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/6/b/8/6b8cfcc86d8c9c2f015fd7b9a741b0c36edde193.png)
![\[r^2 - s^2 + 4r + 3s + 49 = 0 (*)\]](http://latex.artofproblemsolving.com/c/7/a/c7a471ac457ad6f819543d98943b7d8d274d7fb1.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/a/a/1/aa1d06b275da81b54ac926c27f499bcc8a8bb908.png)
which eventually simplifies to
![\[s = \frac{13 + 5r}{2}.\]](http://latex.artofproblemsolving.com/1/f/e/1feee0bcb4227bb3e939bd670c43f597f3dcb9d4.png)
Substitution into (*) should give 
and 
, corresponding to 
and 
, and 
, for an answer of 
.
