SOLUTION: i have a math problem that says my yard is 20 by 30 feet and i want to put a flower bed that boarders the yard. if the width is constant and i have 187ft^2 of flowers what is the w

Algebra ->  Rectangles -> SOLUTION: i have a math problem that says my yard is 20 by 30 feet and i want to put a flower bed that boarders the yard. if the width is constant and i have 187ft^2 of flowers what is the w      Log On


   

Question 239304: i have a math problem that says my yard is 20 by 30 feet and i want to put a flower bed that boarders the yard. if the width is constant and i have 187ft^2 of flowers what is the width of the flower bed
Answer by edjones(8007) About Me  (Show Source):
You can put this solution on YOUR website!
Let x=how much you must increase the length of each side to have the area of the flower bed be 187'
(x+30)(x+20)-(30*20)=187 new area-old area=187
x^2+50x+600-600=187
x^2+50x-187=0
x=3.5' Quadratic formula below.
x/2 = 3.5/2 = 1.75' width of the flower bed is 1/2 of the increase in length of each side.
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Ed
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Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B50x%2B-187+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%2850%29%5E2-4%2A1%2A-187=3248.

Discriminant d=3248 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-50%2B-sqrt%28+3248+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%2850%29%2Bsqrt%28+3248+%29%29%2F2%5C1+=+3.49561369755001
x%5B2%5D+=+%28-%2850%29-sqrt%28+3248+%29%29%2F2%5C1+=+-53.49561369755

Quadratic expression 1x%5E2%2B50x%2B-187 can be factored:
1x%5E2%2B50x%2B-187+=+1%28x-3.49561369755001%29%2A%28x--53.49561369755%29
Again, the answer is: 3.49561369755001, -53.49561369755. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B50%2Ax%2B-187+%29