The area of a rectangle is 63 yd^2, and the length of the rectangle is 5 yd more than twice the width. Find the dimensions of the rectangle.

To find the dimensions of the rectangle, we need to set up an equation based on the given information.

Let's denote:
- \( W \) as the width of the rectangle (in yards).
- \( L \) as the length of the rectangle (in yards).

From the problem, we know two things:
1. The area of the rectangle is 63 square yards.
2. The length of the rectangle is 5 yards more than twice the width.

We can express these relationships with the following equations:

\[
L \times W = 63
\]

\[
L = 2W + 5
\]

We can substitute the second equation into the first equation to solve for \( W \):

\[
(2W + 5) \times W = 63
\]

\[
2W^2 + 5W = 63
\]

This is a quadratic equation, and we can solve it by setting it to zero:

\[
2W^2 + 5W - 63 = 0
\]

We can solve this quadratic equation using the quadratic formula \( W = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = 5 \), and \( c = -63 \).

First, we calculate the discriminant:

\[
b^2 - 4ac = 5^2 - 4 \cdot 2 \cdot (-63) = 25 + 504 = 529
\]

Next, we take the square root of the discriminant:

\[
\sqrt{529} = 23
\]

Now, we can find the solutions for \( W \):

\[
W = \frac{-5 \pm 23}{2 \cdot 2} = \frac{-5 \pm 23}{4}
\]

This gives us two solutions:

\[
W = \frac{-5 + 23}{4} = \frac{18}{4} = 4.5 \quad \text{(valid width)}
\]

\[
W = \frac{-5 - 23}{4} = \frac{-28}{4} = -7 \quad \text{(not valid as width cannot be negative)}
\]

So, the width \( W \) is \( 4.5 \) yards.

Now, we can find the length \( L \):

\[
L = 2W + 5 = 2(4.5) + 5 = 9 + 5 = 14 \text{ yards}
\]

Therefore, the dimensions of the rectangle are:

- Width: \( 4.5 \) yards
- Length: \( 14 \) yards

Would you like more details or have any questions?

Here are some related questions:

1. What is the perimeter of the rectangle with the given dimensions?
2. How would the area change if both the length and width were doubled?
3. What if the length were 10 yards more than twice the width? How would that affect the dimensions?
4. How can you derive the quadratic formula used in solving the problem?
5. Can you verify the dimensions by recalculating the area with the found values?
6. What is the ratio of the length to the width for this rectangle?
7. If the width were increased by 2 yards, what would be the new area?
8. How do you determine if a quadratic equation has real solutions?

**Tip:** When solving quadratic equations, always check the discriminant first to determine the nature of the roots.

Discover how to find the dimensions of a rectangle with an area of 63 square yards. This problem involves solving quadratic equations and applying algebraic concepts. Suitable for Grades 7-9 students.

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