2014 AIME II: Cubic 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 $|b|$.

I assumed that the other root of p(x) would be t. From here, I used Vieta's formulas to get $-rst=b$ and $rs+st+rt=a$. Then, I assumed that the third root of q(x) would be y. Then, I used vieta's formulas to get that $-(r+4)(s-3)y=b+240$ and that $(r+4)(s-3)+y(s-3)+y(r+4)=a$. I am not sure how to proceed from here.

• Any ideas on how to proceed? – Dude156 May 31 '18 at 5:42

Let $r$, $s$, and $-r-s$ be the roots of $p(x)$ (per Vieta's). Then $r^3 + ar + b = 0$ and similarly for $s$. Also, $q(r+4) = (r+4)^3 + a(r+4) + b + 240 = 12r^2 + 48r + 304 + 4a = 0$. Set up a similar equation for $s$:
$q(s-3) = (s-3)^3 + a(s-3) + b + 240 = -9s^2 + 27s + 213 - 3a = 0.$ Simplifying and adding the equations gives $3r^2 - 3s^2 + 12r + 9s + 147 = 0$ and $r^2 - s^2 + 4r + 3s + 49 = 0 (1)$ Plugging the roots $r$, $s$, and $-r-s$ into $p(x)$ yields a long polynomial, and plugging the roots $r+4$, $s-3$, and $-1-r-s$ into $q(x)$ yields another long polynomial. Equating the coefficients of x in both polynomials: $rs + (-r-s)(r+s) = (r+4)(s-3) + (-r-s-1)(r+s+1)$, which eventually simplifies to
$s = \frac{13 + 5r}{2}$ Substitution into (1) should give $r = -5$ and $r = 1$, corresponding to $s = -6$ and $s = 9$, and $|b| = 330, 90$, for an answer of $\boxed{420}$.