Difference between revisions of "2007 AIME I Problems/Problem 8"
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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 = 30</math>. |
== See also == | == See also == |
Revision as of 12:51, 17 January 2020
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 .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.