2014 AIME II Problems/Problem 5
Problem 5
Real numbers 
and 
are roots of 
, and 
and 
are roots of 
. Find the sum of all possible values of 
.
Solution
Let r, s, -r-s be the roots of p(x) (per Vieta's). Then 
and similarly for s. Also, ![\[q(r+4) = (r+4)^3 + a(r+4) + b + 240 = 12r^2 + 48r + 304 + 4a = 0\]](http://latex.artofproblemsolving.com/b/5/1/b519125521b0a1ac6c2dd3381dab2e2c664b6b7c.png)
Set up a similar equation for s: ![\[q(s-3) = (s-3)^3 + a(s-3) + b + 240 = -9s^2 + 27s + 213 - 3a = 0\]](http://latex.artofproblemsolving.com/0/3/4/034672706f6a607dd7cd2d61f24a6e64cea3a61a.png)
. Simplifying and adding the equations gives ![\[3r^2 - 3s^2 + 12r + 9s + 147 = 0 (*)\]](http://latex.artofproblemsolving.com/e/0/c/e0cbe38522438f2bbd4b17d7cea6bf793ae9280b.png)
Now, let's deal with the a*x. Equating the a in both equations (per Vieta) ![\[rs + (-r-s)(r+s) = (r+4)(s-3) + (-r-s-1)(r+s+1)\]](http://latex.artofproblemsolving.com/5/1/0/51078910683e93bfd6c2dad4a38ec3987a6b3521.png)
, which eventually simplifies to ![\[s = \frac{13 + 5r}{2}\]](http://latex.artofproblemsolving.com/0/1/f/01feb9ac5c07ed2951e38eb46807fc301f8964f4.png)
Substitution into (*) should give r = -5 and r = 1, corresponding to s = -6 and s = 9, and |b| = 330 and 90, for an answer of 
.
