Difference between revisions of "2014 AIME II Problems/Problem 5"
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==Solution 2== | ==Solution 2== | ||
− | As above, we know from Vieta's that the roots of <math>p(x)</math> are <math>r</math>, <math>s</math>, and <math>-r-s</math>. Similarly, the roots of <math>q(x)</math> are <math>r + 4</math>, <math>s - 3</math>, and <math>-r-s-1</math>. Then <math>rs+r(-r-s)+s(-r-s) = rs-(r+s)^2 = a</math> and <math>rs(-r-s) = -b</math> from <math>p(x)</math> and <math>(r+4)(s-3)+(r+4)(-r-s-1)+(s-3)(-r-s-1) = (r+4)(s-3)-(r+s+1)^2 = a</math> and <math>(r+4)(s-3)(-r-s-1)=b</math> from <math>q(x)</math>. | + | As above, we know from Vieta's that the roots of <math>p(x)</math> are <math>r</math>, <math>s</math>, and <math>-r-s</math>. Similarly, the roots of <math>q(x)</math> are <math>r + 4</math>, <math>s - 3</math>, and <math>-r-s-1</math>. Then <math>rs+r(-r-s)+s(-r-s) = rs-(r+s)^2 = a</math> and <math>rs(-r-s) = -b</math> from <math>p(x)</math> and <math>(r+4)(s-3)+(r+4)(-r-s-1)+(s-3)(-r-s-1) = (r+4)(s-3)-(r+s+1)^2 = a</math> and <math>(r+4)(s-3)(-r-s-1)=-(b + 240)</math> from <math>q(x)</math>. |
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+ | From these equations, we can write that <math>rs-(r+s)^2 = (r+4)(s-3)-(r+s+1)^2</math>, and simplifying gives us <math>2s-5r-13=0</math> or <math>s = \frac{5r+13}{2}</math>. | ||
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+ | We know move to the other two equations. We see that we can cancel a negative from both sides to get <math>rs(r+s) = b</math> and <math>(r+4)(s-3)(r+s+1)=b + 240</math>. Subtracting the first from the second equation gives us <math>(r+4)(s-3)(r+s+1) - rs(r+s) = 240</math>. Expanding and simplifying, substituting <math>s = \frac{5r+13}{2}</math> and simplifying some more yields the simple quadratic <math>r^2 + 4r - 5 = 0</math>, so <math>r = -5, 1</math>. Then <math>s = -6, 9</math>. | ||
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+ | Finally, we substitute back in to get <math>b = (-5)(-6)(-5-6) = -330</math> or <math>b = (1)(9)(1 + 9) = 90</math>. Then the answer is <math>|-330|+|90| = \boxed{420}</math>. | ||
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== See also == | == See also == | ||
{{AIME box|year=2014|n=II|num-b=4|num-a=6}} | {{AIME box|year=2014|n=II|num-b=4|num-a=6}} |
Revision as of 19:56, 12 March 2016
Contents
[hide]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
.
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 know 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.