Difference between revisions of "2014 AMC 10B Problems/Problem 20"
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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 2. If we try <math>1</math>, we get <math>1^4-51(1)^4+50 = -50+50 = 0</math>, and 0 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' parallel 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 2. If we try <math>1</math>, we get <math>1^4-51(1)^4+50 = -50+50 = 0</math>, and 0 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' parallel counterparts work, <math>6\cdot2=\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 12:40, 4 January 2019
Contents
Problem
For how many integers 
is the number 
negative?
Solution 1
First, note that 
, which motivates us to factor the polynomial as 
. 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 
. Solving this inequality, we find 
. There are exactly 12 integers that satisfy this inequality, 
.
Thus our answer is 
Solution 2
Since the 
part of has to be less than 
(because we want to be negative), we have the inequality 
--> 
. 
has to be positive, so 
is negative. Then we have 
. 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 
, we get 
, and 0 therefore doesn't work. Test two on your own, and then proceed. Since two works, all numbers above 
that satisfy work, that is the set {
}. That equates to 
numbers. Since each numbers' parallel counterparts work, 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
