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Title: Math Question! "Find two consecutive integers such that the square of the smaller is 10 more than the larger."? Post by: irina on Sep 4, 2012 Please help! Can you also show me how you did the problem? Thanks!
Title: Math Question! "Find two consecutive integers such that the square of the smaller is 10 more than the larger."? Post by: julies on Sep 4, 2012 Lets say the first integer is x, and the second is (x+1) since they're consecutive.
now the smaller one is x, to x^2 - 10 = x+1 add 10 to both sides x^2 = x + 11 subtract x x^2 - x = 11 I would solve this like a quadratic and complete the square. So in order to get x^2 - x to be a perfect square, you need to take 2/b then square it. In this case b = 1 (from the standard quadratic form of ax^2+bx+c) 1/2 squared = 1/4 so add 1/4 to both sides x^2 + x + 1/4 = 11 1/4 Then factor (x-1/2)^2 = 11 1/4 Square root x-1/2 = Square root (11 1/4) so x = 1/2 + Square root (11 1/4) Do this in your calculator x = 3.89 Round to 4 Still doesn't give you the right answer though. You get 4 and 5. 4^2 = 16, which is 11 greater than 5, but that's a close as it gets, because as the numbers get larger, their squares get larger, so 4 and 5 are the closest as you can get when you're talking about integers, so sorry. I know I probably wasn't much help to you. But I tried. Sorry. Title: Math Question! "Find two consecutive integers such that the square of the smaller is 10 more than the larger."? Post by: tommychk0235 on Sep 4, 2012 Let the integers be n and n + 1. the information in the question tells you that
n^2 = n + 1 + 10 ----> n^2 - n - 11 = 0 This quadratic does not have integer solutions so there is something wrong with the question. Title: Re: Math Question! Post by: bio_man on Sep 4, 2012 Please mark as solved.
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