Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2014 AMC 10B Problems/Problem 20"
Line 4: Line 4:
 
<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 21:45, 15 October 2015

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 $x^2<51$. Try answers to find $\boxed{\textbf{(C) }12}$.


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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_20&oldid=72488"