Difference between revisions of "2008 iTest Problems/Problem 7"

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==Solution==
 
==Solution==
  
==See also==
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First, we complete the square of the left side of the equation, giving us:
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<math>(n+5)^2 < 2033</math>
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The integers from <math>-50</math> to <math>40</math> satisfy this equation, so the answer is <math> 40- (-50)+1 = \boxed{91}</math>.
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==See Also==
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{{2008 iTest box|num-b=6|num-a=8}}
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[[Category:Introductory Algebra Problems]]

Latest revision as of 23:56, 21 June 2018

Problem

Find the number of integers $n$ for which $n^2 + 10n < 2008$.

Solution

First, we complete the square of the left side of the equation, giving us:

$(n+5)^2 < 2033$

The integers from $-50$ to $40$ satisfy this equation, so the answer is $40- (-50)+1 = \boxed{91}$.

See Also

2008 iTest (Problems)
Preceded by:
Problem 6
Followed by:
Problem 8
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