SOLUTION: what is the length of a rectangle is twice its width if the width is increased by one meter and the length is diminished by three meters the area will be 12 square meters

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Question 223710: what is the length of a rectangle is twice its width if the width is increased by one meter and the length is diminished by three meters the area will be 12 square meters
Answer by drj(1380) About Me  (Show Source):
You can put this solution on YOUR website!
What is the length of a rectangle is twice its width if the width is increased by one meter and the length is diminished by three meters the area will be 12 square meters.

Step 1. Let w be the width

Step 2. Let 2w be the length.

Step 3. Let w+1 be the width increased by one meter.

Step 4. Let 2w-3 be the length diminished by three meters.

Step 5. The Area A=(w+1)(2w-3)=12 since the area will be 12 meters with these changed dimensions.

Step 6. Solving the equation in Step 5 will lead to a quadratic equation as follows:

2w%5E2-3w%2B2w-3=12

Subtract 12 from both sides of the equation

2w%5E2-w-3-12=12-12

2w%5E2-w-15=0

Step 7. To solve, use the quadratic formula given as x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

where a=2, b=-1, and c=-15.

Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 2x%5E2%2B-1x%2B-15+=+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=%28-1%29%5E2-4%2A2%2A-15=121.

Discriminant d=121 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--1%2B-sqrt%28+121+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-1%29%2Bsqrt%28+121+%29%29%2F2%5C2+=+3
x%5B2%5D+=+%28-%28-1%29-sqrt%28+121+%29%29%2F2%5C2+=+-2.5

Quadratic expression 2x%5E2%2B-1x%2B-15 can be factored:
2x%5E2%2B-1x%2B-15+=+2%28x-3%29%2A%28x--2.5%29
Again, the answer is: 3, -2.5. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B-1%2Ax%2B-15+%29



Selecting the positive solution given as w=3, then 2w=6. Check area A with width 3+1=4 and length 2w-3=3. So the area with these dimensions is 12 as given by the problem statement.

Step 8. ANSWER: The length is 6 meters.

I hope the above steps were helpful.

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Good luck in your studies!

Respectfully,
Dr J