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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 <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>.
  
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Thus our answer is <math>\boxed{\textbf {(C) } 12}</math>
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Thus our answer is <math>\boxed{\textbf {(C) } 12}</math>.
  
 
==Solution 2==
 
==Solution 2==

Revision as of 18:55, 27 December 2019

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 1

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 $\boxed{\textbf {(C) } 12}$.

Solution 2

Since the $x^4-51x^2$ part of $x^4-51x^2+50$ has to be less than $-50$ (because we want $x^4-51x^2+50$ to be negative), we have the inequality $x^4-51x^2<-50$ --> $x^2(x^2-51) <-50$. $x^2$ has to be positive, so $(x^2-51)$ is negative. Then we have $x^2<51$. 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 $2$. If we try $1$, we get $1^4-51(1)^4+50 = -50+50 = 0$, and $0$ therefore doesn't work. Test two on your own, and then proceed. Since two works, all numbers above $2$ that satisfy $x^2<51$ work, that is the set {${2,3,4,5,6,7}$}. That equates to $6$ numbers. Since each numbers' negative counterparts work, $6\cdot2=\boxed{\textbf{(C) }12}$.

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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