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

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We then know that <math>a</math> is a root of
 
We then know that <math>a</math> is a root of
\[
+
<math>
 
Q_{2}(x)-2Q_{1}(x) = 2x^{2}+2kx-43x+k-2x^{2}-2kx+58x+2k = 15x+3k = 0
 
Q_{2}(x)-2Q_{1}(x) = 2x^{2}+2kx-43x+k-2x^{2}-2kx+58x+2k = 15x+3k = 0
\]
+
</math>
 
, so <math>x = \frac{-k}{5}</math>.
 
, so <math>x = \frac{-k}{5}</math>.
  
 
We then know that <math>\frac{-k}{5}</math> is a root of <math>Q_{1}</math> so we get:
 
We then know that <math>\frac{-k}{5}</math> is a root of <math>Q_{1}</math> so we get:
\[
+
<math>
 
\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
 
\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
\]
+
</math>
 
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>\boxed{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>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 09:26, 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 $030$.

See also

2007 AIME I (ProblemsAnswer KeyResources)
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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