https://file.aiquickdraw.com/mathbot/file/1727476191219xuh889m5.jpg

The image contains two mathematical problems, which I'll transcribe below:

1. **Problem 1**:  
   Given \( a, b \in \mathbb{R} \), show that:  
   \[
   a + b + ab + 1 = 0 \iff a = -1 \text{ or } b = -1
   \]

2. **Problem 2**:  
   Show that for all \( x \in \mathbb{R} \), the equation:  
   \[
   \sqrt{x + 6} = x \iff x = 3
   \]

Finally, there is a note:
5. **Raisonnement par absurde** (Proof by contradiction).

### Let's address the first problem:

We need to prove the equivalence:  
\[
a + b + ab + 1 = 0 \iff a = -1 \text{ or } b = -1
\]

#### **Solution:**

1. Start with the equation \( a + b + ab + 1 = 0 \).
   
   This can be rewritten as:  
   \[
   (a + 1)(b + 1) = 0
   \]
   Expanding the left-hand side:
   \[
   (a + 1)(b + 1) = ab + a + b + 1
   \]
   This simplifies exactly to the original equation \( a + b + ab + 1 = 0 \), confirming the factorization is correct.

2. Now, \( (a + 1)(b + 1) = 0 \) implies that either \( a + 1 = 0 \) or \( b + 1 = 0 \), which gives:
   - \( a = -1 \) or
   - \( b = -1 \).

Thus, the equation holds if and only if \( a = -1 \) or \( b = -1 \).

---

### Problem 2:

We need to solve the equation:
\[
\sqrt{x + 6} = x
\]
and show that this holds if and only if \( x = 3 \).

#### **Solution:**

1. Square both sides of the equation to eliminate the square root:
   \[
   (\sqrt{x + 6})^2 = x^2
   \]
   This gives:
   \[
   x + 6 = x^2
   \]

2. Rearrange the equation to form a quadratic:
   \[
   x^2 - x - 6 = 0
   \]

3. Factor the quadratic equation:
   \[
   (x - 3)(x + 2) = 0
   \]

4. This gives two possible solutions:
   - \( x = 3 \)
   - \( x = -2 \)

5. However, substitute \( x = -2 \) back into the original equation:
   \[
   \sqrt{-2 + 6} = \sqrt{4} = 2 \quad \text{but} \quad x = -2
   \]
   This results in a contradiction, so \( x = -2 \) is not a valid solution.

6. Therefore, the only solution is \( x = 3 \).

---

Would you like more details on any of these steps, or have any questions?

### Related questions:
1. How do we solve general quadratic equations using factorization?
2. Can the equation \( a + b + ab + 1 = 0 \) have other types of solutions for non-real numbers?
3. What are other methods to solve the equation \( \sqrt{x + 6} = x \) without squaring both sides?
4. How does "proof by contradiction" (raisonnement par absurde) work in different contexts?
5. Can we use graphical methods to verify the solution of \( \sqrt{x + 6} = x \)?

**Tip**: When solving radical equations, always check potential solutions by substituting them back into the original equation, as squaring can introduce extraneous solutions.

1. Given a, b ∈ ℝ, show that: a + b + ab + 1 = 0 ⇔ a = -1 or b = -1. 2. Show that for all x ∈ ℝ, the equation √(x + 6) = x ⇔ x = 3. 5. Raisonnement par absurde (Proof by contradiction).

This lesson covers solving the equation a + b + ab + 1 = 0 and proving it is equivalent to a = -1 or b = -1. It also explains how to solve the radical equation √(x + 6) = x, showing that the only solution is x = 3. This guide is suitable for students in Grades 10-12.

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