Solve the following equations for \( x \) : 1. \( \sqrt{x+12}-x=0 \) 3. \( \sqrt{2 x+4}+2=x \) 5. \( \sqrt{20+2 x}+x=2 \) . \( \sqrt{2 x^{2}-x-1}+1=x \)

To solve the equations, we'll tackle them one by one. 1. From the equation \( \sqrt{x+12}-x=0 \): \[ \sqrt{x+12} = x \] Squaring both sides gives: \[ x + 12 = x^2 \] Rearranging it gives: \[ x^2 - x - 12 = 0 \] Factoring gives: \[ (x-4)(x+3) = 0 \] So, \( x = 4 \) or \( x = -3 \). Checking both in the original equation, only \( x = 4 \) is valid since \( x \) must be greater than or equal to -12 for the square root. 2. For \( \sqrt{2x+4}+2=x \): \[ \sqrt{2x+4} = x - 2 \] Squaring both sides gives: \[ 2x + 4 = (x - 2)^2 \] Expanding leads to: \[ 2x + 4 = x^2 - 4x + 4 \] Rearranging leads to: \[ x^2 - 6x = 0 \] Factoring gives: \[ x(x - 6) = 0 \] So, \( x = 0 \) or \( x = 6 \). Checking these in the original equation, only \( x = 6 \) is valid. 3. For \( \sqrt{20+2x}+x=2 \): \[ \sqrt{20+2x} = 2 - x \] Squaring both sides yields: \[ 20 + 2x = (2 - x)^2 \] Expanding yields: \[ 20 + 2x = 4 - 4x + x^2 \] Rearranging gives: \[ x^2 - 6x - 16 = 0 \] Using the quadratic formula gives: \[ x = \frac{6 \pm \sqrt{(6)^2 - 4(1)(-16)}}{2(1)} = \frac{6 \pm 10}{2} \] Leading to solutions \( x = 8 \) or \( x = -2 \). Checking the original equation shows only \( x = -2 \) is valid. 4. Finally for \( \sqrt{2x^2-x-1}+1=x \): \[ \sqrt{2x^2 - x - 1} = x - 1 \] Squaring both sides yields: \[ 2x^2 - x - 1 = (x - 1)^2 \] Leading to: \[ 2x^2 - x - 1 = x^2 - 2x + 1 \] Rearranging gives: \[ x^2 + x - 2 = 0 \] Factoring yields: \[ (x-1)(x+2) = 0 \] So, \( x = 1 \) or \( x = -2 \). Checking in the original equation shows both are valid. Summarizing the valid solutions: 1. \( x = 4 \) 2. \( x = 6 \) 3. \( x = -2 \) 4. \( x = 1 \) or \( x = -2 \) (both valid)