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

Revision as of 20:55, 19 November 2022

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

@@ Line 21: / Line 21: @@
 or <math>k^{2}=30k</math>, so <math>k=30</math> is the highest.
-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>\boxed{030}</math>.
 === 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>.
-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 = \boxed{030}</math>.
 === Solution 3 ===
@@ Line 41: / Line 41: @@
 \end{align}</cmath>
-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>.
+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

2007 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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