2021 AIME II Problem 4

Question

There are real numbers

a,

b,

c,

and

d

such that

-20

is a root of

x^3 + ax + b

and

-21

is a root of

x^3 + cx^2 + d.

These two polynomials share a complex root

m + \sqrt{n} \cdot i,

where

m

and

n

are positive integers and

i = \sqrt{-1}.

Find

m + n.

Answer

Real coefficients force non-real roots to come in conjugate pairs, so each cubic has roots consisting of its real root and

m \pm \sqrt{n}\,i

The first cubic has no

x^2

term, so its roots sum to

0;

the second has no

x

term, so its pairwise products of roots sum to

0

Both cubics have real coefficients, so their non-real roots come in conjugate pairs: the roots of the first are

-20

and

m \pm \sqrt{n}\,i,

and the roots of the second are

-21

and

m \pm \sqrt{n}\,i.

The first cubic

x^3 + ax + b

has no

x^2

term, so its roots sum to

0:

-20 + 2m = 0,

giving

m = 10.

The second cubic

x^3 + cx^2 + d

has no

x

term, so the sum of pairwise products of its roots is

0:

\begin{aligned} &(m + \sqrt{n}\,i)(m - \sqrt{n}\,i) \\ &\quad {}+ (-21)(2m) \\ &= m^2 + n - 42m = 0, \end{aligned}

so

n = 420 - 100 = 320.

Then

m + n = 10 + 320 = 330.

Problem 4 in other years