Solve for \( x \), if possible: (a) \( \sqrt{x-4}-5=0 \) (d) \( \sqrt{3 x-2}-x=0 \) (g) \( \sqrt{4-2 x}-\sqrt{x+1}=0 \)

Let's solve each equation step by step. (a) Starting with \( \sqrt{x-4} - 5 = 0 \): 1. Isolate the square root: \[ \sqrt{x-4} = 5 \] 2. Square both sides: \[ x - 4 = 25 \] 3. Solve for \( x \): \[ x = 25 + 4 = 29 \] Now, let’s verify by substituting back into the original equation: \[ \sqrt{29-4} - 5 = \sqrt{25} - 5 = 5 - 5 = 0 \] So, \( x=29 \) is valid. (d) Now for \( \sqrt{3x-2} - x = 0 \): 1. Isolate the square root: \[ \sqrt{3x-2} = x \] 2. Square both sides: \[ 3x - 2 = x^2 \] 3. Rearrange it into standard form: \[ x^2 - 3x + 2 = 0 \] 4. Factor the quadratic: \[ (x-1)(x-2) = 0 \] Thus, \( x = 1 \) or \( x = 2 \). Verify both solutions: - For \( x=1 \): \( \sqrt{3(1)-2} - 1 = \sqrt{1} - 1 = 0 \) (valid) - For \( x=2 \): \( \sqrt{3(2)-2} - 2 = \sqrt{4} - 2 = 0 \) (valid) (g) Lastly, solve \( \sqrt{4-2x} - \sqrt{x+1} = 0 \): 1. Isolate one square root: \[ \sqrt{4-2x} = \sqrt{x+1} \] 2. Square both sides: \[ 4 - 2x = x + 1 \] 3. Rearrange: \[ 4 - 1 = x + 2x \] \[ 3 = 3x \] \[ x = 1 \] Finally, verify: \[ \sqrt{4-2(1)} - \sqrt{1+1} = \sqrt{2} - \sqrt{2} = 0 \] (valid) To summarize: - (a) \( x = 29 \) - (d) \( x = 1 \) or \( x = 2 \) - (g) \( x = 1 \)