A garden;s length is 8 meters more than its width. If both the width and the length of the garden are doubled, the area is 192m^2. What are the dimensions of the length and width?

Let's denote the width of the garden as \( w \) meters. Since the length is 8 meters more than the width, the length can be expressed as \( l = w + 8 \).

Now, according to the problem, both the width and the length are doubled, and the area of the garden becomes 192 square meters. The area of a rectangle is given by the formula:

\[
\text{Area} = \text{Length} \times \text{Width}
\]

After doubling the dimensions, the width becomes \( 2w \) and the length becomes \( 2(w + 8) \). The new area is 192 square meters, so:

\[
2w \times 2(w + 8) = 192
\]

Simplify the equation:

\[
4w(w + 8) = 192
\]

Divide both sides by 4:

\[
w(w + 8) = 48
\]

Now, expand the equation:

\[
w^2 + 8w = 48
\]

Rearrange it into a standard quadratic form:

\[
w^2 + 8w - 48 = 0
\]

Now, solve this quadratic equation using the quadratic formula. The quadratic formula is:

\[
w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For the equation \( w^2 + 8w - 48 = 0 \), the coefficients are:

- \( a = 1 \)
- \( b = 8 \)
- \( c = -48 \)

Substitute these values into the quadratic formula:

\[
w = \frac{-8 \pm \sqrt{8^2 - 4(1)(-48)}}{2(1)}
\]
\[
w = \frac{-8 \pm \sqrt{64 + 192}}{2}
\]
\[
w = \frac{-8 \pm \sqrt{256}}{2}
\]
\[
w = \frac{-8 \pm 16}{2}
\]

Now, solve for \( w \):

1. \( w = \frac{-8 + 16}{2} = \frac{8}{2} = 4 \)
2. \( w = \frac{-8 - 16}{2} = \frac{-24}{2} = -12 \)

Since the width cannot be negative, \( w = 4 \) meters.

Now, substitute \( w = 4 \) into the expression for the length:

\[
l = w + 8 = 4 + 8 = 12 \, \text{meters}
\]

Therefore, the original dimensions of the garden are:

- **Width = 4 meters**
- **Length = 12 meters**

Would you like further details or explanations?

### Related Questions:
1. What would the area of the garden be before doubling the dimensions?
2. How would the problem change if the length was 10 meters more than the width?
3. What if the area of the garden became 240 m² instead of 192 m² after doubling the dimensions?
4. How do you solve a quadratic equation using the factorization method?
5. Can you solve for the dimensions if the length is given as 2 meters more than twice the width?

### Tip:
When solving quadratic equations, always check for common factors before applying the quadratic formula—it might simplify the process!

A garden's length is 8 meters more than its width. If both the width and the length of the garden are doubled, the area is 192m². What are the dimensions of the length and width?

Find the dimensions of a garden where the length is 8 meters more than the width, and the area becomes 192 m² after doubling both dimensions. This problem uses algebra, geometry, and the quadratic formula, suitable for students in Grades 8-10.

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