2014 AMC 10B Problems/Problem 20
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 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 . If we try , we get , and 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' negative counterparts work, .
Solution 3 (Graph)
As with Solution , note that the quartic factors to , which means that it has roots at , , , and . 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 -axis between and as well as and . is a bit more than () and therefore means that all give negative values.
Solution 4
Let . Then the expression becomes which can be factored as . Since the expression is negative, one of and need to be negative. , so and , which yields the inequality . Remember, since where is an integer, this means that is a perfect square between and . There are values of that satisfy this constraint, namely , , , , , and . Solving each of these values for yields values (as can be negative or positive) . ~JH. L
See Also
2014 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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