SOLUTION: university of phoenix elementary and intermediate algebra solving quadric equations by factoring . while finding the amount of seed needed to plant his three square wheat fields. h

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Question 95237: university of phoenix elementary and intermediate algebra solving quadric equations by factoring . while finding the amount of seed needed to plant his three square wheat fields. hank observed the side of one field was 1 kilometer longer than the side of the smallest field and that the side of the largest field was 3 kilometers longer than the side of the smallest field . if the total area of the three fields is 38 square kilometers, then what is the area of each field?
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
Draw the picture of the three unequal squares.
Label the side of the smallest as "x"
Label the side of the largest as "x+3"
Label the side of the other as "x+1"
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The area of a square is the square of its side.
EQUATION:
x^2 + (x+1)^2 + (x+3)^2 = 38 km^2
x^2 + x^2+3x+1 + x^2+6x+9= 38
3x^2 + 9x + 10 = 38
3x^2+9x-28 = 0
x = [-9+- sqrt(81-4*3*-28)]/6
x = [-9 +-sqrt(417)]/6
x = [-9+-20.4206]/6
x = [11.4206]/6
x = 1.9034 km; Area of this field = 3.623..km^2
x+1 = 2.903429 km.; Area of this field = 8.4299..km^2
x+3 = 4.903429 km. ; 24.0436... km^2
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Adding these you will not get 38 km^2. I think area
is lost in the restricted decimals but you should
check the arithmetic of the problem.
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Cheers,
Stan H.