2014 AMC 10B Problems/Problem 20
Problem
For how many integers is the number
negative?
Solution
First, note that , which motivates us to factor the polynomial as
. Using the difference of squares factorization
, this can be simplified into
. For this expression to be negative, either one of the terms or three of the terms must be negative. We split into these two cases:
. Note that
, so if exactly one of these is negative it must be
. However,
must also be positive, and thus
. Since
,
, and so
. This case gives exactly
solutions.
. Using the inequality comparing the terms from the above case, we can see that
or
. Using the approximation for
from above, we can see that
, so this case also has exactly
values of
.
Thus our answer is
See Also
2014 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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All AMC 10 Problems and Solutions |
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