1990 AIME Problems/Problem 4

Problem

Find the positive solution to

$\frac 1{x^2-10x-29}+\frac1{x^2-10x-45}-\frac 2{x^2-10x-69}=0$

Solution

We could multiply the entire equation by all of the denominators, though that would obviously be unnecessarily tedious.

To simplify some of the word, a substitution can be used. Define $a$ as the denominator of the first fraction. We can rewrite it as $\frac{1}{a} + \frac{1}{a - 16} - \frac{2}{a - 40} = 0$ . Multiplying out the denominators now, we get:

$\displaystyle (a - 16)(a - 40) + a(a - 40) - 2(a)(a - 16) = 0$

Simplifying, we get that $-64a + 40 \cdot 16 = 0$ , so $a = 10$ . Re-substituting the value of $a$ , we get that $\displaystyle 10 = x^2 - 10x - 29$ . Thus, $\displaystyle 0 = (x - 13)(x + 3)$ . The positive root is $013$ .

1990 AIME (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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1990 AIME Problems/Problem 4

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