SOLUTION: Can anyone help with this problem? Find two consecutive odd positive integers such that the twice the square of the larger minus the square of the smaller is 41 (check your answ

Question 122798: Can anyone help with this problem?
Find two consecutive odd positive integers such that the twice the square of the larger minus the square of the smaller is 41 (check your answer)
larger = x
smaller = y

Answer by oscargut(2103) (Show Source): You can put this solution on YOUR website!
If x is the larger and y the smallest then y=x-2 because are consecutive odds
then 2x^2-(x-2)^2=41
then x^2+4x-4=41 then x^2+4x-45=0 then x=5 or x=-9
so x=5 and y=x-2=3
The numbers are 5 and 3
Check:
twice the square of the larger is 2(5^2)=50
the square of the smaller is 3^2=9
so the twice the square of the larger minus the square of the smaller is 41
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