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> | + | Now, let's deal with the <math>ax</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 |
Revision as of 01:42, 30 January 2016
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.