Difference between revisions of "2007 AIME I Problems/Problem 8"
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== Solution == | == Solution == | ||
{{solution}} | {{solution}} | ||
− | + | 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 <math>a</math>. | |
− | < | + | We then know that <math>a</math> is a root of |
+ | \[ | ||
+ | Q_{2}(x)-2Q_{1}(x) = 2x^{2}+2kx-43x+k-2x^{2}-2kx+58x+2k = 15x+3k = 0 | ||
+ | \] | ||
+ | , 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: | |
+ | \[ | ||
+ | \frac{k^{2}}{25}+(k-29)\left(\frac{-k}{5}\right)-k = 0 = k^{2}-5(k-29)(k)-25k = k^{2}-5k^{2}+145k-25k | ||
+ | \] | ||
+ | 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 is cubic. What is the largest value of for which the polynomials and are both factors of ?
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 .
We then know that is a root of \[ Q_{2}(x)-2Q_{1}(x) = 2x^{2}+2kx-43x+k-2x^{2}-2kx+58x+2k = 15x+3k = 0 \] , so .
We then know that is a root of 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 , so is the highest.
We can trivially check into the original equations to find that produces a root in common, so the answer is .
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 |