# 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 $x^4-51x^2<-50 \rightarrow x^2(x^2-51) <-50$. $x^2$ 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, $6\cdot2=\boxed{\textbf{(C) }12}$.

## Solution 3 (Graph)

As with Solution $1$, note that the quartic factors to $(x^2-50)\cdot(x^2-1)$, which means that it has roots at $-5\sqrt{2}$, $-1$, $1$, and $5\sqrt{2}$. Now, because the original equation is of an even degree and has a positive leading coefficient, both ends of the graph point upwards, meaning that the graph dips below the $x$-axis between $-5\sqrt{2}$ and $-1$ as well as $1$ and $5\sqrt{2}$. $5\sqrt{2}$ is a bit more than $7$ ( $1.4\cdot 5=7$) and therefore means that $-7,-6,-5,-4,-3,-2,2,3,4,5,6,7$ all give negative values.

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