Difference between revisions of "2014 AIME II Problems/Problem 5"
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<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Real numbers <math>r</math> and <math>s</math> are roots of <math>p(x)=x^3+ax+b</math>, and <math>r+4</math> and <math>s-3</math> are roots of <math>q(x)=x^3+ax+b+240</math>. Find the sum of all possible values of <math>|b|</math>.<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude> | <!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Real numbers <math>r</math> and <math>s</math> are roots of <math>p(x)=x^3+ax+b</math>, and <math>r+4</math> and <math>s-3</math> are roots of <math>q(x)=x^3+ax+b+240</math>. Find the sum of all possible values of <math>|b|</math>.<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude> | ||
| − | ==Solution== | + | ==Solution 1== |
Let <math>r</math>, <math>s</math>, and <math>-r-s</math> be the roots of <math>p(x)</math> (per Vieta's). Then <math>r^3 + ar + b = 0</math> and similarly for <math>s</math>. Also, | Let <math>r</math>, <math>s</math>, and <math>-r-s</math> be the roots of <math>p(x)</math> (per Vieta's). Then <math>r^3 + ar + b = 0</math> and similarly for <math>s</math>. Also, | ||
<cmath>q(r+4) = (r+4)^3 + a(r+4) + b + 240 = 12r^2 + 48r + 304 + 4a = 0</cmath> | <cmath>q(r+4) = (r+4)^3 + a(r+4) + b + 240 = 12r^2 + 48r + 304 + 4a = 0</cmath> | ||
Revision as of 23:06, 27 December 2016
Contents
Problem 5
Real numbers 
and 
are roots of 
, and 
and 
are roots of 
. Find the sum of all possible values of 
.
Solution 1
Let , , and 
be the roots of 
(per Vieta's). Then 
and similarly for . 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 :
![\[q(s-3) = (s-3)^3 + a(s-3) + b + 240 = -9s^2 + 27s + 213 - 3a = 0.\]](http://latex.artofproblemsolving.com/8/d/8/8d8ede3c0cf58050757bb900e13139da55c52043.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 
terms. Plugging the roots , , and into yields a long polynomial, and plugging the roots , , and 
into 
yields another long polynomial. Equating the coefficients of x in both polynomials:
![\[rs + (-r-s)(r+s) = (r+4)(s-3) + (-r-s-1)(r+s+1),\]](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 
.
Solution 2
As above, we know from Vieta's that the roots of are , , and . Similarly, the roots of are 
, 
, and 
. Then 
and 
from and 
and 
from .
From these equations, we can write that 
, and simplifying gives us 
or 
.
We now move to the other two equations. We see that we can cancel a negative from both sides to get 
and 
. Subtracting the first from the second equation gives us 
. Expanding and simplifying, substituting and simplifying some more yields the simple quadratic 
, so 
. Then 
.
Finally, we substitute back in to get 
or 
. Then the answer is 
.
See also
| 2014 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
