2000 AMC 10 Problem 9

Attempt Problem 9 of the 2000 AMC 10 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 2000 AMC 10 solutions, or check the answer key.

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9.

If x2=p,|x - 2| = p, where x<2,x \lt 2, then xp=x - p =

2-2

22

22p2 - 2p

2p22p - 2

2p2|2p - 2|

Answer: C
Concepts:absolute valuealgebraic manipulation
Difficulty rating: 1170
Solution:

Because x<2,x \lt 2, we have x2=2x=p,|x - 2| = 2 - x = p, so x=2p.x = 2 - p.

Then xp=(2p)p=22p.x - p = (2 - p) - p = 2 - 2p.

Thus, the correct answer is C.

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