Difference between revisions of "2007 AIME I Problems/Problem 8"
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== Problem == | == Problem == | ||
The [[polynomial]] <math>P(x)</math> is [[cubic polynomial | cubic]]. What is the largest value of <math>k</math> for which the polynomials <math>Q_1(x) = x^2 + (k-29)x - k</math> and <math>Q_2(x) = 2x^2+ (2k-43)x + k</math> are both [[factor]]s of <math>P(x)</math>? | The [[polynomial]] <math>P(x)</math> is [[cubic polynomial | cubic]]. What is the largest value of <math>k</math> for which the polynomials <math>Q_1(x) = x^2 + (k-29)x - k</math> and <math>Q_2(x) = 2x^2+ (2k-43)x + k</math> are both [[factor]]s of <math>P(x)</math>? | ||
− | |||
== Solution == | == Solution == | ||
=== Solution 1 === | === Solution 1 === | ||
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or <math>k^{2}=30k</math>, so <math>k=30</math> is the highest. | or <math>k^{2}=30k</math>, so <math>k=30</math> is the highest. | ||
− | We can trivially check into the original equations to find that <math>k=30</math> produces a root in common, so the answer is <math>030</math>. | + | We can trivially check into the original equations to find that <math>k=30</math> produces a root in common, so the answer is <math>\boxed{030}</math>. |
=== Solution 2 === | === Solution 2 === | ||
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Therefore, we can write four equations (and we have four [[variable]]s), <math>a + m = 29 - k</math>, <math>a + n = \frac{43}{2} - k</math>, <math>am = -k</math>, and <math>an = \frac{k}{2}</math>. | Therefore, we can write four equations (and we have four [[variable]]s), <math>a + m = 29 - k</math>, <math>a + n = \frac{43}{2} - k</math>, <math>am = -k</math>, and <math>an = \frac{k}{2}</math>. | ||
− | The first two equations show that <math>m - n = 29 - \frac{43}{2} = \frac{15}{2}</math>. The last two equations show that <math>\frac{m}{n} = -2</math>. Solving these show that <math>m = 5</math> and that <math>n = -\frac{5}{2}</math>. Substituting back into the equations, we eventually find that <math>k = | + | The first two equations show that <math>m - n = 29 - \frac{43}{2} = \frac{15}{2}</math>. The last two equations show that <math>\frac{m}{n} = -2</math>. Solving these show that <math>m = 5</math> and that <math>n = -\frac{5}{2}</math>. Substituting back into the equations, we eventually find that <math>k = \boxed{030}</math>. |
+ | |||
+ | === Solution 3 === | ||
+ | |||
+ | Since <math>Q_1(x)</math> and <math>Q_2(x)</math> are both factors of <math>P(x)</math>, which is cubic, we know the other factors associated with each of <math>Q_1(x)</math> and <math>Q_2(x)</math> must be linear. Let <math>Q_1(x)R(x) = Q_2(x)S(x) = P(x)</math>, where <math>R(x) = ax + b</math> and <math>S(x) = cx + d</math>. Then we have that <math>((x^2 + (k-29)x - k))(ax + b) = ((2x^2+ (2k-43)x + k))(cx + d)</math>. Equating coefficients, we get the following system of equations: | ||
+ | |||
+ | <cmath>\begin{align} | ||
+ | a = 2c \\ | ||
+ | b = -d \\ | ||
+ | 2c(k - 29) - d = c(2k - 43) + 2d \\ | ||
+ | -d(k - 29) - 2ck = d(2k - 43) + ck | ||
+ | \end{align}</cmath> | ||
+ | |||
+ | Using equations <math>(1)</math> and <math>(2)</math> to make substitutions into equation <math>(3)</math>, we see that the <math>k</math>'s drop out and we're left with <math>d = -5c</math>. Substituting this expression for <math>d</math> into equation <math>(4)</math> and solving, we see that <math>k</math> must be <math>\boxed {030}</math>. | ||
+ | |||
+ | ~ anellipticcurveoverq | ||
+ | |||
+ | === Solution 4 === | ||
+ | |||
+ | Notice that if the roots of <math>Q_1(x)</math> and <math>Q_2(x)</math> are all distinct, then <math>P(x)</math> would have four distinct roots, which is a contradiction since it's cubic. Thus, <math>Q_1(x)</math> and <math>Q_2(x)</math> must share a root. Let this common value be <math>r.</math> | ||
+ | |||
+ | Thus, we see that we have | ||
+ | <cmath>r^2 + (k - 29)r - k = 0,</cmath> | ||
+ | <cmath>2r^2 + (2k - 43)r + k = 0.</cmath> Adding the two equations gives us <cmath>3r^2 + (3k - 72)r = 0 \implies r = 0, 24 - k.</cmath> Now, we have two cases to consider. If <math>r = 0,</math> then we have that <math>Q_1(r) = 0 = r^2 + (k - 29)r - k \implies k = 0.</math> On the other hand, if <math>r = 24 - k,</math> we see that <cmath>Q_1(r) = 0 = (24 - k)^2 + (k - 29)(24 - k) - k \implies k = \boxed{030}.</cmath> This can easily be checked to see that it does indeed work, and we're done! | ||
+ | |||
+ | ~Ilikeapos | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=bsRQZwO7n84&t=64s | ||
== See also == | == See also == |
Latest revision as of 00:11, 29 November 2023
Contents
Problem
The polynomial is cubic. What is the largest value of
for which the polynomials
and
are both factors of
?
Solution
Solution 1
We can see that and
must have a root in common for them to both be factors of the same cubic.
Let this root be .
We then know that is a root of
, so
.
We then know that is a root of
so we get:
or
, so
is the highest.
We can trivially check into the original equations to find that produces a root in common, so the answer is
.
Solution 2
Again, let the common root be ; let the other two roots be
and
. We can write that
and that
.
Therefore, we can write four equations (and we have four variables), ,
,
, and
.
The first two equations show that . The last two equations show that
. Solving these show that
and that
. Substituting back into the equations, we eventually find that
.
Solution 3
Since and
are both factors of
, which is cubic, we know the other factors associated with each of
and
must be linear. Let
, where
and
. Then we have that
. Equating coefficients, we get the following system of equations:
Using equations and
to make substitutions into equation
, we see that the
's drop out and we're left with
. Substituting this expression for
into equation
and solving, we see that
must be
.
~ anellipticcurveoverq
Solution 4
Notice that if the roots of and
are all distinct, then
would have four distinct roots, which is a contradiction since it's cubic. Thus,
and
must share a root. Let this common value be
Thus, we see that we have
Adding the two equations gives us
Now, we have two cases to consider. If
then we have that
On the other hand, if
we see that
This can easily be checked to see that it does indeed work, and we're done!
~Ilikeapos
Video Solution
https://www.youtube.com/watch?v=bsRQZwO7n84&t=64s
See also
2007 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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