7.69 If the equation $\sqrt{x-p}=x$ has two distinct real roots, then the range of the real number $p$ is
(A) $p \leqslant 0$.
(B) $p<\frac{1}{4}$.
(C) $0 \leqslant p<\frac{1}{4}$.
(D) $p \geqslant \frac{1}{4}$.
(China Junior High School Mathematics League, 1994)
[Solution] Let's take $p=-1$, substituting into the original equation yields $\sqrt{x+1}=x$, which simplifies to $x^{2}-x-1=0$. This equation has one negative root, thus we can eliminate (A), (B). Taking $p=1$, substituting into the original equation yields $x^{2}-x+1=0$, which has no real solutions, hence we eliminate (D). Therefore, the answer is $(C)$.