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

Revision as of 21:43, 15 October 2015

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

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

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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 <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>.