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

First, note that $50+1=51$ , which motivates us to factor the polynomial as $(x^2-50)(x^2-1)$ . Using the difference of squares factorization $a^2-b^2=(a-b)(a+b)$ , this can be simplified into $(x-\sqrt{50})(x+\sqrt{50})(x-1)(x+1)$ . For this expression to be negative, either one of the terms or three of the terms must be negative. We split into these two cases:

$\textbf{Case 1: One term}$ . Note that $x-\sqrt{50}<x-1<x+1<x+\sqrt{50}$ , so if exactly one of these is negative it must be $x-\sqrt{50}$ . However, $x-1$ must also be positive, and thus $x-\sqrt{50}<0<x-1\Rightarrow 1<x<\sqrt{50}$ . Since $7^2=49<50<64=8^2$ , $\lfloor\sqrt{50}\rfloor=7$ , and so $1<x\le7$ . This case gives exactly $6$ solutions.

$\textbf{Case 2: Three terms}$ . Using the inequality comparing the terms from the above case, we can see that $x-\sqrt{50},x-1,x+1<0<x+\sqrt{50}$ or $-\sqrt{50}<x<-1$ . Using the approximation for $\sqrt{50}$ from above, we can see that $-7\le x < -1$ , so this case also has exactly $6$ values of $x$ .

Thus our answer is $6+6=\boxed{\textbf {(C) } 12}$

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

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