SOLUTION: The length of a rectangle is 2 m more that the width. The area is 48 m2. Find the dimensions of the rectangle.

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Question 939578: The length of a rectangle is 2 m more that the width. The area is 48 m2. Find the dimensions of the
rectangle.
Answer by srinivas.g(540) About Me  (Show Source):
You can put this solution on YOUR website!
let width be x meters
length = 2+x
area = 48 m^2
formula: area = length x width
48 = (2+x)*x
48= 2*x+x*x
48=+2x%2Bx%5E2
move 48 to the right
0=2x%2Bx%5E2-48
+x%5E2%2B2x-48=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B2x%2B-48+=+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=%282%29%5E2-4%2A1%2A-48=196.

Discriminant d=196 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-2%2B-sqrt%28+196+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%282%29%2Bsqrt%28+196+%29%29%2F2%5C1+=+6
x%5B2%5D+=+%28-%282%29-sqrt%28+196+%29%29%2F2%5C1+=+-8

Quadratic expression 1x%5E2%2B2x%2B-48 can be factored:
1x%5E2%2B2x%2B-48+=+1%28x-6%29%2A%28x--8%29
Again, the answer is: 6, -8. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B2%2Ax%2B-48+%29

x cannot be negative
hence x= 6
width (x)= 6 m
length = 6+2 = 8 m