The length of a rectangle is 2 ft less than three times the width. The area is 60 ft SQUARED. Find the length of the rectangle
w(3w-2) = 60
solve for w, then get length
To find the length of the rectangle, we need to set up an equation using the given information.
Let's start by assigning variables:
Let L be the length of the rectangle.
Let W be the width of the rectangle.
According to the problem, the length L is 2 feet less than three times the width W. We can represent this information as an equation:
L = 3W - 2
Additionally, we know that the area of the rectangle is 60 square feet. The area of a rectangle is given by the formula:
Area = Length × Width
Plugging in the values we have, we can write the equation:
60 = L × W
Now, we have a system of two equations:
L = 3W - 2
60 = L × W
To solve this system of equations, we can substitute the value of L from the first equation into the second equation:
60 = (3W - 2) × W
Expanding the equation:
60 = 3W^2 - 2W
Rearranging the equation to standard quadratic form:
3W^2 - 2W - 60 = 0
To solve a quadratic equation, we can factor, use the quadratic formula, or complete the square. In this case, let's solve it by factoring:
(3W + 10)(W - 6) = 0
Setting each factor equal to zero:
3W + 10 = 0 or W - 6 = 0
From the first equation:
3W = -10
W = -10/3
Since the width cannot be negative, we discard the negative solution. Therefore, the width of the rectangle is 6 feet.
Now, substitute the value of W back into one of the original equations to find the length L:
L = 3W - 2
L = 3(6) - 2
L = 18 - 2
L = 16
Therefore, the length of the rectangle is 16 feet.