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Question 889419: find two consecutive psitive odd numbers which are such that the he sum square of their sum exceeds the sum of their square by 26.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! I could not get an integer solution with a difference of 26
a difference of 30 or of 70, yields integer solutions
Solving this problem
find two consecutive positive odd integers such that their sum squared, exceeds the sum of their squares by 30.
:
Let x = the odd number
then
(x+2) = the next odd number
therefore
(2x+2) = the sum of these two numbers
:
(2x+2)^2 - (x^2 + (x+2)^2) = 30
FOIL
4x^2 + 8x + 4 - (x^2 + x^2 + 4x + 4) = 30
4x^2 + 8x + 4 - x^2 - x^2 - 4x - 4 = 30
combine like terms, to form a quadratic equation
2x^2 + 4x - 30 = 0
simplify, divide by 2
x^2 + 2x - 30 = 0
Factors to
(x+5)(x-3) = 0
The positive solution
x = 3
therefore the two consecutive odd numbers
3, 5
:
:
Check this
8^2 - (3^2 + 5^2) =
64 - 34 = 30
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