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

To solve the equation \( \sqrt{2x - 5} + 3 = x \), first, isolate the square root: \[ \sqrt{2x - 5} = x - 3 \] Next, square both sides to eliminate the square root: \[ 2x - 5 = (x - 3)^2 \] Expanding the right side: \[ 2x - 5 = x^2 - 6x + 9 \] Rearranging the equation to set it to 0: \[ 0 = x^2 - 8x + 14 \] Now, use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -8, c = 14 \): \[ x = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 1 \cdot 14}}{2 \cdot 1} \] \[ x = \frac{8 \pm \sqrt{64 - 56}}{2} \] \[ x = \frac{8 \pm \sqrt{8}}{2} \] \[ x = \frac{8 \pm 2\sqrt{2}}{2} \] \[ x = 4 \pm \sqrt{2} \] Thus, the two potential solutions are: \[ x_1 = 4 + \sqrt{2} \quad \text{and} \quad x_2 = 4 - \sqrt{2} \] Now, let's check for extraneous solutions by substituting back into the original equation. For \( x_1 = 4 + \sqrt{2} \): \[ \sqrt{2(4 + \sqrt{2}) - 5} + 3 = 4 + \sqrt{2} \] \[ \sqrt{8 + 2\sqrt{2} - 5} + 3 = 4 + \sqrt{2} \] \[ \sqrt{3 + 2\sqrt{2}} + 3 \stackrel{?}{=} 4 + \sqrt{2} \] Calculating both sides yields a true statement. For \( x_2 = 4 - \sqrt{2} \): \[ \sqrt{2(4 - \sqrt{2}) - 5} + 3 = 4 - \sqrt{2} \] \[ \sqrt{8 - 2\sqrt{2} - 5} + 3 = 4 - \sqrt{2} \] \[ \sqrt{3 - 2\sqrt{2}} + 3 \stackrel{?}{=} 4 - \sqrt{2} \] Since \( 3 - 2\sqrt{2} < 0 \), the left side results in a non-real number. Therefore, \( x_2 \) is extraneous. The only valid solution is: \[ x = 4 + \sqrt{2} \]