The length of a rectangle is 1 ft more than twice the width, and the area of the rectangle is 66ft. find the dimensions of the rectangle

12 ft long by 5½ ft wide  

Step-by-step explanation:

1. Set up an expression for the area.

   Let l = the length of the rectangle

and w = the width. Then

     2w = twice the width and

2w + 1 = 1 more than twice the width. Then

        l = 2w + 1

The formula for the area of a rectangle is  

                 A = length × width

                 A = lw

               66 = (2w +1)w

               66 = 2w² + w

2w² + w - 66 = 0

2. Solve the quadratic for w

2w² + w - 66 = 0

(a) Multiply the first and last terms

2 × (-66) = -132

(b) List all the factors of 132

 1  132

2   66

3    42

4    33

6    22

11     12

(c) Find a pair of factors whose product is -132 and whose sum is 1.

After some trial and error, you will choose -11 and +12,

-11 × 12 = -132 and -11 + 12 = 1.

(d) Rewrite w as -11w + 12w

2w² - 11w + 12w - 66 = 0

(e) Factor by grouping

w(2w - 11) + 6(2w - 11) = 0

(w + 6)(2w - 11) = 0

(f) Use the zero product theorem

w + 6 = 0     2w - 11 = 0

     w = -6          2w = 11

                           w = 5½

We reject the negative answer, so w = 5½ ft

3. Calculate l

l = 2w + 1 = 2 × 5½ + 1 = 11 + 1 = 12 ft

The rectangle is 12 ft long and 5½ ft wide.

i am unable to solve this problem

step-by-step explanation: