The area of a rectangle is 28 square inches. The length is 8 more than thrice the width. Find the length and the

width.

Thanks

w(3w+8) = 28

3w^2+8w-28 = 0
(3w+14)(w-2) = 0
w = 2

Thank you Steve.

To find the length and width of the rectangle, we can set up two equations based on the given information.

Let's represent the width as "w" and the length as "l".

We are given two pieces of information:

1. The area of the rectangle is 28 square inches, which means that l * w = 28.
2. The length is 8 more than thrice the width, which means that l = 3w + 8.

To solve this system of equations, we can substitute the second equation into the first equation:

(3w + 8) * w = 28

Now we can simplify and solve for w:

3w^2 + 8w = 28

Rearranging this equation:

3w^2 + 8w - 28 = 0

Now we can factor or use the quadratic formula to solve for w.

Factoring would give us:

(3w - 4)(w + 7) = 0

Setting each factor equal to zero, we have two possible solutions:

1. 3w - 4 = 0
3w = 4
w = 4/3

2. w + 7 = 0
w = -7

Since the width cannot be negative, we discard the second solution.

Therefore, the width of the rectangle is w = 4/3.

To find the length, we substitute the value of w into the second equation:

l = 3w + 8
l = 3(4/3) + 8
l = 4 + 8
l = 12

Therefore, the length of the rectangle is l = 12 inches and the width is w = 4/3 inches.