What is the solution to √(2x+3) = x, after rejecting any extraneous candidate?

Square both sides of $\sqrt{2x+3} = x$ to get $2x + 3 = x^2$. Rearrange: $x^2 - 2x - 3 = 0$, which factors as $(x-3)(x+1) = 0$, so $x = 3$ or $x = -1$. Squaring can introduce extraneous solutions, so check both. For $x = 3$: $\sqrt{9} = 3$, which is true. For $x = -1$: $\sqrt{1} = 1$, but the right side is $-1$, so it fails. The only solution is 3, choice B. Choice A ($-1$) is the extraneous root, choice C includes both without checking, and choice D (9) is the value under the radical, not $x$.