SOLUTION: A rectangle sign must have an area of 34 square yards. Its length must be 2 yards more than its width. Find the exact dimensions of the sign.

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Question 902943: A rectangle sign must have an area of 34 square yards. Its length must be 2 yards more than its width. Find the exact dimensions of the sign.
Answer by Stitch(470) About Me  (Show Source):
You can put this solution on YOUR website!
Let L = length of the sign
Let W = width of the sign
Since you are given two unknowns, 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+2 (The length of the sign is 2yds more than the width)
Given: A = 34 yds^2
Plug 34 into equation 1 for A
Equation 1: A+=+L+%2A+W
34+=+L+%2A+W
Notice that equation 2 is already solved for L
Plug (W + 2) into equation 1 for L
34+=+%28W+%2B+2%29+%2A+W
Multiply the W through on the right hand side
34+=+W%5E2+%2B+2W
Subtract 34 from both sides
0+=+W%5E2+%2B+2W+-+34
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%2B2W%2B-34+=+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=%282%29%5E2-4%2A1%2A-34=140.

Discriminant d=140 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-2%2B-sqrt%28+140+%29%29%2F2%5Ca.

W%5B1%5D+=+%28-%282%29%2Bsqrt%28+140+%29%29%2F2%5C1+=+4.91607978309962
W%5B2%5D+=+%28-%282%29-sqrt%28+140+%29%29%2F2%5C1+=+-6.91607978309962

Quadratic expression 1W%5E2%2B2W%2B-34 can be factored:
1W%5E2%2B2W%2B-34+=+1%28W-4.91607978309962%29%2A%28W--6.91607978309962%29
Again, the answer is: 4.91607978309962, -6.91607978309962. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B2%2Ax%2B-34+%29

The quadratic equation tells us that W = 4.916 & -6.916
Since we can not have a negative distance we will only use W = 4.916
Now plug 4.916 into equation 2 for W
Equation 2: L+=+W+%2B+2
L+=+%284.916%29+%2B+2
highlight%28L+=+6.916%29
The Length of the lawn is 6.196 yards and the Width is 4.916 yards