Difference between revisions of "2014 AMC 10B Problems/Problem 20"

Revision as of 21:01, 28 January 2020

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 $x^4-51x^2<-50 \rightarrow x^2(x^2-51) <-50$ . $x^2$ 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.

@@ Line 10: / Line 10: @@
 ==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 \yields 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>. 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 <math>2</math>. If we try <math>1</math>, we get <math>1^4-51(1)^4+50 = -50+50 = 0</math>, and <math>0</math> therefore doesn't work. Test two on your own, and then proceed. Since two works, all numbers above <math>2</math> that satisfy <math>x^2<51</math> work, that is the set {<math>{2,3,4,5,6,7}</math>}. That equates to <math>6</math> numbers. Since each numbers' negative counterparts work, <math>6\cdot2=\boxed{\textbf{(C) }12} </math>.
+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 \rightarrow 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>. 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 <math>2</math>. If we try <math>1</math>, we get <math>1^4-51(1)^4+50 = -50+50 = 0</math>, and <math>0</math> therefore doesn't work. Test two on your own, and then proceed. Since two works, all numbers above <math>2</math> that satisfy <math>x^2<51</math> work, that is the set {<math>{2,3,4,5,6,7}</math>}. That equates to <math>6</math> numbers. Since each numbers' negative counterparts work, <math>6\cdot2=\boxed{\textbf{(C) }12} </math>.
 ==Solution 3 (Graph)==

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