Let's denote the length of the rectangle as L and the width as W.

From the problem, we know that:

1) W = 4L + 2 (since the width is 2 feet longer than four times its length)

2) The area of a rectangle is given by the formula A = L * W. We know that A = 42, so we can write this as 42 = L * W.

We can substitute the first equation into the second to solve for L:

42 = L * (4L + 2)

This simplifies to:

42 = 4L^2 + 2L

Rearranging terms gives us a quadratic equation:

4L^2 + 2L - 42 = 0

We can divide through by 2 to simplify:

2L^2 + L - 21 = 0

This factors to:

(2L - 7)(L + 3) = 0

Setting each factor equal to zero gives the solutions L = 7/2 = 3.5 and L = -3. Since length cannot be negative, we discard the second solution.

Substituting L = 3.5 into the first equation gives W = 4*3.5 + 2 = 16.

So, the length of the rectangle is 3.5 feet and the width is 16 feet.

Question

Let's denote the length of the rectangle as L and the width as W.

From the problem, we know that:

1) W = 4L + 2 (since the width is 2 feet longer than four times its length)

2) The area of a rectangle is given by the formula A = L * W. We know that A = 42, so we can write this as 42 = L * W.

We can substitute the first equation into the second to solve for L:

42 = L * (4L + 2)

This simplifies to:

42 = 4L^2 + 2L

Rearranging terms gives us a quadratic equation:

4L^2 + 2L - 42 = 0

We can divide through by 2 to simplify:

2L^2 + L - 21 = 0

This factors to:

(2L - 7)(L + 3) = 0

Setting each factor equal to zero gives the solutions L = 7/2 = 3.5 and L = -3. Since length cannot be negative, we discard the second solution.

Substituting L = 3.5 into the first equation gives W = 4*3.5 + 2 = 16.

So, the length of the rectangle is 3.5 feet and the width is 16 feet.

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