To solve this problem, let's define the variables and set up an equation.

Step 1: Define Variables

• Let the width of the rectangle be $w$ (in units).
• Then, the length of the rectangle is $w + 4$ (since the length is 4 units more than the width).

Step 2: Use the Area Formula

The area of a rectangle is given by:

$\text{Area} = \text{Length} \times \text{Width}$

Substitute the values: $48 = (w + 4) \cdot w$

Step 3: Solve the Quadratic Equation

Expand the equation: $48 = w^2 + 4w$

Rearrange it into standard form: $w^2 + 4w - 48 = 0$

Step 4: Factor or Use the Quadratic Formula

We solve this quadratic equation using the quadratic formula: $w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 1$, $b = 4$, and $c = -48$.

$w = \frac{-4 \pm \sqrt{4^2 - 4(1)(-48)}}{2(1)}$ $w = \frac{-4 \pm \sqrt{16 + 192}}{2}$ $w = \frac{-4 \pm \sqrt{208}}{2}$ $w = \frac{-4 \pm 14.42}{2}$

Step 5: Calculate the Roots

$w = \frac{-4 + 14.42}{2} \quad \text{or} \quad w = \frac{-4 - 14.42}{2}$

1. First root: $w = \frac{10.42}{2} = 5.21$

2. Second root: $w = \frac{-18.42}{2} = -9.21 \quad (\text{Not valid, as width cannot be negative.})$

Final Answer:

$w = 5.21 \, \text{units (approx.)}$

If exactness is required: $w = \frac{\sqrt{208} - 4}{2}$

Do you need further clarifications, or should I proceed with related questions? 😊

---

Related Questions:

1. How would the dimensions change if the area was doubled?
2. Can the same problem be solved graphically? How?
3. What are some examples of real-life situations involving rectangles?
4. How do quadratic equations apply to other geometric shapes?
5. If the length increased to 8 more than the width, how would that change the width?

Tip: For quick checks, always estimate roots when solving quadratic equations involving areas.