Difference between revisions of "2007 AIME I Problems/Problem 8"
(+ my solution (I have a feeling I did something wrong though, since I only got one possible solution)) |
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== Problem == | == Problem == | ||
− | The polynomial <math>P(x)</math> is cubic. What is the largest value of <math>k</math> for which the polynomials <math>Q_1(x) = x^2 + (k-29)x - k</math> and <math>Q_2(x) = 2x^2+ (2k-43)x + k</math> are both | + | The [[polynomial]] <math>P(x)</math> is [[cubic]]. What is the largest value of <math>k</math> for which the polynomials <math>\displaystyle Q_1(x) = x^2 + (k-29)x - k</math> and <math>\displaystyle Q_2(x) = 2x^2+ (2k-43)x + k</math> are both [[factor]]s of <math>P(x)</math>? |
+ | __TOC__ | ||
== Solution == | == Solution == | ||
− | + | === Solution 1 === | |
− | We can see that | + | We can see that <math>Q_1</math> and <math>Q_2</math> must have a [[root]] in common for them to both be [[factor]]s of the same cubic. |
Let this root be <math>a</math>. | Let this root be <math>a</math>. | ||
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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> | <math> | ||
− | Q_{2}(x)-2Q_{1}(x) = 2x^{2}+2kx-43x+k-2x^{2}-2kx+58x+2k = 15x+3k = 0 | + | \displaystyle Q_{2}(x)-2Q_{1}(x) = 2x^{2}+2kx-43x+k-2x^{2}-2kx+58x+2k = 15x+3k = 0 |
</math> | </math> | ||
, so <math>x = \frac{-k}{5}</math>. | , so <math>x = \frac{-k}{5}</math>. | ||
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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>. | 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>. | ||
+ | |||
+ | === Solution 2 === | ||
+ | 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>\displaystyle (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>. | ||
+ | |||
+ | Therefore, we can write four equations (and we have four [[variable]]s), <math>\displaystyle a + m = 29 - k</math>, <math>a + n = \frac{43}{2} - k</math>, <math>\displaystyle 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>\displaystyle n = -\frac{5}{2}</math>. Substituting back into the equations, we eventually find that <math>k = 30</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}} | ||
+ | |||
+ | [[Category:Intermediate Algebra Problems]] |
Revision as of 18:13, 15 March 2007
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 |