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

(Solution 2)
m (Solution 2)
 
Line 28: Line 28:
 
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>.  
  
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>
+
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 ==

Latest revision as of 12:51, 17 January 2020

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

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 $k = 30$.

See also

2007 AIME I (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS