SOLUTION: The length of a rectangle is 2 inches less than twice the width. The diagonal is 5 inches. Find the length and width of the rectangle.

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Question 145993This question is from textbook
: The length of a rectangle is 2 inches less than twice the width. The diagonal is 5 inches. Find the length and width of the rectangle. This question is from textbook

Found 2 solutions by Alan3354, nerdybill:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
L = 2W-2
W%5E2+%2B+%282W-2%29%5E2+=+5%5E2
W%5E2+%2B+%284W%5E2+-8W+%2B+4%29+=+25
5W%5E2+-8W+%2B+4+=+25
5W%5E2+-8W+-+21+=+0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 5x%5E2%2B-8x%2B-21+=+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%2A5%2A-21=484.

Discriminant d=484 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--8%2B-sqrt%28+484+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-8%29%2Bsqrt%28+484+%29%29%2F2%5C5+=+3
x%5B2%5D+=+%28-%28-8%29-sqrt%28+484+%29%29%2F2%5C5+=+-1.4

Quadratic expression 5x%5E2%2B-8x%2B-21 can be factored:
5x%5E2%2B-8x%2B-21+=+%28x-3%29%2A%28x--1.4%29
Again, the answer is: 3, -1.4. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+5%2Ax%5E2%2B-8%2Ax%2B-21+%29

Ignore the negative length.
W = 3
L = 2*3 -2
L = 4

Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
The length of a rectangle is 2 inches less than twice the width. The diagonal is 5 inches. Find the length and width of the rectangle.
Let w = width of rectangle
then from "length of a rectangle is 2 inches less than twice the width" we have
2w-2 = length of rectangle
.
the length, width and diagonal forms a right triangle allowing you to use Pythagorean theorem:
(2w-2)^2 + w^2 = 5^2
4w^2 - 8w + 4 + w^2 = 25
5w^2 - 8w + 4 = 25
5w^2 - 8w - 21 = 0
.
At this point, you could "factor it" or use the "quadratic equation".
Using the quadratic equation produces:
3 and -1.4
You can't have a negative width therefore could you check whether you had a typo when submitted your problem?