# 2014 AMC 10B Problems/Problem 20

## 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. Solving this inequality, we find $1. 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 $\boxed{\textbf{(C) }12}$.

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