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# If xy≠0, and √x−y=√x−√y, which of the following must be

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If xy≠0, and √x−y=√x−√y, which of the following must be  [#permalink]

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23 Sep 2019, 09:07
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Question Stats:

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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

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Re: If xy≠0, and √x−y=√x−√y, which of the following must be  [#permalink]

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23 Sep 2019, 09:45
(√x-y)²=x+y-2√xy
x-y=x+y-2√xy
2y=2√xy
y=√xy
y=x
So, option E

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Re: If xy≠0, and √x−y=√x−√y, which of the following must be  [#permalink]

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23 Sep 2019, 09:45
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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

$$\sqrt{x-y}=\sqrt{x}-\sqrt{y}$$
Squaring on both sides,
—> x - y = x + y - $$2\sqrt{xy}$$
—> $$2\sqrt{xy}$$ = 2y
Squaring on both sides again
—> xy = y^2
—> y(x - y) = 0
Since xy≠0,
—> y≠0 —> x - y = 0 definitely
—> x = y
—> x/y = 1

Also, x is positive. If x is 0, then y = 0 which is not possible and x can’t be negative as x is inside a square root

Since x - y = 0
—> (x - y)^2 = 0
—> x^2 - 2xy + y^2 = 0

IMO Option E

Pls Hit kudos if you like the solution

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Re: If xy≠0, and √x−y=√x−√y, which of the following must be  [#permalink]

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23 Sep 2019, 22:07
I had a different approach, I just plugged in some numbers and figured out the only way that the Equation work's is when all three are true, hence E)

For example try 4 for x and 4 for y ; left side = 0 , right side = 0
XY =! 0 so the condition in the problem text is met as well

X/Y = 1
X > 0
iii) -> 0

I tried it with a few different numbers on top of that and it was sufficient enough for me to justify it as a correct solution, the main thought approach basically was "what possibilities are there that the square root of a difference can be the same as the difference of the sqare roots of each number" -> and the only logical conclusion at which I arrived was that it has to be 0 = 0
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If xy≠0, and √x−y=√x−√y, which of the following must be  [#permalink]

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24 Sep 2019, 08:07
Top Contributor
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

Cheers,
Brent
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Re: If xy≠0, and √x−y=√x−√y, which of the following must be  [#permalink]

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29 Sep 2019, 23:25
GMATPrepNow wrote:
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: $$x-y=x-2\sqrt{xy}+y$$

Right here, Should this not be $$|x-y|=x-2\sqrt{xy}+y$$

and then two equation be evaluated ? sqrt((x-y)^2) = |x-y| correct ?
Re: If xy≠0, and √x−y=√x−√y, which of the following must be   [#permalink] 29 Sep 2019, 23:25
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