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put this solution on YOUR website! find 3 consecutive positive odd integers such that the sum of the squares of the first and second integers is equal to the square of the third integer plus 33
Smallest odd integer = x-2
Middle-sized odd integer = x
Largest odd integer = x+2
(x-2)² + x² = (x+2)² + 33
(x-2)² - (x+2)² = 33 - x²
[(x-2)-(x+2)][(x-2)+(x+2)] = 33-x²
[x-2-x-2][x-2+x+2] = 33-x²
[-4][2x] = 33-x²
-8x = 33-x²
x²-8x-33 = 0
(x-11)(x+3) = 0
x-11 = 0 x+3 = 0
x = 11 x = -3
We only want the positive odd integers so we
ignore the negative answer.
Smallest odd integer = x-2 = 11-2 = 9
Middle-sized odd integer = x = 11
Largest odd integer = x+2 = 11+2 = 13
Edwin