The length of a rectangle is 3ft more than twice its width, and the area of the rectangle is 54ft^2 . Find the dimensions of the rectangle.
The other problem worked but how do you SOLVE it. I need to be able to do similar problems. I know the factors multiply to get the area but what if they're are more than one numbers that can equal to that number?
To solve this problem, we can use algebraic equations to represent the given information and find the dimensions of the rectangle. Let's start step by step.
1. Let's assume the width of the rectangle as "w" feet.
2. According to the problem, the length of the rectangle is 3 feet more than twice its width. So, the length can be expressed as "2w + 3" feet.
3. The area of the rectangle is given as 54 square feet. The formula for the area of a rectangle is length × width. So, we can set up the equation: (2w + 3) × w = 54.
4. Simplify the equation by multiplying: 2w^2 + 3w = 54.
5. Rearrange the equation to form a quadratic equation in standard form: 2w^2 + 3w - 54 = 0.
6. We can now solve this quadratic equation by factoring, completing the square, or using the quadratic formula.
In this case, let's use factoring. The equation can be factored as: (2w - 9)(w + 6) = 0.
By equating each factor to zero, we get two solutions: 2w - 9 = 0 or w + 6 = 0.
Solving these equations, we find two possibilities for the width: w = 9/2 or w = -6.
However, since width cannot be negative, we ignore the second solution.
7. Now that we have the width, we can substitute it back into the expression for the length to find the dimensions. Using w = 9/2, we have: length = 2w + 3 = 2(9/2) + 3 = 9 + 3 = 12 feet.
Therefore, the dimensions of the rectangle are width = 9/2 feet (or 4.5 feet) and length = 12 feet.