Let's solve the problem in the provided image using the quadratic equation.

Problem Statement

We have two plane figures:

1. Half a circle of radius $x$ units.
2. A quarter of a circle of radius $x + 2$ units.

The areas of these two figures are equal.

We need to:

1. Construct a quadratic equation in terms of $x$.
2. Solve for $x$ correct to the first decimal place using $\sqrt{2} \approx 1.41$.

Step-by-Step Solution

1. Area Calculation:

   • Area of the half-circle: $\text{Area} = \frac{1}{2} \pi x^2$

   • Area of the quarter-circle: $\text{Area} = \frac{1}{4} \pi (x + 2)^2$

2. Equating the areas: $\frac{1}{2} \pi x^2 = \frac{1}{4} \pi (x + 2)^2$

   Simplify by multiplying both sides by 4: $2 \pi x^2 = \pi (x + 2)^2$

   Cancel $\pi$ from both sides: $2 x^2 = (x + 2)^2$

   Expanding the right side: $2 x^2 = x^2 + 4x + 4$

   Rearrange to form a quadratic equation: $2 x^2 - x^2 - 4x - 4 = 0$ $x^2 - 4x - 4 = 0$

3. Solving the quadratic equation: We solve the quadratic equation $x^2 - 4x - 4 = 0$ using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ For our equation $x^2 - 4x - 4 = 0$: $a = 1, \quad b = -4, \quad c = -4$ $x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1}$ $x = \frac{4 \pm \sqrt{16 + 16}}{2}$ $x = \frac{4 \pm \sqrt{32}}{2}$ $x = \frac{4 \pm 4\sqrt{2}}{2}$ $x = 2 \pm 2\sqrt{2}$ Since $\sqrt{2} \approx 1.41$: $x = 2 + 2(1.41)$ [