In the provided image, the dimensions of the rectangle are expressed in terms of $x$ and $x + 8$, where:

• The width is $x$,
• The length is $x + 8$.

You want to find $x$ and $x + 8$, given that the area of the rectangle is 144 square meters. The area $A$ of a rectangle is calculated by multiplying its length and width, so we have:

$A = \text{length} \times \text{width}$

Substituting the given expressions for the length and width:

$144 = x(x + 8)$

Simplifying this:

$144 = x^2 + 8x$

Now, we rearrange this equation into standard quadratic form:

$x^2 + 8x - 144 = 0$

We'll solve this quadratic equation using factoring. First, we need to find two numbers that multiply to give $-144$ and add to give $8$. The numbers $12$ and $-12$ satisfy this, so we factor the quadratic as follows:

$x^2 + 8x - 144 = (x - 12)(x + 12) = 0$

Thus, we have two solutions for $x$:

$x = 12 \quad \text{or} \quad x = -12$

Since $x$ represents the width of a rectangle, it must be positive, so:

$x = 12$

Now that we have $x = 12$, we can find the length:

$\text{Length} = x + 8 = 12 + 8 = 20$

Therefore, the dimensions of the rectangle are:

• Width: $12$ meters
• Length: $20$ meters

Would you like further details or have any additional questions?

Here are 5 related questions for further practice:

1. How would you solve the same problem if the area was given as 200 square meters?
2. Can you use the quadratic formula to solve $x^2 + 8x - 144 = 0$ instead of factoring?
3. What are the units of measurement for area, length, and width in this context?
4. How would the dimensions change if the length was increased by 5 units?
5. What would happen to the area if the length and width were both doubled?

Tip: Factoring is a faster method to solve quadratics if the numbers are manageable. For more complex cases, try using the quadratic formula.