Find the positive solution to &1x^2-10x-29 & +1x^2-10x-45 & -2x^2-10x-69=0.

1990 AIME Problem 4

Attempt Problem 4 of the 1990 AIME below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1990 AIME solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

Find the positive solution to

\begin{aligned}&\frac1{x^2-10x-29}\\&\quad+\frac1{x^2-10x-45}\\&\quad-\frac2{x^2-10x-69}=0.\end{aligned}

Answer

Set

y=x^2-10x

so the three denominators differ only by constants

Combine the first two fractions before clearing denominators

Set

y=x^2-10x.

Combining the first two fractions and clearing the nonzero denominators gives an equality between

(y-37)(y-69)

and

(y-29)(y-45).

Expanding and canceling

y^2

yields

y=39.

Thus

x^2-10x-39=0,

(x-13)(x+3)=0.

The positive solution is

13.