\( \left. \begin{array} { l l } { 2 } & { \sqrt { x - 3 } + 3 = 0 } \\ { 4 } & { \frac { \sqrt { x } } { \sqrt { 2 } } = 3 \sqrt { 2 } } \\ { 6 } & { \sqrt { x ^ { 2 } - 9 } = 4 } \\ { 8 } & { \sqrt { x + 5 } - x = 3 } \\ { 10 } & { x + \sqrt { 5 - x } + 1 = 0 } \\ { 12 } & { x + 2 \sqrt { x - 1 } = 4 } \end{array} \right. \)

Let's solve each of the given equations for \( x \): --- ### **Problem 2** \[ \sqrt{x - 3} + 3 = 0 \] **Solution:** 1. Isolate the square root: \[ \sqrt{x - 3} = -3 \] 2. **Observation:** The square root of a real number is always non-negative. Therefore, \( \sqrt{x - 3} \) cannot be equal to \(-3\). **Conclusion:** **No real solution** exists for this equation. --- ### **Problem 4** \[ \frac{\sqrt{x}}{\sqrt{2}} = 3\sqrt{2} \] **Solution:** 1. Multiply both sides by \( \sqrt{2} \): \[ \sqrt{x} = 3\sqrt{2} \times \sqrt{2} = 3 \times 2 = 6 \] 2. Square both sides to solve for \( x \): \[ x = 6^2 = 36 \] 3. **Verification:** \[ \frac{\sqrt{36}}{\sqrt{2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2} \] The solution satisfies the original equation. **Conclusion:** **\( x = 36 \)** --- ### **Problem 6** \[ \sqrt{x^2 - 9} = 4 \] **Solution:** 1. Square both sides: \[ x^2 - 9 = 16 \Rightarrow x^2 = 25 \] 2. Solve for \( x \): \[ x = \pm 5 \] 3. **Verification:** - For \( x = 5 \): \[ \sqrt{5^2 - 9} = \sqrt{25 - 9} = \sqrt{16} = 4 \] - For \( x = -5 \): \[ \sqrt{(-5)^2 - 9} = \sqrt{25 - 9} = \sqrt{16} = 4 \] Both solutions satisfy the original equation. **Conclusion:** **\( x = 5 \) and \( x = -5 \)** --- ### **Problem 8** \[ \sqrt{x + 5} - x = 3 \] **Solution:** 1. Let \( y = \sqrt{x + 5} \). Then: \[ y - x = 3 \Rightarrow x = y - 3 \] 2. Substitute \( x = y - 3 \) into \( y^2 = x + 5 \): \[ y^2 = (y - 3) + 5 \Rightarrow y^2 = y + 2 \Rightarrow y^2 - y - 2 = 0 \] 3. Solve the quadratic equation: \[ y = \frac{1 \pm \sqrt{1 + 8}}{2} = \frac{1 \pm 3}{2} \Rightarrow y = 2 \text{ or } y = -1 \] Since \( y \) represents a square root, \( y \geq 0 \). Thus, \( y = 2 \). 4. Find \( x \): \[ x = y - 3 = 2 - 3 = -1 \] 5. **Verification:** \[ \sqrt{-1 + 5} - (-1) = \sqrt{4} + 1 = 2 + 1 = 3 \] The solution satisfies the original equation. **Conclusion:** **\( x = -1 \)** --- ### **Problem 10** \[ x + \sqrt{5 - x} + 1 = 0 \] **Solution:** 1. Rearrange the equation: \[ x + 1 = -\sqrt{5 - x} \] Since \( \sqrt{5 - x} \geq 0 \), it follows that \( x + 1 \leq 0 \) (i.e., \( x \leq -1 \)). 2. Let \( t = \sqrt{5 - x} \). Then: \[ x = -t - 1 \] 3. Substitute into \( t = \sqrt{5 - x} \): \[ t = \sqrt{5 - (-t - 1)} = \sqrt{6 + t} \] 4. Square both sides: \[ t^2 = 6 + t \Rightarrow t^2 - t - 6 = 0 \] 5. Solve the quadratic equation: \[ t = \frac{1 \pm \sqrt{1 + 24}}{2} = \frac{1 \pm 5}{2} \Rightarrow t = 3 \text{ or } t = -2 \] Since \( t \geq 0 \), \( t = 3 \). 6. Find \( x \): \[ x = -3 - 1 = -4 \] 7. **Verification:** \[ -4 + \sqrt{5 - (-4)} + 1 = -4 + \sqrt{9} + 1 = -4 + 3 + 1 = 0 \] The solution satisfies the original equation. **Conclusion:** **\( x = -4 \)** --- ### **Problem 12** \[ x + 2\sqrt{x - 1} = 4 \] **Solution:** 1. **Domain:** \( x - 1 \geq 0 \Rightarrow x \geq 1 \) 2. Rearrange the equation: \[ 2\sqrt{x - 1} = 4 - x \] Since \( 2\sqrt{x - 1} \geq 0 \), it follows that \( 4 - x \geq 0 \Rightarrow x \leq 4 \). Thus, \( x \in [1, 4] \). 3. Let \( t = \sqrt{x - 1} \). Then: \[ x = t^2 + 1 \] 4. Substitute into the equation: \[ 2t = 4 - (t^2 + 1) \Rightarrow 2t = 3 - t^2 \Rightarrow t^2 + 2t - 3 = 0 \] 5. Solve the quadratic equation: \[ t = \frac{-2 \pm \sqrt{4 + 12}}{2} = \frac{-2 \pm 4}{2} \Rightarrow t = 1 \text{ or } t = -3 \] Since \( t \geq 0 \), \( t = 1 \). 6. Find \( x \): \[ x = 1^2 + 1 = 2 \] 7. **Verification:** \[ 2 + 2\sqrt{2 - 1} = 2 + 2 \times 1 = 4 \] The solution satisfies the original equation. **Conclusion:** **\( x = 2 \)** --- ### **Summary of Solutions** - **Problem 2:** No real solution. - **Problem 4:** \( x = 36 \). - **Problem 6:** \( x = 5 \) and \( x = -5 \). - **Problem 8:** \( x = -1 \). - **Problem 10:** \( x = -4 \). - **Problem 12:** \( x = 2 \).