Difference between revisions of "2014 AMC 10B Problems/Problem 20"
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<math> \textbf {(A) } 8 \qquad \textbf {(B) } 10 \qquad \textbf {(C) } 12 \qquad \textbf {(D) } 14 \qquad \textbf {(E) } 16</math> | <math> \textbf {(A) } 8 \qquad \textbf {(B) } 10 \qquad \textbf {(C) } 12 \qquad \textbf {(D) } 14 \qquad \textbf {(E) } 16</math> | ||
+ | ==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 12 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> | ||
+ | |||
+ | ==Solution 2== | ||
+ | Since the <math>x^4-51x^2</math> part of <math>x^4-51x^2+50</math> has to be less than <math>-50</math> (because we want <math>x^4-51x^2+50</math> to be negative), we have the inequality <math>x^4-51x^2<-50</math> --> <math>x^2(x^2-51) <-50</math>. <math>x^2</math> has to be positive, so <math>(x^2-51)</math> is negative. Then we have <math>x^2<51</math>. Try answers to find <math> \boxed{\textbf{(C) }12} </math>. | ||
Revision as of 22:45, 15 October 2015
Contents
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 12 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 . Try answers to find .
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 |
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