Question 280766
Each field is square, so their areas are simply l*w = l^2 = w^2.
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x = sides of square field 1
y = sides of square field 2
z = sides of square field 3
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x^2 + y^2 + z^2 = 38  :: the sum of the areas of the three fields is 38
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x = sides of the smallest field
y = x+1 = sides of the middle-sized field
z = x+3 = sides of the largest field
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x^2 + (x+1)^2 + (x+3)^2 = 38
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x^2 + (x^2 +2x +1) + (x^2  +6x +9) = 38
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collecting terms and simplifying
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3x^2 + 8x + 10 - 38 = 0
3x^2 + 8x -28 = 0
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factoring
(3x +14 )(x - 2) = 0
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The roots are x=-14/3 and x=2.
Negative values are the sides of fields are nonsensical, so we suspect x=2.
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x = 2
y = x+1 = 3
z = x+3 = 5
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Checking the proposed solution...
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Does x^2 +y^2 +z^2 = 38?
2^2 = 4
3^2 = 9
5^2 = 25
4+9+25 = 38
YES!
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Answer:  The areas of each of the three square fields are:  4, 9, and 25.
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Done