1990 AIME Problems/Problem 4

Revision as of 20:43, 2 March 2007 by Azjps (talk | contribs) (solution)

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 $013$.

See also

1990 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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All AIME Problems and Solutions
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