SOLUTION: Winter wheat. While finding the amount of seed needed to plant his three square wheat fields, Hank observed that the side of one field was 1 kilometer longer than the side of th

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: Winter wheat. While finding the amount of seed needed to plant his three square wheat fields, Hank observed that the side of one field was 1 kilometer longer than the side of th      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 200495This question is from textbook Elementary and Intermediate Algebra
: Winter wheat. While finding the amount of seed needed
to plant his three square wheat fields, Hank observed that
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? This question is from textbook Elementary and Intermediate Algebra

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Let x = the side of the smallest field so its area is: A%5B1%5D+=+x%5E2
The side of the next largest field is x+1 so its area is is: A%5B2%5D+=+%28x%2B1%29%5E2
The side of the largest field is x+3 so its area is: A%5B3%5D+=+%28x%2B3%29%5E2
Add the three areas and set the sum = 38 sq.km.
x%5E2%2B%28x%5E2%2B2x%2B1%29%2B%28x%5E2%2B6x%2B9%29+=+38 Simplify the left side.
3x%5E2%2B8x%2B10+=+38 Subtract 38 from both sides.
3x%5E2%2B8x-28+=+0 Using the quadratic formula to solve: x+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a where a = 3, b = 8, and c = -28, we get...
x+=+%28-8%2B-sqrt%288%5E2-4%283%29%28-28%29%29%29%2F2%283%29 Simplifying, we get...
x+=+%28-8%2B-sqrt%2864-%28-336%29%29%29%2F6
x+=+%28-8%2B-sqrt%28400%29%29%29%2F6
x+=+%28-8%2B20%29%2F6 or x+=+%28-8-20%29%2F6 Discard the negative solution as the side of the square is a positive value.
x+=+%28-8%2B20%2F6%29
highlight%28x+=+2%29
The area of the smallest field is A%5B1%5D+=+x%5E2 or highlight%28A%5B1%5D+=+4%29sq.km.
The area of the middle field is A%5B2%5D+=+%28x%2B1%29%5E2 or highlight_green%28A%5B2%5D+=+9%29sq.km.
The area of the largest field is A%5B3%5D+=+%28x%2B3%29%5E2 or highlight%28A%5B3%5D+=+25%29sq.km.