2 Given that $\sqrt{2 x+y}+\sqrt{x^{2}-9}=0$, find the value(s) of $y-x$.
(A) -9
(B) -6
(C) -9 or 9
(D) -3 or 3
(E) None of the above
2. Answer: (C)
For $\sqrt{2 x+y}+\sqrt{x^{2}-9}=0, \sqrt{2 x+y}=0$ and $\sqrt{x^{2}-9}=0$. So we have $x=3$ or -3 and $y=-2 x=-6$ or $6 \Rightarrow y-x=-9$ or 9 .