SOLUTION: Find the dimensions of a rectangle if the perimeter is 72m and area is 288m^2

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Question 968372: Find the dimensions of a rectangle if the perimeter is 72m and area is 288m^2
Answer by amarjeeth123(569) About Me  (Show Source):
You can put this solution on YOUR website!
Let the length be x.
Let the breadth be y.
Perimeter=2(length+breadth)=2(x+y)
2(x+y)=72
x+y=36
y=36-x
Area=length*breadth
x(36-x)=288
36x-x^2=288
x^2-36x+288=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-36x%2B288+=+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=%28-36%29%5E2-4%2A1%2A288=144.

Discriminant d=144 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--36%2B-sqrt%28+144+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-36%29%2Bsqrt%28+144+%29%29%2F2%5C1+=+24
x%5B2%5D+=+%28-%28-36%29-sqrt%28+144+%29%29%2F2%5C1+=+12

Quadratic expression 1x%5E2%2B-36x%2B288 can be factored:
1x%5E2%2B-36x%2B288+=+1%28x-24%29%2A%28x-12%29
Again, the answer is: 24, 12. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-36%2Ax%2B288+%29

x=24 m
y=12 m
The dimensions of the rectangle are 24m and 12m.