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Difference between revisions of "2007 AIME I Problems/Problem 8"

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== Problem ==
 
== Problem ==
 
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>?
 
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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== Solution ==
 
== Solution ==
 
=== Solution 1 ===
 
=== Solution 1 ===
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or <math>k^{2}=30k</math>, so <math>k=30</math> is the highest.
 
or <math>k^{2}=30k</math>, so <math>k=30</math> is the highest.
  
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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>.
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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>\boxed{030}</math>.
  
 
=== Solution 2 ===
 
=== Solution 2 ===
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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>.  
  
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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>.
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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 = \boxed{030}</math>.
  
 
=== Solution 3 ===
 
=== Solution 3 ===
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\end{align}</cmath>
 
\end{align}</cmath>
  
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Using equations <math>(1)</math> and <math>(2)</math> to make substitutions into equation <math>(3)</math>, we see that the <math>k</math>'s drop out and we're left with <math>d = -5c</math>. Substituting this expression for <math>d</math> into equation <math>(4)</math> and solving, we see that <math>k</math> must be <math>\boxed {30}</math>.   
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Using equations <math>(1)</math> and <math>(2)</math> to make substitutions into equation <math>(3)</math>, we see that the <math>k</math>'s drop out and we're left with <math>d = -5c</math>. Substituting this expression for <math>d</math> into equation <math>(4)</math> and solving, we see that <math>k</math> must be <math>\boxed {030}</math>.   
  
 
~ anellipticcurveoverq
 
~ anellipticcurveoverq
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Notice that if the roots of <math>Q_1(x)</math> and <math>Q_2(x)</math> are all distinct, then <math>P(x)</math> would have four distinct roots, which is a contradiction since it's cubic. Thus, <math>Q_1(x)</math> and <math>Q_2(x)</math> must share a root. Let this common value be <math>r.</math>  
 
Notice that if the roots of <math>Q_1(x)</math> and <math>Q_2(x)</math> are all distinct, then <math>P(x)</math> would have four distinct roots, which is a contradiction since it's cubic. Thus, <math>Q_1(x)</math> and <math>Q_2(x)</math> must share a root. Let this common value be <math>r.</math>  
  
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Thus, we see that we have <math>\begin{align*}
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Thus, we see that we have  
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& r^2 + (k - 29)r - k = 0, \\
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<cmath>r^2 + (k - 29)r - k = 0,</cmath>
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& 2r^2 + (2k - 43)r + k = 0. \end{align*}</math> Adding the two equations gives us <cmath>3r^2 + 3k - 72r = 0 \implies r = 0, 24 - k.</cmath> Now, we have two cases to consider. If <math>r = 0,</math> then we have that <math>Q_1(r) = 0 = r^2 + (k - 29)r - k \implies k = 0.</math> On the other hand, if <math>r = 24 - k,</math> we see that <cmath>Q_1(r) = 0 = (24 - k)^2 + (k - 29)(24 - k) - k \implies k = \boxed{030}.</cmath> This can easily be checked to see that it does indeed work, and we're done!  
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<cmath>2r^2 + (2k - 43)r + k = 0.</cmath> Adding the two equations gives us <cmath>3r^2 + (3k - 72)r = 0 \implies r = 0, 24 - k.</cmath> Now, we have two cases to consider. If <math>r = 0,</math> then we have that <math>Q_1(r) = 0 = r^2 + (k - 29)r - k \implies k = 0.</math> On the other hand, if <math>r = 24 - k,</math> we see that <cmath>Q_1(r) = 0 = (24 - k)^2 + (k - 29)(24 - k) - k \implies k = \boxed{030}.</cmath> This can easily be checked to see that it does indeed work, and we're done!  
  
 
~Ilikeapos
 
~Ilikeapos

Latest revision as of 23:11, 28 November 2023

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 $\boxed{030}$.

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 $2(x - a)(x - n) = 2\left(x^2 + \left(k - \frac{43}{2}\right)x + \frac{k}{2}\right)$.

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 $k = \boxed{030}$.

Solution 3

Since $Q_1(x)$ and $Q_2(x)$ are both factors of $P(x)$, which is cubic, we know the other factors associated with each of $Q_1(x)$ and $Q_2(x)$ must be linear. Let $Q_1(x)R(x) = Q_2(x)S(x) = P(x)$, where $R(x) = ax + b$ and $S(x) = cx + d$. Then we have that $((x^2 + (k-29)x - k))(ax + b) = ((2x^2+ (2k-43)x + k))(cx + d)$. Equating coefficients, we get the following system of equations:

\begin{align} a = 2c \\ b = -d \\ 2c(k - 29) - d = c(2k - 43) + 2d \\ -d(k - 29) - 2ck = d(2k - 43) + ck \end{align}

Using equations $(1)$ and $(2)$ to make substitutions into equation $(3)$, we see that the $k$'s drop out and we're left with $d = -5c$. Substituting this expression for $d$ into equation $(4)$ and solving, we see that $k$ must be $\boxed {030}$.

~ anellipticcurveoverq

Solution 4

Notice that if the roots of $Q_1(x)$ and $Q_2(x)$ are all distinct, then $P(x)$ would have four distinct roots, which is a contradiction since it's cubic. Thus, $Q_1(x)$ and $Q_2(x)$ must share a root. Let this common value be $r.$

Thus, we see that we have \[r^2 + (k - 29)r - k = 0,\] \[2r^2 + (2k - 43)r + k = 0.\] Adding the two equations gives us \[3r^2 + (3k - 72)r = 0 \implies r = 0, 24 - k.\] Now, we have two cases to consider. If $r = 0,$ then we have that $Q_1(r) = 0 = r^2 + (k - 29)r - k \implies k = 0.$ On the other hand, if $r = 24 - k,$ we see that \[Q_1(r) = 0 = (24 - k)^2 + (k - 29)(24 - k) - k \implies k = \boxed{030}.\] This can easily be checked to see that it does indeed work, and we're done!

~Ilikeapos

Video Solution

https://www.youtube.com/watch?v=bsRQZwO7n84&t=64s

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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