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Solve the equation using substitution and a Z equation

Let's solve the given equation step-by-step using substitution.

### Problem
The equation given is:
\[
(x + 2)^2 - 4(x + 2) + 3 = 0
\]

We are asked to use substitution, defining \( z = x + 2 \).

### Steps to Solve

1. **Substitute \( z = x + 2 \)** into the equation:
   - This substitution turns the equation into:
     \[
     z^2 - 4z + 3 = 0
     \]

2. **Solve the equation in terms of \( z \)**:
   - Now we have a quadratic equation in \( z \):
     \[
     z^2 - 4z + 3 = 0
     \]
   - To solve this quadratic equation, we can factor it:
     \[
     (z - 3)(z - 1) = 0
     \]
   - Setting each factor equal to zero gives:
     \[
     z - 3 = 0 \quad \Rightarrow \quad z = 3
     \]
     \[
     z - 1 = 0 \quad \Rightarrow \quad z = 1
     \]

3. **Back-substitute to find \( x \)**:
   - Since \( z = x + 2 \), we can substitute back for each value of \( z \):
     - If \( z = 3 \):
       \[
       x + 2 = 3 \quad \Rightarrow \quad x = 3 - 2 = 1
       \]
     - If \( z = 1 \):
       \[
       x + 2 = 1 \quad \Rightarrow \quad x = 1 - 2 = -1
       \]

### Solution
The values of \( x \) are:
\[
x = 1 \quad \text{and} \quad x = -1
\]

Would you like more details on the factoring or any other part of the solution?

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Here are five related questions to deepen your understanding:

1. How can you determine the correct substitution variable in similar problems?
2. What other methods can be used to solve a quadratic equation aside from factoring?
3. Why does back-substitution work when solving equations with substitution?
4. How can you verify the solutions you found for \( x \) are correct?
5. Can you use this substitution technique in equations that aren’t quadratic?

**Tip:** Always double-check your solutions by substituting them back into the original equation to ensure they satisfy it.

Solve the equation using substitution and a 'z equation': (x + 2)^2 - 4(x + 2) + 3 = 0

Learn how to solve the quadratic equation (x + 2)^2 - 4(x + 2) + 3 = 0 using substitution with a 'z equation'. This detailed step-by-step solution covers the substitution method and quadratic factorization, ideal for Grade 10 algebra students.

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