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2007 AIME I Problems/Problem 8

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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)?

Contents

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 + (k - \frac{43}{2})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 (ProblemsResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Want to learn how to tackle those tough MATHCOUNTS and AMC counting and probability problems? Check out Art of Problem Solving's Introduction to Counting & Probability by David Patrick.
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