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

m (→‎Solution 1)
(Added solution using end behavior)
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==Solution 2==
 
==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</math> --> <math>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</math> --> <math>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)==
 +
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.
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2014|ab=B|num-b=19|num-a=21}}
 
{{AMC10 box|year=2014|ab=B|num-b=19|num-a=21}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:36, 28 December 2019

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' 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.

See Also

2014 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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