Solve the following equation. (Enter your answers as a comma-separated list.) \[ 2 x+1=\sqrt{8-8 x} \] \( x=\square \)

To solve the equation \( 2x + 1 = \sqrt{8 - 8x} \), we first square both sides to eliminate the square root: \[ (2x + 1)^2 = 8 - 8x \] Expanding the left side gives: \[ 4x^2 + 4x + 1 = 8 - 8x \] Next, we bring all terms to one side of the equation: \[ 4x^2 + 4x + 1 + 8x - 8 = 0 \] This simplifies to: \[ 4x^2 + 12x - 7 = 0 \] Now, we can apply the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 4 \), \( b = 12 \), and \( c = -7 \): \[ x = \frac{-12 \pm \sqrt{12^2 - 4 \cdot 4 \cdot (-7)}}{2 \cdot 4} \] Calculating the discriminant: \[ 12^2 = 144 \] \[ 4 \cdot 4 \cdot (-7) = -112 \Rightarrow 144 + 112 = 256 \] So: \[ x = \frac{-12 \pm \sqrt{256}}{8} = \frac{-12 \pm 16}{8} \] This gives two potential solutions: 1. \( x = \frac{4}{8} = \frac{1}{2} \) 2. \( x = \frac{-28}{8} = -\frac{7}{2} \) Now we should check for extraneous solutions by substituting back into the original equation. For \( x = \frac{1}{2} \): \[ 2(\frac{1}{2}) + 1 = \sqrt{8 - 8(\frac{1}{2})} \] This simplifies to: \[ 2 = \sqrt{4} \quad \text{(True)} \] For \( x = -\frac{7}{2} \): \[ 2(-\frac{7}{2}) + 1 = \sqrt{8 - 8(-\frac{7}{2})} \] \[ -7 + 1 = \sqrt{8 + 28} \Rightarrow -6 = \sqrt{36} \quad \text{(False)} \] Thus, the only valid solution is: \[ x = \frac{1}{2} \] Final answer: \( x = \frac{1}{2} \)