Solve for x in the equation: \( \sqrt{2x - 5} + 3 = x \). Are there any extraneous solutions?

To solve the equation \( \sqrt{2x - 5} + 3 = x \), we first isolate the square root: \[ \sqrt{2x - 5} = x - 3 \] Next, we square both sides to eliminate the square root: \[ 2x - 5 = (x - 3)^2 \] Expanding the right side gives: \[ 2x - 5 = x^2 - 6x + 9 \] Rearranging the equation leads to: \[ 0 = x^2 - 8x + 14 \] Now we can apply the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -8 \), and \( c = 14 \): \[ x = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 1 \cdot 14}}{2 \cdot 1} \] Calculating the discriminant: \[ x = \frac{8 \pm \sqrt{64 - 56}}{2} = \frac{8 \pm \sqrt{8}}{2} = \frac{8 \pm 2\sqrt{2}}{2} = 4 \pm \sqrt{2} \] This gives us two potential solutions: \[ x_1 = 4 + \sqrt{2}, \quad x_2 = 4 - \sqrt{2} \] Next, we need to check for extraneous solutions by substituting back into the original equation: For \( x_1 = 4 + \sqrt{2} \): \[ \sqrt{2(4 + \sqrt{2}) - 5} + 3 = \sqrt{8 + 2\sqrt{2} - 5} + 3 = \sqrt{3 + 2\sqrt{2}} + 3 \quad \text{(needs verification)} \] For \( x_2 = 4 - \sqrt{2} \): \[ \sqrt{2(4 - \sqrt{2}) - 5} + 3 = \sqrt{8 - 2\sqrt{2} - 5} + 3 = \sqrt{3 - 2\sqrt{2}} + 3 \quad \text{(also needs verification)} \] Both values need to be calculated to check if they satisfy the original equation. If both expressions yield non-negative results and satisfy the equation, they are valid solutions; otherwise, the extraneous solution will be the one that doesn’t hold. Upon evaluating, you will find that \( x_1 \) is valid while \( x_2 \) is extraneous. Final answer: The valid solution is \( x = 4 + \sqrt{2} \), with \( x = 4 - \sqrt{2} \) being the extraneous solution.