Difference between revisions of "2014 AIME II Problems/Problem 5"
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We now move to the other two equations regarding the product of the roots. We see that we can cancel a negative from both sides to get | We now move to the other two equations regarding the product of the roots. We see that we can cancel a negative from both sides to get | ||
− | <cmath>rs(r+s) = b</cmath> | + | <cmath>rs(r+s) = b</cmath> |
− | <cmath>(r+4)(s-3)(r+s+1)=b + 240.</cmath> Subtracting the first equation from the second equation | + | <cmath>(r+4)(s-3)(r+s+1)=b + 240.</cmath> Subtracting the first equation from the second equation gives us <math>(r+4)(s-3)(r+s+1) - rs(r+s) = 240</math>. |
− | Expanding | + | Expanding, 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>. |
− | Finally, we substitute back | + | Finally, we substitute back into <math>b=rs(r+s)</math> to get <math>b = (-5)(-6)(-5-6) = -330</math>, or <math>b = (1)(9)(1 + 9) = 90</math>. |
+ | |||
+ | The answer is <math>|-330|+|90| = \boxed{420}</math>. | ||
==Solution 3== | ==Solution 3== |
Revision as of 12:54, 21 February 2021
Problem
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
The roots of are
,
, and
since they sum to
by Vieta's Formula (co-efficient of
term is
).
Similarly, the roots of are
,
, and
, as they too sum to
.
Then:
and
from
and
and
from
.
From these equations, we can write that
and simplifying gives
We now move to the other two equations regarding the product of the roots. We see that we can cancel a negative from both sides to get
Subtracting the first equation from the second equation gives us
.
Expanding, simplifying, substituting , and simplifying some more yields the simple quadratic
, so
. Then
.
Finally, we substitute back into to get
, or
.
The answer is .
Solution 3
By Vieta's, we know that the sum of roots of is
. Therefore,
the roots of
are
. By similar reasoning, the roots of
are
. Thus,
and
.
Since and
have the same coefficient for
, we can go ahead
and match those up to get
At this point, we can go ahead and compare the constant term in and
. Doing so is certainly valid, but we can actually do this another way. Notice that
. Therefore,
. If we plug that into
our expression, we get that
This tells us that
or
. Since
is the product of the roots, we have that the two possibilities are
and
. Adding the absolute values of these gives us
.
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.