Difference between revisions of "2008 iTest Problems/Problem 7"
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==Problem== | ==Problem== | ||
Find the number of integers <math>n</math> for which <math>n^2 + 10n < 2008</math>. | Find the number of integers <math>n</math> for which <math>n^2 + 10n < 2008</math>. | ||
| + | |||
| + | ==Solution== | ||
| + | |||
| + | First, we complete the square of the left side of the equation, giving us: | ||
| + | |||
| + | <math>(n+5)^2 < 2033</math> | ||
| + | |||
| + | The integers from <math>-50</math> to <math>40</math> satisfy this equation, so the answer is <math> 40- (-50)+1 = \boxed{91}</math>. | ||
| + | |||
| + | ==See Also== | ||
| + | {{2008 iTest box|num-b=6|num-a=8}} | ||
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
Latest revision as of 23:56, 21 June 2018
Problem
Find the number of integers 
for which 
.
Solution
First, we complete the square of the left side of the equation, giving us:
The integers from 
to 
satisfy this equation, so the answer is 
.
See Also
| 2008 iTest (Problems) | ||
| Preceded by: Problem 6 |
Followed by: Problem 8 | |
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