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 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 $\boxed{013}$ .

Video Solution

https://www.youtube.com/watch?v=_vslL2YcEaE

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
All AIME Problems and Solutions

Page

Toolbox

Search

1990 AIME Problems/Problem 4

Contents

Problem

Solution

Video Solution

See also