Which of the following is an extraneous solution to the following equation? \( \sqrt{x-1}+4=x-3 \) Both are extraneous. 5 Neither are extraneous. 10

To determine if either of the proposed solutions (5 or 10) is an extraneous solution to the equation \[ \sqrt{x-1} + 4 = x - 3, \] we will first solve the equation step by step and then check each proposed solution. ### Step 1: Isolate the square root Subtract 4 from both sides: \[ \sqrt{x-1} = x - 3 - 4 \] This simplifies to: \[ \sqrt{x-1} = x - 7. \] ### Step 2: Square both sides To eliminate the square root, we square both sides: \[ (\sqrt{x-1})^2 = (x - 7)^2. \] This gives us: \[ x - 1 = (x - 7)(x - 7). \] ### Step 3: Expand the right side Expanding the right side: \[ x - 1 = x^2 - 14x + 49. \] ### Step 4: Rearrange the equation Rearranging gives us: \[ 0 = x^2 - 15x + 50. \] ### Step 5: Solve the quadratic equation Now we will solve the quadratic equation \(x^2 - 15x + 50 = 0\) using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \] where \(a = 1\), \(b = -15\), and \(c = 50\). Calculating the discriminant: \[ b^2 - 4ac = (-15)^2 - 4 \cdot 1 \cdot 50 = 225 - 200 = 25. \] Now substituting into the quadratic formula: \[ x = \frac{15 \pm \sqrt{25}}{2} = \frac{15 \pm 5}{2}. \] This gives us two potential solutions: \[ x = \frac{20}{2} = 10 \quad \text{and} \quad x = \frac{10}{2} = 5. \] ### Step 6: Check for extraneous solutions Now we need to check both solutions in the original equation. 1. **For \(x = 10\)**: \[ \sqrt{10 - 1} + 4 = 10 - 3, \] \[ \sqrt{9} + 4 = 7, \] \[ 3 + 4 = 7 \quad \text{(True)}. \] 2. **For \(x = 5\)**: \[ \sqrt{5 - 1} + 4 = 5 - 3, \] \[ \sqrt{4} + 4 = 2, \] \[ 2 + 4 = 2 \quad \text{(False)}. \] ### Conclusion The solution \(x = 5\) is an extraneous solution, while \(x = 10\) is valid. Therefore, the answer is: **5 is an extraneous solution.**