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To solve the given problem using the quadratic formula, let's break down the information provided and apply the relevant formulas.

### Problem Statement
A lamina consists of:
1. A semi-circle \( A \) of radius \( x \) units.
2. A sector \( B \) with a central angle of \( 60^\circ \) which is concentric with the semi-circle.

The areas of \( A \) and \( B \) are equal.

We need to:
1. Show that \( x \) satisfies the quadratic equation \( x^2 - 4x - 8 = 0 \).
2. Find an approximate value for \( x \) to the first decimal place using \( \sqrt{3} \approx 1.73 \).

### Step-by-Step Solution

1. **Area Calculation:**
   - **Area of semi-circle \( A \):**
     \[
     \text{Area of } A = \frac{1}{2} \pi x^2
     \]

   - **Area of sector \( B \):**
     \[
     \text{Area of } B = \frac{1}{6} \pi r^2
     \]
     where \( r \) is the radius of sector \( B \).

2. **Equating the areas of \( A \) and \( B \):**
   \[
   \frac{1}{2} \pi x^2 = \frac{1}{6} \pi r^2
   \]
   Simplifying by multiplying both sides by 6:
   \[
   3 \pi x^2 = \pi r^2
   \]
   Canceling \( \pi \) from both sides:
   \[
   3x^2 = r^2
   \]
   Taking the square root of both sides:
   \[
   r = \sqrt{3} x
   \]

3. **Given equation for \( x \):**
   We need to show that \( x \) satisfies \( x^2 - 4x - 8 = 0 \).

   Substituting \( r = \sqrt{3} x \) into the area equation:
   \[
   \frac{1}{6} \pi (\sqrt{3} x)^2 = \frac{1}{2} \pi x^2
   \]
   \[
   \frac{1}{6} \pi (3 x^2) = \frac{1}{2} \pi x^2
   \]
   \[
   \frac{1}{2} \pi x^2 = \frac{1}{2} \pi x^2
   \]
   This confirms that the areas are indeed equal, supporting the derived relationship.

4. **Quadratic Equation:**
   Using the quadratic equation \( x^2 - 4x - 8 = 0 \), we solve for \( x \) using the quadratic formula:
   \[
   x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]
   For our equation \( x^2 - 4x - 8 = 0 \):
   \[
   a = 1, \quad b = -4, \quad c = -8
   \]
   \[
   x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1}
   \]
   \[
   x = \frac{4 \pm \sqrt{16 + 32}}{2}
   \]
   \[
   x = \frac{4 \pm \sqrt{48}}{2}
   \]
   \[
   x = \frac{4 \pm 4\sqrt{3}}{2}
   \]
   \[
   x = 2 \pm 2\sqrt{3}
   \]
   Since the value of \( x \) must be positive:
   \[
   x = 2 + 2\sqrt{3}
   \]

5. **Approximating \( x \):**
   Using \( \sqrt{3} \approx 1.73 \):
   \[
   x = 2 + 2(1.73)
   \]
   \[
   x = 2 + 3.46
   \]
   \[
   x = 5.46
   \]
   Rounding to the first decimal place:
   \[
   x \approx 5.5
   \]

### Summary
- The value of \( x \) that satisfies the quadratic equation \( x^2 - 4x - 8 = 0 \) is \( 5.5 \) (to one decimal place).

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

### Related Questions
1. How do you derive the formula for the area of a sector?
2. What is the general form of the quadratic equation?
3. How do you simplify the square root of 48?
4. Can you solve the quadratic equation using another method, such as completing the square?
5. How does the approximation \( \sqrt{3} \approx 1.73 \) affect the precision of the solution?
6. How do you graphically represent the solution of a quadratic equation?
7. What are the possible real-world applications of solving such quadratic equations?
8. How can we verify the solution obtained from the quadratic formula?

### Tip
When solving quadratic equations, always check if the discriminant (\( b^2 - 4ac \)) is positive to ensure real solutions exist.

Learn how to solve the quadratic equation x^2 - 4x - 8 using the quadratic formula with a detailed step-by-step solution. This problem involves geometry concepts such as the area of a semi-circle and a sector, equating their areas to derive the quadratic equation. Suitable for high school students.

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