Question 232923: Find 2 consecutive numbers such that when you square the smaller number, the result is 8 less than the product of the numbers. Answer by checkley77(12844) (Show Source):
You can put this solution on YOUR website! Let x & x+1 be the 2 numbers.
x^2=x(x+1)-8
x^2=x^2+x-8
x^2-x^2-x=-8
-x=-8
x=8 ans. for the smaller number.
8^2=8(8+1)-8
64=8*9-8
64=72-8
64=64