$2 \cdot 126$ The number of roots of the equation $2 \sqrt{x-3}+6=x$ is
(A) 3.
(B) 2.
(C) 1.
(D) No real roots.
(2nd China Tianjin "Chinese Youth Cup" Mathematics Invitational, 1988)
［Solution］Given $2 \sqrt{x-3}=x-6$, squaring both sides yields

Solving, we get
$$
x^{2}-16 x+48=0 \text {, }
$$
$x_{1}=12, x_{2}=4$ (discard).
Therefore, the answer is $(C)$.