Solving Multi-Step Equations worksheet with handwritten solutions.

It looks like you've uploaded a handwritten worksheet containing several problems related to solving radical equations. Let's go through each problem step by step, solve them, and explain the solutions clearly.

---

🔹 Problem 1:

$$
\sqrt{2x + 1} = 3
$$

#### Step-by-step solution:
1. Square both sides to eliminate the square root:
$$
(\sqrt{2x + 1})^2 = 3^2 \Rightarrow 2x + 1 = 9
$$

2. Solve for x:
$$
2x = 8 \Rightarrow x = 4
$$

3. Check for extraneous solutions (always check when solving radical equations):
$$
\sqrt{2(4) + 1} = \sqrt{8 + 1} = \sqrt{9} = 3 \quad \text{✓ Valid}
$$

✔ Answer: $ x = 4 $

---

🔹 Problem 2:

$$
\sqrt{5x - 2} = 3
$$

1. Square both sides:
$$
(\sqrt{5x - 2})^2 = 3^2 \Rightarrow 5x - 2 = 9
$$

2. Solve:
$$
5x = 11 \Rightarrow x = \frac{11}{5}
$$

3. Check:
$$
\sqrt{5 \cdot \frac{11}{5} - 2} = \sqrt{11 - 2} = \sqrt{9} = 3 \quad \text{✓ Valid}
$$

✔ Answer: $ x = \frac{11}{5} $

---

🔹 Problem 3:

$$
\sqrt{2x - 3} = \sqrt{x + 4}
$$

1. Square both sides:
$$
(\sqrt{2x - 3})^2 = (\sqrt{x + 4})^2 \Rightarrow 2x - 3 = x + 4
$$

2. Solve:
$$
2x - x = 4 + 3 \Rightarrow x = 7
$$

3. Check:
$$
\sqrt{2(7) - 3} = \sqrt{14 - 3} = \sqrt{11}, \quad \sqrt{7 + 4} = \sqrt{11} \quad \text{✓ Valid}
$$

✔ Answer: $ x = 7 $

---

🔹 Problem 4:

$$
\sqrt{3x + 6} = x
$$

1. Square both sides:
$$
(\sqrt{3x + 6})^2 = x^2 \Rightarrow 3x + 6 = x^2
$$

2. Rearrange into standard quadratic form:
$$
x^2 - 3x - 6 = 0
$$

3. Use quadratic formula:
$$
x = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(-6)}}{2(1)} = \frac{3 \pm \sqrt{9 + 24}}{2} = \frac{3 \pm \sqrt{33}}{2}
$$

So,
$$
x = \frac{3 + \sqrt{33}}{2} \quad \text{or} \quad x = \frac{3 - \sqrt{33}}{2}
$$

4. Check for extraneous solutions:

- First, approximate values:
- $ \sqrt{33} \approx 5.74 $
- $ x_1 \approx \frac{3 + 5.74}{2} = 4.37 $
- $ x_2 \approx \frac{3 - 5.74}{2} = -1.37 $

- Check $ x = 4.37 $:
$$
\sqrt{3(4.37) + 6} = \sqrt{13.11 + 6} = \sqrt{19.11} \approx 4.37 \quad \text{✓ Valid}
$$

- Check $ x = -1.37 $:
$$
\sqrt{3(-1.37) + 6} = \sqrt{-4.11 + 6} = \sqrt{1.89} \approx 1.37 \neq -1.37
$$
But LHS is positive, RHS is negative → Invalid.

✘ Reject $ x = \frac{3 - \sqrt{33}}{2} $

✔ Answer: $ x = \frac{3 + \sqrt{33}}{2} $

---

🔹 Problem 5:

$$
\sqrt{x + 2} = x - 2
$$

1. Square both sides:
$$
(\sqrt{x + 2})^2 = (x - 2)^2 \Rightarrow x + 2 = x^2 - 4x + 4
$$

2. Rearrange:
$$
0 = x^2 - 5x + 2
$$

3. Use quadratic formula:
$$
x = \frac{5 \pm \sqrt{(-5)^2 - 4(1)(2)}}{2} = \frac{5 \pm \sqrt{25 - 8}}{2} = \frac{5 \pm \sqrt{17}}{2}
$$

So,
$$
x = \frac{5 + \sqrt{17}}{2} \quad \text{or} \quad x = \frac{5 - \sqrt{17}}{2}
$$

