Difference between revisions of "2014 AIME II Problems/Problem 5"
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==Solution 2== | ==Solution 2== | ||
− | + | The roots of <math>p(x)</math> are <math>r</math>, <math>s</math>, and <math>-r-s</math> since they sum to <math>0</math> by Vieta's Formula (co-efficient of <math>x^2</math> term is <math>0</math>). | |
− | + | Similarly, the roots of <math>q(x)</math> are <math>r + 4</math>, <math>s - 3</math>, and <math>-r-s-1</math>, as they too sum to <math>0</math>. | |
− | We now move to the other two equations. We see that we can cancel a negative from both sides to get < | + | Then: |
+ | |||
+ | <math>a = rs+r(-r-s)+s(-r-s) = rs-(r+s)^2</math> and <math>-b = rs(-r-s)</math> from <math>p(x)</math> and | ||
+ | |||
+ | <math>a=(r+4)(s-3)+(r+4)(-r-s-1)+(s-3)(-r-s-1) = (r+4)(s-3)-(r+s+1)^2</math> and <math>-(b+240)=(r+4)(s-3)(-r-s-1)</math> from <math>q(x)</math>. | ||
+ | |||
+ | From these equations, we can write that | ||
+ | <cmath>rs-(r+s)^2 = (r+4)(s-3)-(r+s+1)^2 = a</cmath> | ||
+ | and simplifying gives | ||
+ | <cmath>2s-5r-13=0 \Rightarrow s = \frac{5r+13}{2}.</cmath> | ||
+ | |||
+ | |||
+ | |||
+ | 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> and | ||
+ | <cmath>(r+4)(s-3)(r+s+1)=b + 240.</cmath> Subtracting the first equation from the second equation yields <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>. | ||
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>. | 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>. |
Revision as of 19:19, 20 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
and
Subtracting the first equation from the second equation yields
.
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
.
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.