2014 AIME II Problem 5

Question

Real numbers

r

and

s

are roots of

p(x) = x^3 + ax + b,

and

r + 4

and

s - 3

are roots of

q(x) = x^3 + ax + b + 240.

Find the sum of all possible values of

|b|.

Answer

Answer: 420

Concepts:polynomial Vieta’s Formulas system of equations

Difficulty rating: 2560

Small hint:

Neither cubic has an

x^2

term, so the third roots are

t = -r - s

and

t - 1.

Equate the coefficients of

x

in the two cubics.

Big hint:

Comparing constant terms and substituting

t = 4r - 3s + 13

yields

(r-s)^2 + 7(r-s) - 8 = 0;

finish each case using

r + s + t = 0.

Solution:

Both cubics have zero

x^2

coefficient, so their roots sum to

0:

the third root of

p

is

t = -r - s,

and the third root of

q

is

-(r+4) - (s-3) = t - 1.

The coefficient of

x

is

a

in both, so

\begin{aligned} &rs + st + tr \\ &= (r+4)(s-3) \\ &\quad {}+ (s-3)(t-1) \\ &\quad {}+ (t-1)(r+4), \end{aligned}

which simplifies to

t = 4r - 3s + 13.

The constant terms give

b = -rst

and

b + 240 =

-(r+4)(s-3)(t-1),

so

240 =

rst - (r+4)(s-3)(t-1),

i.e.

rs - 4st + 3tr - 3r

+ 4s + 12t - 252 = 0.

Substituting

t = 4r - 3s + 13

reduces this to

12\left[(r-s)^2 + 7(r-s) - 8\right] = 0,

so

r - s = 1

or

r - s = -8.

If

r - s = 1,

then

t = 4r - 3s + 13 = r + 16

and

t = -r - s = -2r + 1,

so

r = -5:

the roots are

-5,

-6,

11,

and

b = -rst = -330.

If

r - s = -8,