Difference between revisions of "2014 AMC 10B Problems/Problem 20"
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==Solution 1== | ==Solution 1== | ||
− | First, note that <math>50+1=51</math>, which motivates us to factor the polynomial as <math>(x^2-50)(x^2-1)</math>. Since this expression is negative, one term must be negative and the other positive. Also, the first term is obviously smaller than the second, so <math>x^2-50<0<x^2-1</math>. Solving this inequality, we find <math>1<x^2<50</math>. There are exactly <math>12</math> integers <math>x</math> that satisfy this inequality, <math>\pm 2,3,4,5,6,7</math>. | + | First, note that <math>50+1=51</math>, which motivates us to factor the polynomial as <math>(x^2-50)(x^2-1)</math>. Since this expression is negative, one term must be negative and the other positive. Also, the first term is obviously smaller than the second, so <math>x^2-50<0<x^2-1</math>. Solving this inequality, we find <math>1<x^2<50</math>. There are exactly <math>12</math> integers <math>x</math> that satisfy this inequality, <math>\pm \{2,3,4,5,6,7\}</math>. |
Thus our answer is <math>\boxed{\textbf {(C) } 12}</math>. | Thus our answer is <math>\boxed{\textbf {(C) } 12}</math>. |
Revision as of 23:31, 24 October 2020
Problem
For how many integers is the number
negative?
Solution 1
First, note that , which motivates us to factor the polynomial as
. Since this expression is negative, one term must be negative and the other positive. Also, the first term is obviously smaller than the second, so
. Solving this inequality, we find
. There are exactly
integers
that satisfy this inequality,
.
Thus our answer is .
Solution 2
Since the part of
has to be less than
(because we want
to be negative), we have the inequality
.
has to be positive, so
is negative. Then we have
. We know that if we find a positive number that works, it's parallel negative will work. Therefore, we just have to find how many positive numbers work, then multiply that by
. If we try
, we get
, and
therefore doesn't work. Test two on your own, and then proceed. Since two works, all numbers above
that satisfy
work, that is the set {
}. That equates to
numbers. Since each numbers' negative counterparts work,
.
Solution 3 (Graph)
As with Solution , note that the quartic factors to
, which means that it has roots at
,
,
, and
. Now, because the original equation is of an even degree and has a positive leading coefficient, both ends of the graph point upwards, meaning that the graph dips below the
-axis between
and
as well as
and
.
is a bit more than
(
) and therefore means that
all give negative values.
See Also
2014 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.