$1 \cdot 87$ If $x<0, y<0$, and $3 x-2 y=\sqrt{x y}$, then the value of $\frac{y}{x}$ is
(A) $-\frac{9}{4}$.
(B) 1 .
(C) $\frac{9}{4}$.
(D) None of the above answers is correct.
(China Beijing Junior High School Mathematics Competition, 1985)
[Solution] Given $3 x-2 y=\sqrt{x y}$, squaring both sides yields
$$
9 x^{2}-13 x y+4 y^{2}=0 \text {, }
$$

which can be factored as
$$
(9 x-4 y)(x-y)=0 \text {. }
$$

From $9 x-4 y=0$, we get $\frac{y}{x}=\frac{9}{4}$;
from $x-y=0$, we get $\frac{y}{x}=1$.
Upon substitution and verification, $\frac{y}{x}=1$ does not satisfy the condition, so $\frac{y}{x}=\frac{9}{4}$. Therefore, the answer is $(C)$.