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 $x^2-50<0<x^2-1$ . 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' parallel counterparts work, $6\cdot2=\boxed{\textbf{(C) }12}$ .

2014 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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2014 AMC 10B Problems/Problem 20

Contents

Problem

Solution 1

Solution 2

See Also