Difference between revisions of "2014 AMC 10B Problems/Problem 20"

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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>.
 
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>6+6=\boxed{\textbf {(C) } 12}</math>
+
Thus our answer is <math>\boxed{\textbf {(C) } 12}</math>
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2014|ab=B|num-b=19|num-a=21}}
 
{{AMC10 box|year=2014|ab=B|num-b=19|num-a=21}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:31, 21 February 2014

Problem

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

$\textbf {(A) } 8 \qquad \textbf {(B) } 10 \qquad \textbf {(C) } 12 \qquad \textbf {(D) } 14 \qquad \textbf {(E) } 16$

Solution

First, note that $50+1=51$, which motivates us to factor the polynomial as $(x^2-50)(x^2-1)$. 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 $x^2-50<0<x^2-1$. Solving this inequality, we find $1<x^2<50$. There are exactly 12 integers $x$ that satisfy this inequality, $\pm 2,3,4,5,6,7$.

Thus our answer is $\boxed{\textbf {(C) } 12}$

See Also

2014 AMC 10B (ProblemsAnswer KeyResources)
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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