Math Question! "Find two consecutive integers such that the square of the smaller is 10 more than the larger."?

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.

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.

Please mark as solved.