SOLUTION: The width of a rectangular picture is 8in less than the length. The area of the picture is 20in^2. what are the dimensions of the picture?

Algebra ->  Rectangles -> SOLUTION: The width of a rectangular picture is 8in less than the length. The area of the picture is 20in^2. what are the dimensions of the picture?       Log On


   

Question 939572: The width of a rectangular picture is 8in less than the length. The area of the picture is 20in^2. what are the dimensions of the picture?
Answer by srinivas.g(540) About Me  (Show Source):
You can put this solution on YOUR website!
let length be x in
width = x-8
area = 20 in^2
formula: area = length x width
20 = x*(x-8)
20 = x*x-x*8
20+=+x%5E2-8x
move 20 to the right
+0=x%5E2-8x-20+
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-8x%2B-20+=+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=%28-8%29%5E2-4%2A1%2A-20=144.

Discriminant d=144 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--8%2B-sqrt%28+144+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-8%29%2Bsqrt%28+144+%29%29%2F2%5C1+=+10
x%5B2%5D+=+%28-%28-8%29-sqrt%28+144+%29%29%2F2%5C1+=+-2

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

but x cannot be negative .
it should be positive value
x=10
length = 10 in
width = 10-8 = 2 inches
Result: length = 10 in
width = 2 in