If (xy)^)(1/2) = xy , what is the value of x + y?

[quote="dhushan"]If sqrt(xy)=xy , what is the value of x + y?
(1) x = -1/2
(2) y is not equal to zero


XY=(XY)^2........ie: x,y have the same sign and they could be (0,anything)(1,1),(-1,-1) receprocals

from 1

no info about y......x,y could be (0,-1/2) or receprocals
from 2
insuff

both

receprocals..........suff

Thanks guys, I see how your answers work, but I was wondering what is wrong with solving the problem this way.

sqrt(xy) = xy

xy = x^2y^2
1/x = y

therefore if y = 1/x, and from the info in (1), couldn't we deduce that y = =-2 by substituting -1/2 in for x?

Therefore A would be the answer?

You cancelled out a root of the equation which is incorrect..you should consider each and every real root of the equation..

Sorry, I still don't follow. what do you mean by "cancelled out a root of the equation" - I still have x and y in the equation.

for a function the square root of xy is only equal to xy if the function is equal to 0 or 1, you can do the math and find the roots by squaring but I just accept that it can only equal 0 or 1. So if we know X is not 0 and is in fact a #, we know Y can only be 0 or the multiplicative reciporcal of X so that XY=1 or 0. When we get statement 2 we know that X*Y can not be equal to 0 so we know that XY= 1 and if we know what X is we can calculate what Y is.

Hope that made sense

x,y are both interrelated ( roots ie: the solution is a unique combination of x,y value.s but not any of them on its own),

you can never cancell out a VARIABLE, because its unique value in the combination xy or x^2y^2 makes the equation valid.

think of it as if there are 2 conditions for the equation to be true the first is the value of x and the second is values of y but not any of them alone.

hope am clear

Thanks, I finally get it in this situation. However, does the same hold true in other questions, for example

x^2y^2 = x^2 (so here I would have determine whether, x = 0 and y = 0, or x and y = 1)

Thanks for everyone's help, it is greatly appreciated.

Here you'll have:

x^2y^2 -x^2 = 0 --> x^2(y^2-1)=0 --> x^2(y-1)(y+1)=0

One of the multiples must be zero --> x=0, y=1 or y=-1.

Here is the catch .....

In mathematics ... division by zero is not allowed ...
so
xy = x^2Y^2 => x^2y^2 - xy = 0 => xy(xy - 1) = => xy = 0 or xy = 1 => x is not zero therefor y = 0 or y =1/x

in ur second case .....

x^2y^2 = x^2 => x^2y^2 - x^2 = 0 => x^2(y^2 - 1) = 0 => x^2 = 0 or y^2 = 1 => x = 0 or y = 1 or y = -1...

statement 1:
==========
x = -1/2 .so y can be 0 or -2.Nt suff

Statement 2:
==========
y is not equal to zero. Nt suff

Combining both we can get x = -1/2 y = -2.

sqrt(xy)=xy -> two solutions: xy = 0 or xy = 1.

1: insufficient: y = -2 or y = 0
2: insufficient: x = 1/y

1+2: sufficient y = -2, x = -1/2 -> x+y = -2.5 -> C

Hey guys - Just a quick clarification

Why can't we divide the equation by \sqrt{xy} to yield the following:

1 = \sqrt{xy}

Based on this .. just option 1 would be sufficient because if x is -1/2 y has to be -2 to satisfy this above equation.

On combining both the statements, we still wouldn't know the value of x. What if x is 0?

Your question is confusing. On one hand you are assuming that x=0 and on the other hand you are saying that you dont know the value of x. Can you rephrase your question?

As for this question, when you combine both the statements, x=-0.5 and as y \(\neq\) 0 ---> this thus rules out the case when xy=0, leaving you with unique values of y and x. Hence C is the correct answer.

Also, S2 alone does leave the door open for assuming x=0 but then again it can very well be \(\neq\) 0, making statement 2 not sufficient.

Hope this helps.

When I first looked at this question, especially the bit on √xy=xy , I knew that plugging in the values of 1 and 0 for xy would be the fastest approach since these are the only two values that satisfy the equation given.
Remember that if the question says √xy=xy or something similar, it’s telling you that the values under the root can be 1 or 0.

So if √xy=xy , it means xy = 1 or xy = 0. If xy=1, x=y=1 or x=y=-1; if xy=0, at least one of the numbers is 0.
With this, let us analyse the statements.

From statement I alone, we know x = -\(\frac{1}{2}\). If xy=1, value of y will be -2 and if xy=0, value of y will be 0. Since we do not know the exact value of xy, statement I alone is insufficient to find the value of y and hence the value of x+y.
Answer options A and D can be eliminated. Possible answer options are B, C or E.

From statement II alone, we know that y is not equal to 0. This eliminates the possibility of y being 0. However, we still do not know the exact values of x and y and hence cannot find the value of x+y. Statement II alone is insufficient.
Answer option B can be eliminated. Possible answer options are C or E.

Combining both statements I and II, we have the following:
From statement II alone, we know that xy≠0. Coupling this with the data given in the question, we can say that xy HAS TO be equal to 1.
From statement I alone, we know that x = -\(\frac{1}{2}\).
Since xy=1 and x=-\(\frac{1}{2}\), we can say that y=-2. Since we know x and y uniquely, we can calculate the value of x+y. The combination of statements is sufficient.
The correct answer option is C.

Hope that helps!

Hello Pleonasm,

Consider that the value of sq.root(xy) can be 0. As such, you cannot just divide both sides by sq.root(xy) since that would tantamount to division by ZERO, which is not allowed on the GMAT.

In general, you should avoid cancellations and cross-multiplications (yes, even when there's an equation) when you know that certain quantities can be ZERO. A better approach would be to take all terms on to the LHS, keeping the RHS as zero and simplifying the expression on the LHS to yield roots, as Bunuel has done in his response to this query.

Hope that helps!

Hi Bunuel when we have xy=(xy)^2 why we cannot assume that (xy)^2/xy=1 why should we do (xy)^2 -xy=o

My question is how do we know when the power on two side of equality should cancel out to simplify and when should we perform subtract as done above

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