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.

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1.

What is the sum of the solutions to the equation x4=127x4?\sqrt[4]{x}=\frac{12}{7-\sqrt[4]{x}}?

Answer: 337
Concepts:quadraticradicalsubstitution
Difficulty rating: 1660
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 t0.t\geq0. The equation becomes t(7t)=12, t(7-t)=12, or t27t+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.

Full Exam

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