Difference between revisions of "2014 AIME II Problems/Problem 5"
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==Solution== | ==Solution== | ||
| − | Let r, s, -r-s be the roots of p(x) (per Vieta's). Then <math>r^3 + ar + b = 0</math> and similarly for s. Also, | + | Let <math>r</math>, <math>s</math>, <math>-r</math>, and <math>-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> | ||
| − | Set up a similar equation for s: | + | Set up a similar equation for <math>s</math>: |
<cmath>q(s-3) = (s-3)^3 + a(s-3) + b + 240 = -9s^2 + 27s + 213 - 3a = 0.</cmath> | <cmath>q(s-3) = (s-3)^3 + a(s-3) + b + 240 = -9s^2 + 27s + 213 - 3a = 0.</cmath> | ||
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<cmath>r^2 - s^2 + 4r + 3s + 49 = 0 (*)</cmath> | <cmath>r^2 - s^2 + 4r + 3s + 49 = 0 (*)</cmath> | ||
| − | Now, let's deal with the a*x. Equating the a in both equations (per Vieta) | + | Now, let's deal with the <math>a*x</math>. Equating the a in both equations (per Vieta) |
<cmath>rs + (-r-s)(r+s) = (r+4)(s-3) + (-r-s-1)(r+s+1),</cmath> | <cmath>rs + (-r-s)(r+s) = (r+4)(s-3) + (-r-s-1)(r+s+1),</cmath> | ||
which eventually simplifies to | which eventually simplifies to | ||
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Substitution into (*) should give <math>r = -5</math> and <math>r = 1</math>, corresponding to <math>s = -6</math> and <math>s = 9</math>, and <math>|b| = 330, 90</math>, for an answer of <math>\boxed{420}</math>. | Substitution into (*) should give <math>r = -5</math> and <math>r = 1</math>, corresponding to <math>s = -6</math> and <math>s = 9</math>, and <math>|b| = 330, 90</math>, for an answer of <math>\boxed{420}</math>. | ||
| + | |||
| + | == See also == | ||
| + | {{AIME box|year=2014|n=II|num-b=4|num-a=6}} | ||
| + | |||
| + | [[Category:Intermediate Algebra Problems]] | ||
Revision as of 21:30, 20 May 2014
Problem 5
Real numbers 
and 
are roots of 
, and 
and 
are roots of 
. Find the sum of all possible values of 
.
Solution
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 
. 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 
.
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 | ||
