GMATPrepNow wrote:
If \(xy≠0\), and \(\sqrt{x-y}=\sqrt{x}-\sqrt{y}\), which of the following must be true?
i) \(\frac{x}{y}=1\)
ii) \(x>0\)
iii) \(x^2-2xy+y^2=0\)
A) i only
B) i & ii only
C) i & iii only
D) ii & iii only
E) i, ii & iii
Given: \(\sqrt{x-y}=\sqrt{x}-\sqrt{y}\)
Square both sides: \((\sqrt{x-y})^2=(\sqrt{x}-\sqrt{y})^2\)
Expand and simplify both sides: \(x-y=x-2\sqrt{xy}+y\)
Add \(y\) to both sides: \(x=x-2\sqrt{xy}+2y\)
Subtract \(x\) from both sides: \(0=-2\sqrt{xy}+2y\)
Add \(2\sqrt{xy}\) to both sides: \(2\sqrt{xy}=2y\)
Divide both sides by 2 to get: \(\sqrt{xy}=y\)
Square both sides to get: \(xy=y^2\)
Rewrite as: \(y^2-xy=0\)
Factor: \(y(x-y)=0\)
So, either \(y = 0\) or \(x-y=0\)
Since y CANNOT equal zero, it must be the case that \(x-y=0\)
i) If \(x-y=0\), then \(x=y\), which means \(\frac{x}{y}=1\)
Statement i is TRUE
ii) In order for \(\sqrt{x}\) and \(\sqrt{y}\) to have REAL values, it must be the case that x and y are greater than or equal to 0.
Since we're told that \(xy≠0\), we can be certain that x and y are both positive
Statement ii is TRUE
iii) If \(x-y=0\), then we can square both sides to get: \((x-y)^2=0^2\)
Expand and simplify left side to get: \(x^2-2xy+y^2=0\)
Statement iii is TRUE
Answer: E
Cheers,
Brent
_________________
Test confidently with gmatprepnow.com