SOLUTION: Find two consecutive odd positive integers such that the twice the square of the larger minus the square of the smaller is 41. Set up an algebraic equation and solve for the two nu

Question 127920: Find two consecutive odd positive integers such that the twice the square of the larger minus the square of the smaller is 41. Set up an algebraic equation and solve for the two numbers.
Answer by oscargut(2103) (Show Source): You can put this solution on YOUR website!
x= first number
y= second number (y=x+2)
so y is larger than x
then
2y^2-x^2=41
then
2(x+2)^2-x^2=41
then
x^2+8x-33=0
then x=[-8+sqrt(196)]/2 or x=[-8-sqrt(196)]/2
then x=3 or x=-11
then x=3 (because x and y are positives)
then
R: x=3 and y=5
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