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>. | + | 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>. |
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==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 13:39, 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 |
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