Answer :
Let's solve this problem step-by-step.
1. Define the unknowns:
- Let \( w \) represent the width of the rectangle.
- Therefore, the length \( l \) can be described as \( l = 2w + 3 \).
2. Express the area of the rectangle:
- The area of a rectangle is given by the product of its length and width.
- We are given that the area is 44 square feet, so we can write the equation:
[tex]\[ w \times (2w + 3) = 44 \][/tex]
3. Solve the quadratic equation:
- Expanding the left side of the equation, we get:
[tex]\[ w \times 2w + w \times 3 = 44 \][/tex]
[tex]\[ 2w^2 + 3w = 44 \][/tex]
- Rearrange this to form a standard quadratic equation:
[tex]\[ 2w^2 + 3w - 44 = 0 \][/tex]
4. Find the roots of the quadratic equation:
- We need to solve \( 2w^2 + 3w - 44 = 0 \) to find the possible values for \( w \). The solutions to this equation are:
[tex]\[ w = -\frac{11}{2} \quad \text{and} \quad w = 4 \][/tex]
5. Determine the feasible width:
- Since the width of a rectangle cannot be negative, we discard \( w = -\frac{11}{2} \).
- Therefore, the width \( w \) is 4 feet.
6. Find the length using the width:
- Substitute the width \( w = 4 \) into the equation for the length:
[tex]\[ l = 2w + 3 \][/tex]
[tex]\[ l = 2(4) + 3 \][/tex]
[tex]\[ l = 8 + 3 \][/tex]
[tex]\[ l = 11 \][/tex]
So, the length of the rectangle is 11 feet.
1. Define the unknowns:
- Let \( w \) represent the width of the rectangle.
- Therefore, the length \( l \) can be described as \( l = 2w + 3 \).
2. Express the area of the rectangle:
- The area of a rectangle is given by the product of its length and width.
- We are given that the area is 44 square feet, so we can write the equation:
[tex]\[ w \times (2w + 3) = 44 \][/tex]
3. Solve the quadratic equation:
- Expanding the left side of the equation, we get:
[tex]\[ w \times 2w + w \times 3 = 44 \][/tex]
[tex]\[ 2w^2 + 3w = 44 \][/tex]
- Rearrange this to form a standard quadratic equation:
[tex]\[ 2w^2 + 3w - 44 = 0 \][/tex]
4. Find the roots of the quadratic equation:
- We need to solve \( 2w^2 + 3w - 44 = 0 \) to find the possible values for \( w \). The solutions to this equation are:
[tex]\[ w = -\frac{11}{2} \quad \text{and} \quad w = 4 \][/tex]
5. Determine the feasible width:
- Since the width of a rectangle cannot be negative, we discard \( w = -\frac{11}{2} \).
- Therefore, the width \( w \) is 4 feet.
6. Find the length using the width:
- Substitute the width \( w = 4 \) into the equation for the length:
[tex]\[ l = 2w + 3 \][/tex]
[tex]\[ l = 2(4) + 3 \][/tex]
[tex]\[ l = 8 + 3 \][/tex]
[tex]\[ l = 11 \][/tex]
So, the length of the rectangle is 11 feet.
