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Let's work through some of the equations from this exercise set. Here are the first few solutions, step by step:

### 1. \( 1 - 2x^2 = 0 \)
- Rearrange: \( 2x^2 = 1 \)
- Divide by 2: \( x^2 = \frac{1}{2} \)
- Take square roots: \( x = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2} \)

### 2. \( -5x^2 = 0 \)
- Divide by -5: \( x^2 = 0 \)
- Solution: \( x = 0 \)

### 3. \( x^2 + 4x + 4 = 0 \)
- This is a perfect square trinomial: \( (x + 2)^2 = 0 \)
- Solution: \( x = -2 \)

### 4. \( 12x - 3x^2 - 12 = 0 \)
- Divide through by 3: \( 4x - x^2 - 4 = 0 \)
- Rearrange: \( x^2 - 4x + 4 = 0 \)
- Factor: \( (x - 2)^2 = 0 \)
- Solution: \( x = 2 \)

### 5. \( 8x + 4x^2 + 3 = 0 \)
- Rearrange: \( 4x^2 + 8x + 3 = 0 \)
- Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 4 \), \( b = 8 \), and \( c = 3 \):
  \[
  x = \frac{-8 \pm \sqrt{8^2 - 4(4)(3)}}{2(4)} = \frac{-8 \pm \sqrt{64 - 48}}{8} = \frac{-8 \pm \sqrt{16}}{8} = \frac{-8 \pm 4}{8}
  \]
- Solutions: \( x = \frac{-8 + 4}{8} = -\frac{1}{2} \), and \( x = \frac{-8 - 4}{8} = -\frac{3}{2} \)

Would you like to continue with more solutions or ask for clarifications? Here are 5 related questions for further exploration:

1. How do you derive the quadratic formula for any quadratic equation?
2. Can you solve equations that do not have real solutions?
3. What are the different methods to solve quadratic equations?
4. How do you solve a quadratic equation using the method of completing the square?
5. What is the difference between rational and irrational solutions in quadratic equations?

**Tip:** Always check if a quadratic equation can be factored easily before applying the quadratic formula; this can save time!

Solve the given set of quadratic equations (e.g., 1 - 2x^2 = 0, -5x^2 = 0, x^2 + 4x + 4 = 0, etc.).

This page provides detailed solutions for solving quadratic equations from Exercise 3.2. Learn key algebraic techniques such as factoring and applying the quadratic formula. Suitable for Grades 8-10 students.

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