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

Latest revision as of 22:18, 18 June 2022

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.

Solution 4

Let $x^{2}=u$ . Then the expression becomes $u^{2}-51u+50$ which can be factored as $\left(u-1\right)\left(u-50\right)$ . Since the expression is negative, one of $\left(u-1\right)$ and $\left(u-50\right)$ need to be negative. $u-1>u-50$ , so $u-1>0$ and $u-50<0$ , which yields the inequality $50>u>1$ . Remember, since $u=x^{2}$ where $x$ is an integer, this means that $u$ is a perfect square between $1$ and $50$ . There are $6$ values of $u$ that satisfy this constraint, namely $4$ , $9$ , $16$ , $25$ , $36$ , and $49$ . Solving each of these values for $x$ yields $12$ values (as $x$ can be negative or positive) $\Longrightarrow \boxed{\textbf {(C) } 12}$ . ~JH. L

@@ Line 5: / Line 5: @@
 ==Solution 1==
-First, note that <math>50+1=51</math>, which motivates us to factor the polynomial as <math>(x^2-50)(x^2-1)</math>. 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 <math>x^2-50<0<x^2-1</math>. Solving this inequality, we find <math>1<x^2<50</math>. There are  exactly <math>12</math> integers <math>x</math> that satisfy this inequality, <math>\pm 2,3,4,5,6,7</math>.
+First, note that <math>50+1=51</math>, which motivates us to factor the polynomial as <math>(x^2-50)(x^2-1)</math>. 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 <math>x^2-50<0<x^2-1</math>. Solving this inequality, we find <math>1<x^2<50</math>. There are  exactly <math>12</math> integers <math>x</math> that satisfy this inequality, <math>\pm \{2,3,4,5,6,7\}</math>.
 Thus our answer is <math>\boxed{\textbf {(C) } 12}</math>.
@@ Line 14: / Line 14: @@
 ==Solution 3 (Graph)==
 As with Solution <math>1</math>, note that the quartic factors to <math>(x^2-50)\cdot(x^2-1)</math>, which means that it has roots at <math>-5\sqrt{2}</math>, <math>-1</math>, <math>1</math>, and <math>5\sqrt{2}</math>. 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 <math>x</math>-axis between <math>-5\sqrt{2}</math> and <math>-1</math> as well as <math>1</math> and <math>5\sqrt{2}</math>. <math>5\sqrt{2}</math> is a bit more than <math>7</math> (<math>1.4\cdot 5=7</math>) and therefore means that <math> -7,-6,-5,-4,-3,-2,2,3,4,5,6,7</math> all give negative values.
+==Solution 4==
+Let <math>x^{2}=u</math>. Then the expression becomes <math>u^{2}-51u+50</math> which can be factored as <math>\left(u-1\right)\left(u-50\right)</math>. Since the expression is negative, one of <math>\left(u-1\right)</math> and <math>\left(u-50\right)</math> need to be negative. <math>u-1>u-50</math>, so <math>u-1>0</math> and <math>u-50<0</math>, which yields the inequality <math>50>u>1</math>. Remember, since <math>u=x^{2}</math> where <math>x</math> is an integer, this means that <math>u</math> is a perfect square between <math>1</math> and <math>50</math>. There are <math>6</math> values of <math>u</math> that satisfy this constraint, namely <math>4</math>, <math>9</math>, <math>16</math>, <math>25</math>, <math>36</math>, and <math>49</math>. Solving each of these values for <math>x</math> yields <math>12</math> values (as <math>x</math> can be negative or positive) <math>\Longrightarrow \boxed{\textbf {(C) } 12}</math>.
+~JH. L
 ==See Also==
 {{AMC10 box|year=2014|ab=B|num-b=19|num-a=21}}
 {{MAA Notice}}

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