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

Question

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

CareerGeek · Accepted Answer

\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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Mohammadmo · Answer

(√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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chrtpmdr · Answer

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

BrentGMATPrepNow · Answer

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

navderm · Answer

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 ?

Fdambro294 · Answer

Just to add to the awesome explanations above, such as  BrentGMATPrepNow

You can utilize the answer choices to your advantage in this question.

Start with Roman Numeral II. One of the expressions on the right hand side of the equation is: sqrt(x)

Since the GMAT only deals with Real Numbers, X must be either 0 or positive. We are given that X and Y can not equal 0. Thus it must be true that X > 0.

Then, if you look at Roman Numeral I and IIi, after some algebraic maneuvers, you can discover that they both give you the same information ————> X = Y

Thus, either both must be true or both can be false.

Since there is no option for “II only” and we know II must be true, we can’t say I and III are wrong.

Therefore, it must be the case that “I , II, and III” must all be true. E

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Fdambro294 · Answer

The rule I believe you are referring to:

Sqrt( (X)^2 ) =

Applies when you are “Taking” the Square Root of a Variable Expression Raised to the 2nd Power (or Taking another Even Positive Integer Root, such as the 4th Root of (X)^4)

Here, instead, the algebra move made is Squaring both sides of the equation.

While squaring both sides of an equation does have the potential to introduce solutions that do not satisfy the equation, we are looking for a general algebraic form here.

We are not looking for a specific numerical value such as X = 6 or Y = 2, in which case it would be smart to plug the answers back into the original equation to ensure they are valid solutions.

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GobaChatro · Answer

Initially chose C for failing to apply the real number/value aspect

Daksh29C · Answer

Took me 51s

Squaring both sides we get,
x-y = x-2sqrt(xy)+y
Simplifying we get
sqrt (x)=sqrt (y)
thus x=y and x has to be greater than 0 as it is a square of a number

bumpbot · Answer

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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

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