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1990 AIME Problems/Problem 4

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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 clear out the denominators by multiplying, though that would be unnecessarily tedious.

To simplify the equation, substitute (the denominator of the first fraction). We can rewrite the equation 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, , so . Re-substituting, 10 = x^2 - 10x - 29 \Longleftrightarrow 0 = (x - 13)(x + 3). The positive root is .

See also

1990 AIME (ProblemsResources)
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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