The area of a rectangle is 28y^2, and the length of the rectangle is 1 more than twice the width. Find the dimensions of the rectangle.

To solve this problem, we will define the width and length of the rectangle in terms of a variable and then use the information provided to set up and solve an equation.

### Step-by-Step Solution:

1. **Define the variables:**
   - Let \( w \) be the width of the rectangle.
   - According to the problem, the length \( l \) is "1 more than twice the width". So, we can write this as:
     \[
     l = 2w + 1
     \]

2. **Set up the equation for the area:**
   - The area \( A \) of a rectangle is given by the product of its length and width:
     \[
     A = l \times w
     \]
   - We know the area is \( 28y^2 \). Substitute the expressions for \( l \) and \( A \) into the equation:
     \[
     28y^2 = (2w + 1) \times w
     \]

3. **Expand and simplify the equation:**
   - Expand the equation:
     \[
     28y^2 = 2w^2 + w
     \]
   - Rearrange the equation to set it equal to 0:
     \[
     2w^2 + w - 28y^2 = 0
     \]

4. **Solve the quadratic equation for \( w \):**
   - To solve for \( w \), use the quadratic formula \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = 1 \), and \( c = -28y^2 \).
   - Plugging in the values:
     \[
     w = \frac{-1 \pm \sqrt{1^2 - 4 \times 2 \times (-28y^2)}}{2 \times 2}
     \]
   - Simplify under the square root:
     \[
     w = \frac{-1 \pm \sqrt{1 + 224y^2}}{4}
     \]

5. **Determine the width \( w \):**
   - Since the width must be a positive number, we take the positive root:
     \[
     w = \frac{-1 + \sqrt{1 + 224y^2}}{4}
     \]

6. **Calculate the length \( l \):**
   - Now, substitute \( w \) back into the expression for \( l \):
     \[
     l = 2w + 1 = 2 \left( \frac{-1 + \sqrt{1 + 224y^2}}{4} \right) + 1
     \]
   - Simplify this expression:
     \[
     l = \frac{-2 + 2\sqrt{1 + 224y^2}}{4} + 1 = \frac{-2 + 2\sqrt{1 + 224y^2} + 4}{4}
     \]
     \[
     l = \frac{2 + 2\sqrt{1 + 224y^2}}{4} = \frac{1 + \sqrt{1 + 224y^2}}{2}
     \]

### Final Dimensions:

- The **width** of the rectangle is:
  \[
  w = \frac{-1 + \sqrt{1 + 224y^2}}{4}
  \]

- The **length** of the rectangle is:
  \[
  l = \frac{1 + \sqrt{1 + 224y^2}}{2}
  \]

Would you like more details or have any questions about this solution?

### Related Questions:
1. How would the dimensions change if the area was a different expression?
2. Can you find the width and length of the rectangle if \( y = 1 \)?
3. What happens to the dimensions if the length was defined differently, such as being 3 times the width?
4. How can we graph the relationship between width and length for different values of \( y \)?
5. What are the properties of the quadratic formula that help us solve this problem?

### Tip:
To solve a quadratic equation, always make sure to rearrange it in the standard form \( ax^2 + bx + c = 0 \) and then use the quadratic formula or factorization for solutions.

Learn how to find the dimensions of a rectangle with an area of 28y^2, where the length is 1 more than twice the width. This problem involves algebraic concepts, including quadratic equations and the quadratic formula. Suitable for students in Grades 8-10.

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