1995 AIME Problem 5

Question

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.

Answer

Because the polynomial has real coefficients, its non-real roots occur in conjugate pairs

Group the six pairwise products into the products within the two groups and the four cross-products

Let the first two roots be

\alpha

and

\beta.

Since

\alpha\beta=13+i

is not real, they are not conjugates, so the other roots are

\overline\alpha

and

\overline\beta.

Hence

\alpha+\beta=3-4i

and

\overline\alpha\,\overline\beta=13-i.

By Vieta’s formulas,

\begin{aligned}b&=\alpha\beta+\overline\alpha\,\overline\beta\\&\quad+(3-4i)(3+4i)\\&=26+25=51.\end{aligned}

Problem 5 in other years