SOLUTION: The length of a picture is 1 inch less than twice its width. The frame around the picture has a uniform width of 2 inches and an area of 96 square inches. What are the dimensions

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Question 40285: The length of a picture is 1 inch less than twice its width. The frame around the picture has a uniform width of 2 inches and an area of 96 square inches. What are the dimensions of the picture?
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
The length of a picture is 1 inch less than twice its width. The frame around the picture has a uniform width of 2 inches and an area of 96 square inches. What are the dimensions of the picture?
OUTSIDE FRAME DATA:
Let the width be "x"
Then the length is "2x-1" inches.
PICTURE DATA:
width = x-2(2)=x-4 inches
length = 2x-1-2(2)= 2x-5 inches
EQUATION:
Picture area = length * width = 96 sq. in.
(x-4)(2x-5)=96
2x^2-13x+20=96
2x^2-13x-76=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 2x%5E2%2B-13x%2B-76+=+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=%28-13%29%5E2-4%2A2%2A-76=777.

Discriminant d=777 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--13%2B-sqrt%28+777+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-13%29%2Bsqrt%28+777+%29%29%2F2%5C2+=+10.2186799323832
x%5B2%5D+=+%28-%28-13%29-sqrt%28+777+%29%29%2F2%5C2+=+-3.71867993238318

Quadratic expression 2x%5E2%2B-13x%2B-76 can be factored:
2x%5E2%2B-13x%2B-76+=+2%28x-10.2186799323832%29%2A%28x--3.71867993238318%29
Again, the answer is: 10.2186799323832, -3.71867993238318. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B-13%2Ax%2B-76+%29

Only x=10.22 is a usuable answer.
So, dimensions of the picture are:
width = x-2=8.22 inches
length = 2x-5=15.44 inches
cheers,
Stan H.