1986 AIME Problem 1

Attempt Problem 1 of the 1986 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 1986 AIME solutions, or check the answer key.

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

1.

What is the sum of the solutions to the equation x4=127−x4?\sqrt[4]{x}=\frac{12}{7-\sqrt[4]{x}}?
Answer: 337
Concepts:quadraticradicalsubstitution
Small hint:
Substitute t=x4t=\sqrt[4]{x}
Big hint:
After clearing the denominator, factor the resulting quadratic in tt
Solution:
Let t=x4,t=\sqrt[4]{x}, so t≥0.t\geq0. The equation becomes t(7−t)=12, t(7-t)=12, or t2−7t+12=0.t^2-7t+12=0. Thus t=3t=3 or t=4,t=4, and both values are valid in the original equation. Hence x=34=81x=3^4=81 or x=44=256,x=4^4=256, and their sum is 81+256=337.81+256=337.
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