real roots. The product of two of these roots is 13+i and the sum of the other two roots is 3+4i, where i=\sqrt{-1}. Find b.

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 m,n. Since m+n is not real, m,n are not conjugates, so the other pair of roots must be the conjugates of m,n. Let m' be the conjugate of m, and n' be the conjugate of n. Then, m\cdot n = 13 + i,m' + n' = 3 + 4i\Longrightarrow m'\cdot n' = 13 - i,m + n = 3 - 4i. By Vieta's formulas, we have that b = mm' + nn' + mn' + nm' + mn + m'n' = (m + n)(m' + n') + mn + m'n' = \boxed{051}.

See also

1995 AIME (ProblemsResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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