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 00:06, 28 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,
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
.
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.