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2007 AIME I Problems/Problem 8: Difference between revisions

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+ 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>\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>?
The [[polynomial]] <math>P(x)</math> is [[cubic polynomial | 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 [[factor]]s of <math>P(x)</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>
\displaystyle 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>
</math>
, so <math>x = \frac{-k}{5}</math>.
, so <math>x = \frac{-k}{5}</math>.
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=== Solution 2 ===
=== 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>.
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>.


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>.  
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>\displaystyle n = -\frac{5}{2}</math>. Substituting back into the equations, we eventually find that <math>k = 30</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 = 30</math>.


== See also ==
== See also ==

Revision as of 18:49, 8 September 2009

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

Solution 1

We can see that $Q_1$ and $Q_2$ 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 $030$.

Solution 2

Again, let the common root be $a$; let the other two roots be $m$ and $n$. We can write that $(x - a)(x - m) = x^2 + (k - 29)x - k$ and that $2(x - a)(x - n) = 2\left(x^2 + (k - \frac{43}{2})x + \frac{k}{2}\right)$.

Therefore, we can write four equations (and we have four variables), $a + m = 29 - k$, $a + n = \frac{43}{2} - k$, $am = -k$, and $an = \frac{k}{2}$.

The first two equations show that $m - n = 29 - \frac{43}{2} = \frac{15}{2}$. The last two equations show that $\frac{m}{n} = -2$. Solving these show that $m = 5$ and that $n = -\frac{5}{2}$. Substituting back into the equations, we eventually find that $k = 30$.

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
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