# Difference between revisions of "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:

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

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