Solve for \( x \), if possible: \( \begin{array}{lll}\text { (a) } \sqrt{x-4}-5=0 & \text { (b) } \sqrt{x-4}+5=0 & \text { (c) } \sqrt{5 x+6}=x \\ \text { (d) } \sqrt{3 x-2}-x=0 & \text { (e) } x+\sqrt{-4 x-3}=0 & \text { (f) } \sqrt{x+2}+4=x\end{array} \)

Let's solve for \( x \) in each of the equations step by step. **(a)** \( \sqrt{x-4}-5=0 \) 1. Add 5 to both sides: \( \sqrt{x-4} = 5 \) 2. Square both sides: \( x-4 = 25 \) 3. Add 4: \( x = 29 \) --- **(b)** \( \sqrt{x-4}+5=0 \) 1. Subtract 5 from both sides: \( \sqrt{x-4} = -5 \) 2. Since the square root cannot be negative, this has **no solution**. --- **(c)** \( \sqrt{5x+6}=x \) 1. Square both sides: \( 5x + 6 = x^2 \) 2. Rearranging gives: \( x^2 - 5x - 6 = 0 \) 3. Factor: \( (x-6)(x+1) = 0 \) 4. Solutions are \( x = 6 \) and \( x = -1 \). 5. Check both solutions: - For \( x = 6 \): \( \sqrt{5(6)+6} = \sqrt{36} = 6 \), valid. - For \( x = -1 \): \( \sqrt{5(-1)+6} = \sqrt{1} = 1 \neq -1 \), invalid. - Final solution: \( x = 6 \). --- **(d)** \( \sqrt{3x-2}-x=0 \) 1. Add \( x \) to both sides: \( \sqrt{3x-2} = x \) 2. Square both sides: \( 3x - 2 = x^2 \) 3. Rearranging gives: \( x^2 - 3x + 2 = 0 \) 4. Factor: \( (x-2)(x-1) = 0 \) 5. Solutions are \( x = 2 \) and \( x = 1 \). 6. Check both solutions: - For \( x = 2 \): \( \sqrt{3(2)-2} = \sqrt{4} = 2 \), valid. - For \( x = 1 \): \( \sqrt{3(1)-2} = \sqrt{1} = 1 \), valid. - Final solutions: \( x = 2 \) and \( x = 1 \). --- **(e)** \( x+\sqrt{-4x-3}=0 \) 1. Rearrange to get \( \sqrt{-4x-3} = -x \). 2. Since \( \sqrt{-4x-3} \geq 0 \), \( -x \) must also be non-negative, so \( x \leq 0 \). 3. Square both sides: \( -4x - 3 = x^2 \). 4. Rearranging gives: \( x^2 + 4x + 3 = 0 \) 5. Factor: \( (x+3)(x+1) = 0 \) 6. Solutions are \( x = -3 \) and \( x = -1 \). 7. Check both solutions: - For \( x = -3 \): \( -3 + \sqrt{-4(-3)-3} = -3 + \sqrt{9} = -3 + 3 = 0 \), valid. - For \( x = -1 \): \( -1 + \sqrt{-4(-1)-3} = -1 + \sqrt{1} = -1 + 1 = 0 \), valid. - Final solutions: \( x = -3 \) and \( x = -1 \). --- **(f)** \( \sqrt{x+2}+4=x \) 1. Subtract 4 from both sides: \( \sqrt{x+2} = x - 4 \). 2. Square both sides: \( x + 2 = (x - 4)^2 \). 3. Expand: \( x + 2 = x^2 - 8x + 16 \). 4. Rearranging gives: \( x^2 - 9x + 14 = 0 \) 5. Factor: \( (x-7)(x-2) = 0 \) 6. Solutions are \( x = 7 \) and \( x = 2 \). 7. Check both solutions: - For \( x = 7 \): \( \sqrt{7+2} + 4 = \sqrt{9} + 4 = 3 + 4 =