A rectangle has a width of x. Its length is 4 feet longer than twice the width. If the area of the rectangle is 30 ft.^2 what is the width, in feet, of the rectangle?

Answer 1

We can use the formula for the area of a rectangle:

Area = length x width

We know that the area is 30 ft^2, so we can plug that in:

30 = length x width

We also know that the length is 4 feet longer than twice the width:

length = 2x + 4

Now we can substitute this expression for length into the area formula:

30 = (2x + 4) x width

Simplifying:

30 = 2x^2 + 4x

Dividing both sides by 2:

15 = x^2 + 2x

Rearranging:

0 = x^2 + 2x - 15

Now we can use the quadratic formula to solve for x:

x = (-b ± sqrt(b^2 - 4ac)) / 2a

In this case, a = 1, b = 2, and c = -15:

x = (-2 ± sqrt(2^2 - 4(1)(-15))) / (2)(1)

x = (-2 ± sqrt(64)) / 2

x = (-2 ± 8) / 2

x = -5 or x = 3

Since the width of a rectangle can't be negative, the only possible solution is x = 3. Therefore, the width of the rectangle is 3 feet.