4. Approximate:
- $ \sqrt{17} \approx 4.12 $
- $ x_1 \approx \frac{5 + 4.12}{2} = 4.56 $
- $ x_2 \approx \frac{5 - 4.12}{2} = 0.44 $

5. Check each:

- $ x = 4.56 $:
$$
\sqrt{4.56 + 2} = \sqrt{6.56} \approx 2.56, \quad x - 2 = 2.56 \quad \text{✓ Valid}
$$

- $ x = 0.44 $:
$$
\sqrt{0.44 + 2} = \sqrt{2.44} \approx 1.56, \quad x - 2 = -1.56 \quad \text{Not equal, invalid}
$$

✔ Answer: $ x = \frac{5 + \sqrt{17}}{2} $

---

🔹 Problem 6:

$$
\sqrt{2x - 1} = x - 3
$$

1. Square both sides:
$$
2x - 1 = (x - 3)^2 = x^2 - 6x + 9
$$

2. Rearranged:
$$
0 = x^2 - 8x + 10
$$

3. Quadratic formula:
$$
x = \frac{8 \pm \sqrt{64 - 40}}{2} = \frac{8 \pm \sqrt{24}}{2} = \frac{8 \pm 2\sqrt{6}}{2} = 4 \pm \sqrt{6}
$$

So,
$$
x = 4 + \sqrt{6} \quad \text{or} \quad x = 4 - \sqrt{6}
$$

4. Approximate:
- $ \sqrt{6} \approx 2.45 $
- $ x_1 \approx 6.45 $
- $ x_2 \approx 1.55 $

5. Check:

- $ x = 6.45 $:
$$
\sqrt{2(6.45) - 1} = \sqrt{12.9 - 1} = \sqrt{11.9} \approx 3.45, \quad x - 3 = 3.45 \quad \text{✓ Valid}
$$

- $ x = 1.55 $:
$$
\sqrt{2(1.55) - 1} = \sqrt{3.1 - 1} = \sqrt{2.1} \approx 1.45, \quad x - 3 = -1.45 \quad \text{Invalid}
$$

✔ Answer: $ x = 4 + \sqrt{6} $

---

🔹 Problem 7:

$$
\sqrt{x + 1} = x - 1
$$

1. Square both sides:
$$
x + 1 = (x - 1)^2 = x^2 - 2x + 1
$$

2. Rearranged:
$$
0 = x^2 - 3x
\Rightarrow x(x - 3) = 0
\Rightarrow x = 0 \quad \text{or} \quad x = 3
$$

3. Check:

- $ x = 0 $:
$$
\sqrt{0 + 1} = 1, \quad x - 1 = -1 \quad \text{Not equal → Invalid}
$$

- $ x = 3 $:
$$
\sqrt{3 + 1} = \sqrt{4} = 2, \quad x - 1 = 2 \quad \text{✓ Valid}
$$

✔ Answer: $ x = 3 $

---

🔹 Problem 8:

$$
\sqrt{3x + 2} = x + 2
$$

1. Square both sides:
$$
3x + 2 = (x + 2)^2 = x^2 + 4x + 4
$$

2. Rearranged:
$$
0 = x^2 + x + 2
$$

3. Discriminant:
$$
D = 1^2 - 4(1)(2) = 1 - 8 = -7 < 0
$$

No real solutions.

✔ Answer: No real solution

---

✔ Summary of Answers:

| Problem | Solution |
|--------|----------|
| 1 | $ x = 4 $ |
| 2 | $ x = \frac{11}{5} $ |
| 3 | $ x = 7 $ |
| 4 | $ x = \frac{3 + \sqrt{33}}{2} $ |
| 5 | $ x = \frac{5 + \sqrt{17}}{2} $ |
| 6 | $ x = 4 + \sqrt{6} $ |
| 7 | $ x = 3 $ |
| 8 | No real solution |

---

🔍 Key Concepts Used:

- Isolate the radical before squaring.
- Square both sides to eliminate radicals.
- Always check for extraneous solutions, because squaring can introduce false solutions.
- Domain restrictions: The expression under the square root must be ≥ 0.
- For equations like $ \sqrt{A} = B $, $ B $ must be ≥ 0 (since square roots are non-negative).

Let me know if you'd like these problems explained in a different format or with graphs!