SOLUTION: find two numbers such that their sum multiplied by the sum of their squares is 5500, and their difference multiplied by the difference of their squares is 352.
Algebra ->
Problems-with-consecutive-odd-even-integers
-> SOLUTION: find two numbers such that their sum multiplied by the sum of their squares is 5500, and their difference multiplied by the difference of their squares is 352.
Log On
Question 770323: find two numbers such that their sum multiplied by the sum of their squares is 5500, and their difference multiplied by the difference of their squares is 352. Answer by oscargut(2103) (Show Source):
You can put this solution on YOUR website! Hi,
Here is a summary of the solution:
(a+b)(a2+b2)= 5500
(a-b) (a2-b2) = 352 then (a-b)2(a+b) = 352
Then:
5500/(a2-b2) = 352/(a-b)2
Then (a-b)2 must be a factor of 352 then (a-b)2 = 4 or (a-b)2 = 16
But it can`t be 4, (a-b)2 = 16
Then: a2+b2=250 (a-b) = 4
So a solution is a = 13 and b = 9
Answer: a = 13 and b = 9
You can ask me more at: mthman@gmail.com
Thanks
