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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 a = x^2 - 10x - 29 (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, -64a + 40 \times 16 = 0, so a = 10. Re-substituting, 10 = x^2 - 10x - 29 \Longleftrightarrow 0 = (x - 13)(x + 3). The positive root is \boxed{013}.

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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