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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 ==
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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 factors of <math>P(x)</math>?
+
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 ==
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{{solution}}
+
=== Solution 1 ===
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We can see that they must have a root in common for them to both be factors of the same cubic.
+
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>
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Q_{2}(x)-2Q_{1}(x) = 2x^{2}+2kx-43x+k-2x^{2}-2kx+58x+2k = 15x+3k = 0
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\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 $P(x)$ is cubic. What is the largest value of $k$ for which the polynomials $\displaystyle Q_1(x) = x^2 + (k-29)x - k$ and $\displaystyle Q_2(x) = 2x^2+ (2k-43)x + k$ 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 $\displaystyle Q_{2}(x)-2Q_{1}(x) = 2x^{2}+2kx-43x+k-2x^{2}-2kx+58x+2k = 15x+3k = 0$ , 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 $\displaystyle (x - a)(x - m) = x^2 + (k - 29)x - k$ 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), $\displaystyle a + m = 29 - k$, $a + n = \frac{43}{2} - k$, $\displaystyle am = -k$, 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 $\displaystyle n = -\frac{5}{2}$. 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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