Question 892521: find two consecutive positive odd numbers which are such that the square of their sum exceeds the sum of their squares by 126. Answer by LinnW(1048) (Show Source):
You can put this solution on YOUR website! set x = the first number
x + 1 = the second number
The square of the sum is (x + (x + 1))^2
The sum of the squares is x^2 + (x + 1)^2
(x + (x + 1))^2 - ( x^2 + (x + 1)^2) = 126
(2x + 1)^2 - ( x^2 + x^2 +2x + 1 ) = 126
(4x^2 + 4x + 1) - ( 2x^2 + 2x + 1 ) = 126
(4x^2 + 4x + 1) - 2x^2 - 2x - 1 ) = 126
2x^2 + 2x = 126
add -126 to each side
2x^2 +2x -126 = 0
divide by 2
x^2 + x -63 = 0
Since there is no integer solution for this,
two positive integers do not exist to solve the problem.
