SOLUTION: the length of a lawn is 10 m more than the width, and the area is 600 m^2. Write the equation and solve it to find the dimensions of the lawn. can someone please explain how y

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: the length of a lawn is 10 m more than the width, and the area is 600 m^2. Write the equation and solve it to find the dimensions of the lawn. can someone please explain how y      Log On


   

Question 902628: the length of a lawn is 10 m more than the width, and the area is 600 m^2. Write the equation and solve it to find the dimensions of the lawn.

can someone please explain how you get the answer and what equation you are going to use i.e how are you going to get the equation with this information provided :):)
Answer by Stitch(470) About Me  (Show Source):
You can put this solution on YOUR website!
Let L = length of the lawn
Let W = width of the lawn
Since you are given two variables, you will need two equations.
Area(A) of a rectangle = Length x Width
Equation 1: A+=+L+%2A+W (Equation for the area of a rectangle)
Equation 2: L+=+W+%2B+10 (The length of the lawn is 10m more than the width)
Given: A = 600
Plug 600 into equation 1 for A
Equation 1: A+=+L+%2A+W
600+=+L+%2A+W
Notice that equation 2 is already solved for L
Plug (W + 10) into equation 1 for L
600+=+%28W+%2B+10%29+%2A+W
Multiply the W through on the right hand side
600+=+W%5E2+%2B+10W
Subtract 600 from both sides
0+=+W%5E2+%2B+10W+-+600
Now you can use the quadratic equation to solve for W
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation aW%5E2%2BbW%2Bc=0 (in our case 1W%5E2%2B10W%2B-600+=+0) has the following solutons:

W%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=%2810%29%5E2-4%2A1%2A-600=2500.

Discriminant d=2500 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-10%2B-sqrt%28+2500+%29%29%2F2%5Ca.

W%5B1%5D+=+%28-%2810%29%2Bsqrt%28+2500+%29%29%2F2%5C1+=+20
W%5B2%5D+=+%28-%2810%29-sqrt%28+2500+%29%29%2F2%5C1+=+-30

Quadratic expression 1W%5E2%2B10W%2B-600 can be factored:
1W%5E2%2B10W%2B-600+=+1%28W-20%29%2A%28W--30%29
Again, the answer is: 20, -30. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B10%2Ax%2B-600+%29

The quadratic equation tells us that W = 20 & -30.
Since we can not have a negative distance we will only use W = 20
Now plug 20 into equation 2 for W
Equation 2: L+=+W+%2B+10
L+=+%2820%29+%2B+10
highlight%28L+=+30%29
The Length of the lawn is 30m and the Width is 20m