SOLUTION: The length of a rectangle is 2 feet more than the width. The area of the rectangle is 2 square feet. What are the dimensions of the rectangle?

Question 1144267: The length of a rectangle is 2 feet more than the width. The area of the rectangle is 2 square feet. What are the dimensions of the rectangle?
Found 3 solutions by Alan3354, Theo, ikleyn:
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
The length of a rectangle is 2 feet more than the width. The area of the rectangle is 2 square feet. What are the dimensions of the rectangle?
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Area = L*W = 2 sq ft
L = W+2
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W*(W+2) = 2
W^2 + 2w - 2 = 0
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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=12 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 0.732050807568877, -2.73205080756888. Here's your graph:

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W = the positive value
Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
l = length
w = width
l = w + 2
area = l * w, and since l = w + 2, then area = (w + 2) * w = w^2 + 2w.
since area = 2, then 2 = w^2 + 2w.
subtract 2 from both sides of the equation to get w^2 + 2w - 2 = 0
factor this quadratic equation to get:
w = .7320508076 or w = -2.7320508076
w has to be positive so w = .7320508076
since area = length * width, and since area = 2 and since width = .7320508076 and since length = width + 2, then length = 2.7320508076.
length * width = .7320508076 * 2.7320508076 = 2
this confirms the solution is correct.

Answer by ikleyn(52803)   (Show Source): You can put this solution on YOUR website!
.

            For this problem, I will show you two ways (two methods) of solution.

            First way is TRADITIONAL. You may find it everywhere, and it is boring.

If x is the width, then the length is (x+2) and the area equation is x*(x+2) = 2 square feet, x^2 + 2x - 2 = 0. Use the quadratic formula = = = = . The width should be positive, so only positive root x = is the solution for the width. ANSWER. The width is W = . The length is L = W+2 = .

            The other method is fresh as a matutinal dawn in May, unexpected and elegant.

            You may learn it only from me at this forum and in this site -- and practically nowhere else.

Let "x" be an unknown value on number line exactly half-way between the length L and the width W values of the rectangle. Then, OBVIOUSLY, x = W + 1 = L - 1, and the area is L*W = (x+1)*(x-1) = 2, or = 2, i.e. = 2 + 1 = 3; hence, x = . Thus the dimensions of the rectangle are W = x-1 = (the width) and L = x+1 = (the length). You got the same answer, in a quick and simple manner.

See the lessons
    - HOW TO solve the problem on quadratic equation mentally and avoid boring calculations

    - Problems on the area and the dimensions of a rectangle
    - Three methods to find the dimensions of a rectangle when its perimeter and the area are given
    - Three methods to find the dimensions of a rectangle when its area and the difference of two dimensions are given
in this site,

where you will find many other similar solved problems (your TEMPLATES) with detailed explanations.

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            Come again soon to this forum to learn more (!)

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