The area of a rectangle is 14m^2, and the length of the rectangle is 3m less than twice the width. Find the dimensions of the rectangle.
L = 2w - 3
L * w = (2w-3) * w = 14
Solve for w then L.
width --- x m
length = 2x-3 m
x(2x-3) = 14
2x^2 - 3x - 14 = 0
Solve using your favourite method of solving quadratics
Hint: it factors
To solve this problem, we'll first set up equations based on the given information.
Let's assume that the width of the rectangle is represented by "w" meters.
Since the length of the rectangle is 3 meters less than twice the width, the length can be represented as "2w - 3" meters.
The formula to find the area of a rectangle is:
Area = Length * Width
We are given that the area is 14 square meters. Therefore, we can set up the equation:
14 = (2w - 3) * w
To solve this equation, we can use the distributive property:
14 = 2w^2 - 3w
Rearranging the equation to set it equal to zero:
2w^2 - 3w - 14 = 0
Now, we can factor or use the quadratic formula to solve for "w". In this case, let's use factoring:
(2w - 7)(w + 2) = 0
Setting each factor equal to zero:
2w - 7 = 0 or w + 2 = 0
Solving each equation:
2w = 7 or w = -2
Dividing both sides by 2:
w = 7/2 or w = -2
Since the width cannot be negative, we can ignore the solution w = -2.
Therefore, the width of the rectangle is 7/2 or 3.5 meters.
To find the length, we can substitute this value back into our expression for the length:
Length = 2w - 3
Length = 2(3.5) - 3
Length = 7 - 3
Length = 4 meters
So, the dimensions of the rectangle are width = 3.5 meters and length = 4 meters.