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.

This is how I solve this problem