Difference between revisions of "2014 AMC 10B Problems/Problem 20"
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==Solution== | ==Solution== | ||
− | 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: | + | 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: |
<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. | <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>. | <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>. | ||
− | Thus our answer is <math>6+6=\boxed{\textbf {(C) } 12 | + | Thus our answer is <math>6+6=\boxed{\textbf {(C) } 12}</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 14:35, 20 February 2014
Problem
For how many integers is the number
negative?
Solution
First, note that , which motivates us to factor the polynomial as
. Using the difference of squares factorization
, this can be simplified into
. 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:
. Note that
, so if exactly one of these is negative it must be
. However,
must also be positive, and thus
. Since
,
, and so
. This case gives exactly
solutions.
. Using the inequality comparing the terms from the above case, we can see that
or
. Using the approximation for
from above, we can see that
, so this case also has exactly
values of
.
Thus our answer is
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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