SOLUTION: find two consecutive positive odd numbers which are such that the square of their sum exceeds the sum of their squares by 126.

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.
RELATED QUESTIONS
find two consecutive positive odd numbers which are such that the square of their sum... (answered by mananth)

Find two consecutive positive odd numbers which are such that the square of their sum... (answered by ikleyn)

find two consecutive psitive odd numbers which are such that the he sum square of their... (answered by ankor@dixie-net.com)

Find two consecutive positive odd numbers such that the sum of their squares is... (answered by rothauserc)

Find two consecutive positive odd numbers such that sum of their squares is... (answered by geetha_rama)

Three consecutive positive integers are such that the square of their sum exceeds the sum (answered by htmentor)

What are two consecutive odd whole numbers such that the sum of their squares is... (answered by ankor@dixie-net.com)

find two consecutive positive even numbers such that the sum of their squares is 164. (answered by Alan3354)

find two consecutive positive numbers such that the sum of their squares is equal to... (answered by Big Poop)