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 $|b|.$

答案

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 $rs + st + tr = (r+4)(s-3) + (s-3)(t-1) + (t-1)(r+4),$ 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 $r - s = -8,$ then $t = r - 11 = -2r - 8,$ so $r = 1:$ the roots are $1,$ $9,$ $-10,$ and $b = 90.$ The requested sum is $330 + 90 = 420.$

2014 AIME II Problem 5

Solution:

Problem 5 in Other Years