Difference between revisions of "2014 AIME II Problems/Problem 5"
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<cmath>\color{red}\boxed{\boxed{\color{blue}\textbf{Use the Vieta Formula!}}}</cmath> | <cmath>\color{red}\boxed{\boxed{\color{blue}\textbf{Use the Vieta Formula!}}}</cmath> | ||
− | == | + | ==Solution== |
− | Because the coefficient of <math>x^2</math> in both <math>p(x)</math> and <math>q(x)</math> is 0, the remaining root of <math>p(x)</math> is <math> | + | Because the coefficient of <math>x^2</math> in both <math>p(x)</math> and <math>q(x)</math> is 0, the remaining root of <math>p(x)</math> is <math>-(r+s)</math>, and the remaining root of <math>q(x)</math> is <math>-(r+s+1)</math>. The coefficients of <math>x</math> in <math>p(x)</math> and <math>q(x)</math> are both equal to <math>a</math>, and equating the two coefficients gives |
+ | <cmath>rs-(r+s)^2 = (r+4)(s-3)-(r+s+1)^2 </cmath>from which <math>s = \tfrac 12 (5r+13)</math>. The product of the roots of <math>p(x)</math> differs from that of <math>q(x)</math> by <math>240</math>, so<cmath>(r+4)\cdot \tfrac 12 (5r+7)\cdot \tfrac 12(7r+15)- r\cdot \tfrac 12 (5r+13)\cdot \tfrac 12(7r+13)=240</cmath>from which <math>r^2+4r-5 =0</math>, with roots <math>r=1</math> and <math>r=-5</math>. | ||
− | If <math>r | + | If <math>r = 1</math>, then the roots of <math>p(x)</math> are <math>r=1</math>, <math>s=9</math>, and <math>-(r+s)=-10</math>, and <math>b=-rst=90</math>. |
− | If <math>r | + | If <math>r = -5</math>, then the roots of <math>p(x)</math> are <math>r=-5</math>, <math>s=-6</math>, and <math>-(r+s)=11</math>, and <math>b=-rst=-330</math>. |
− | Therefore the requested sum is <math>|-330| + |90| = \boxed{420}</math>. | + | Therefore the requested sum is <math>|- 330| + |90| = \boxed{420}</math>. |
==Solution 1== | ==Solution 1== |
Revision as of 18:23, 26 January 2022
Problem
Real numbers and
are roots of
, and
and
are roots of
. Find the sum of all possible values of
.
Hint
Solution
Because the coefficient of in both
and
is 0, the remaining root of
is
, and the remaining root of
is
. The coefficients of
in
and
are both equal to
, and equating the two coefficients gives
from which
. The product of the roots of
differs from that of
by
, so
from which
, with roots
and
.
If , then the roots of
are
,
, and
, and
.
If , then the roots of
are
,
, and
, and
.
Therefore the requested sum is .
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
in both polynomials, we get:
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.