The length of a rectangle exceeds twice its width by 4 feet.
If the area of the rectangle is 286 square feet, find the dimensions.
width --- x
length --- 2x+4
x(2x+4) = 286
2x^2 + 4x - 286 = 0
x^2 + 2x - 143 = 0
(x+13)(x-11) = 0
x = 11 or x = -13, the last part is not possible
the width is 11, the length is 26
check:
26 is 4 more than twice 11
area = 11(26) = 286
Let's assume that the width of the rectangle is represented by "w" feet.
According to the problem, the length of the rectangle exceeds twice its width by 4 feet. Therefore, the length can be expressed as "(2w + 4)" feet.
The formula for the area of a rectangle is given by the product of its length and width. So, we can write the equation:
Area = Length * Width
Plugging in the given area (286 square feet) into the equation, we have:
286 = (2w + 4) * w
Now, let's solve this equation step-by-step to find the value of "w" (width).
1. Expand the equation:
286 = 2w^2 + 4w
2. Rearrange the equation in standard form and set it equal to zero:
2w^2 + 4w - 286 = 0
3. Divide the entire equation by 2 to simplify it:
w^2 + 2w - 143 = 0
4. Factorize the quadratic equation:
(w + 13)(w - 11) = 0
5. Set each factor equal to zero and solve for "w":
w + 13 = 0 or w - 11 = 0
Solving these equations, we get:
w = -13 or w = 11
Since the width of a rectangle cannot be negative, we discard w = -13.
Therefore, the width of the rectangle is 11 feet.
To find the length, we can substitute this value back into the formula "length = 2w + 4":
Length = 2 * 11 + 4 = 22 + 4 = 26 feet
Hence, the dimensions of the rectangle are width = 11 feet and length = 26 feet.
To solve this problem, we need to set up an equation based on the given information and solve for the dimensions of the rectangle.
Let's assume the width of the rectangle is "w" feet.
According to the problem, the length of the rectangle exceeds twice its width by 4 feet. So, the length can be expressed as "2w + 4" feet.
We are also given that the area of the rectangle is 286 square feet. The area of a rectangle is given by the formula: Area = Length × Width.
Therefore, we can write the equation as:
Area = Length × Width
286 = (2w + 4) × w
Now, we can solve this equation to find the value of "w".
Multiplying out the brackets, we have:
286 = 2w^2 + 4w
Rearranging the equation and bringing everything to one side, we have:
2w^2 + 4w - 286 = 0
Now, we can solve this quadratic equation using factoring, completing the square, or using the quadratic formula. Let's use factoring in this case.
Factoring the quadratic equation, we get:
(2w - 22)(w + 13) = 0
Setting each factor equal to zero, we have:
2w - 22 = 0 or w + 13 = 0
Solving each equation separately, we get:
2w = 22 or w = -13
Dividing both sides by 2, we find:
w = 11 or w = -13
We discard the negative value since width cannot be negative.
Therefore, the width of the rectangle is 11 feet.
To find the length, we can substitute this value of "w" back into our expression for the length:
Length = 2w + 4
Length = 2(11) + 4
Length = 22 + 4
Length = 26
Hence, the dimensions of the rectangle are: Width = 11 feet and Length = 26 feet.