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

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==Solution==
 
==Solution==
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First, note that <math>50+1=51</math>, which motivates us to factor the polynomial as <math>(x^2-50)(x^2-1)</math>. Using the difference of squares factorization <math>a^2-b^2=(a-b)(a+b)</math>, this can be simplified into <math>(x-\sqrt{50})(x+\sqrt{50}(x-1)(x+1)</math>. 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:
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<math>\textbf{Case 1: One term}</math>. Note that <math>x-\sqrt{50}<x-1<x+1<x+\sqrt{50}</math>, so if exactly one of these is negative it must be <math>x-\sqrt{50}</math>. However, <math>x-1</math> must also be positive, and thus <math>x-\sqrt{50}<0<x-1\Rightarrow 1<x<\sqrt{50}</math>. Since <math>7^2=49<50<64=8^2</math>, <math>\lfloor\sqrt{50}\rfloor=7</math>, and so <math>1<x\le7</math>. This case gives exactly <math>6</math> solutions.
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<math>\textbf{Case 2: Three terms}</math>. Using the inequality comparing the terms from the above case, we can see that <math>x-\sqrt{50},x-1,x+1<0<x+\sqrt{50}</math> or  <math>-\sqrt{50}<x<-1</math>. Using the approximation for <math>\sqrt{50}</math> from above, we can see that <math>-7\le x < -1</math>, so this case also has exactly <math>6</math> values of <math>x</math>.
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Thus our answer is <math>6+6=\boxed{\textbf {(C) } 12 \qquad }</math>
  
 
==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 13:34, 20 February 2014

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 $6+6=\boxed{\textbf {(C) } 12 \qquad }$

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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