Difference between revisions of "2007 AIME I Problems/Problem 8"
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− | Again, let the common root be <math>a</math>; let the other two roots be <math>m</math> and <math>n</math>. We can write that <math>(x - a)(x - m) = x^2 + (k - 29)x - k</math> and that <math>2(x - a)(x - n) = 2\left(x^2 + (k - \frac{43}{2})x + \frac{k}{2}\right)</math>. | + | Again, let the common root be <math>a</math>; let the other two roots be <math>m</math> and <math>n</math>. We can write that <math>(x - a)(x - m) = x^2 + (k - 29)x - k</math> and that <math>2(x - a)(x - n) = 2\left(x^2 + \left(k - \frac{43}{2}\right)x + \frac{k}{2}\right)</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>. | 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>. |
Revision as of 08:04, 17 August 2017
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.