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.

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) (Show Source): You can put this solution on YOUR website!
L = 2W-2

Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=484 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 3, -1.4. Here's your graph:

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

Answer by nerdybill(7384) (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?

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