Difference between revisions of "2014 AIME II Problems/Problem 5"
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==Solution 1== | ==Solution 1== | ||
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 | 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) | + | <cmath>rs-(r+s)^2 = (r+4)(s-3)-(r+s+1)^2 </cmath>from which <math>s = \tfrac 12 (5r+13)</math>. |
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==Solution 2== | ==Solution 2== | ||
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==Solution 3== | ==Solution 3== | ||
− | 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 ( | + | 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 (coefficient 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>. | 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>. |
Latest revision as of 22:07, 18 August 2024
Problem
Real numbers and
are roots of
, and
and
are roots of
. Find the sum of all possible values of
.
Solution 1
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
.
Solution 2
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 3
The roots of are
,
, and
since they sum to
by Vieta's Formula (coefficient 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 4
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.