2014 AMC 10B Problem 20

Attempt Problem 20 of the 2014 AMC 10B 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 2014 AMC 10B solutions, or check the answer key.

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20.

For how many integers xx is the number x451x2+50x^4-51x^2+50 negative?

8 8

10 10

12 12

14 14

16 16

Answer: C
Concepts:factoringinequalitycounting integers in a range
Difficulty rating: 1280
Solution:

First, note that x451x2+50x^4-51x^2+50 =(x250)(x21).= (x^2-50)(x^2-1).

The product is negative exactly when its two factors have opposite signs. Since x250<x21x^2-50<x^2-1, this requires x250<0<x21x^2-50<0<x^2-1. Thus 1<x2<501<x^2<50, or 2x72\le |x|\le7. There are 66 positive and 66 negative integer solutions, for a total of 1212.

Thus, the correct answer is C .

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