Difference between revisions of "2014 AIME II Problems/Problem 5"
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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 <math>a*x</math> terms. Plugging the roots <math>r</math>, <math>s</math>, and <math>-r-s</math> into <math>p(x)</math> yields a long polynomial, and plugging the roots <math>r+4</math>, <math>s-3</math>, and <math>-1-r-s</math> into <math>q(x) yields another long polynomial. Equating the coefficients of x in both polynomials: | + | Now, let's deal with the <math>a*x</math> terms. Plugging the roots <math>r</math>, <math>s</math>, and <math>-r-s</math> into <math>p(x)</math> yields a long polynomial, and plugging the roots <math>r+4</math>, <math>s-3</math>, and <math>-1-r-s</math> into <math>q(x)</math> yields another long polynomial. Equating the coefficients of x in both polynomials: |
<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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<cmath>s = \frac{13 + 5r}{2}.</cmath> | <cmath>s = \frac{13 + 5r}{2}.</cmath> | ||
− | Substitution into (*) should give < | + | 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 == | == See also == |
Revision as of 16:03, 9 February 2015
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,
Set up a similar equation for :
Simplifying and adding the equations gives
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: which eventually simplifies to
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.