2007 AIME I Problems/Problem 8

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 $k = \boxed30$ .

2007 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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2007 AIME I Problems/Problem 8

Problem

Contents

Solution

Solution 1

Solution 2

See also