1991 AMC 12 Problem 27

Attempt Problem 27 of the 1991 AMC 12 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 1991 AMC 12 solutions, or check the answer key.

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27.

If x+x2−1+1x−x2−1=20, \begin{aligned} x+\sqrt{x^2-1} &+\frac1{x-\sqrt{x^2-1}}\\ &=20, \end{aligned} then x2+x4−1+1x2+x4−1= \begin{aligned} x^2+\sqrt{x^4-1} &+\frac1{x^2+\sqrt{x^4-1}}\\ &= \end{aligned}
5.055.05
2020
51.00551.005
61.2561.25
400400
Answer: C
Concepts:conjugate radicalsreciprocalalgebraic identity
Small hint:
Rationalize 1x−x2−1\frac1{x-\sqrt{x^2-1}}
Big hint:
The last two terms of the requested expression are conjugates
Solution:
Because 1x−x2−1=x+x2−1, \frac1{x-\sqrt{x^2-1}}=x+\sqrt{x^2-1}, the given equation implies x+x2−1=10.x+\sqrt{x^2-1}=10. Its reciprocal is x−x2−1=110,x-\sqrt{x^2-1}=\frac{1}{10}, so adding yields 2x=10.12x=10.1 and x=5.05.x=5.05. Also 1x2+x4−1=x2−x4−1. \frac1{x^2+\sqrt{x^4-1}}=x^2-\sqrt{x^4-1}. Therefore the requested expression is 2x2=2(5.05)2=51.005.2x^2=2(5.05)^2=51.005.
Thus the correct answer is C.
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