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Difference between revisions of "2007 AIME I Problems/Problem 8"

m (incomplete solution)
 
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== Solution ==
 
== Solution ==
 
{{solution}}
 
{{solution}}
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For them both to be factors of <math>P(x)</math>, they must both share a common factor. Denote this factor as <math>m</math>, and the other two factors <math>n</math> and <math>o</math>.
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We can see that they must have a root in common for them to  both be factors of the same cubic.
  
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Therefore, <math>(x - m)(x - n) = \ldots</math> and <math>(x - m)(x - 0) = \ldots</math>.
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Let this root be <math>a</math>.
  
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<!-- solution needed -->
+
We then know that <math>a</math> is a root of
 +
\[
 +
Q_{2}(x)-2Q_{1}(x) = 2x^{2}+2kx-43x+k-2x^{2}-2kx+58x+2k = 15x+3k = 0
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\]
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, so <math>x = \frac{-k}{5}</math>.
  
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The answer is <math>k = 30</math>.
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We then know that <math>\frac{-k}{5}</math> is a root of <math>Q_{1}</math> so we get:
 +
\[
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\frac{k^{2}}{25}+(k-29)\left(\frac{-k}{5}\right)-k = 0 = k^{2}-5(k-29)(k)-25k = k^{2}-5k^{2}+145k-25k
 +
\]
 +
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>\boxed{030}</math>.
 
== See also ==
 
== See also ==
 
{{AIME box|year=2007|n=I|num-b=7|num-a=9}}
 
{{AIME box|year=2007|n=I|num-b=7|num-a=9}}

Revision as of 08:24, 15 March 2007

Problem

The polynomial $P(x)$ is cubic. What is the largest value of $k$ for which the polynomials $Q_1(x) = x^2 + (k-29)x - k$ and $Q_2(x) = 2x^2+ (2k-43)x + k$ are both factors of $P(x)$?

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it. We can see that they must have a root in common for them to both be factors of the same cubic.

Let this root be $a$.

We then know that $a$ is a root of \[ Q_{2}(x)-2Q_{1}(x) = 2x^{2}+2kx-43x+k-2x^{2}-2kx+58x+2k = 15x+3k = 0 \] , so $x = \frac{-k}{5}$.

We then know that $\frac{-k}{5}$ is a root of $Q_{1}$ so we get: \[ \frac{k^{2}}{25}+(k-29)\left(\frac{-k}{5}\right)-k = 0 = k^{2}-5(k-29)(k)-25k = k^{2}-5k^{2}+145k-25k \] or $k^{2}=30k$, so $k=30$ is the highest.

We can trivially check into the original equations to find that $k=30$ produces a root in common, so the answer is $\boxed{030}$.

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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