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\(\sqrt{xy} = xy\) --> \(xy=x^2y^2\) --> \(x^2y^2-xy=0\) --> \(xy(xy-1)=0\) --> either \(xy=0\) or \(xy=1\).

(1) x = -1/2 --> either \(-\frac{1}{2}*y=0\) --> \(y=0\) and \(x+y=-\frac{1}{2}\) OR \(-\frac{1}{2}*y=1\) --> \(y=-2\) and \(x+y=-\frac{5}{2}\). Not sufficient.

(2) y is not equal to zero. Clearly not sufficient.

(1)+(2) Since from (2) \(y\neq{0}\), then from (1) \(y=-2\) and \(x+y=-\frac{5}{2}\). Sufficient.

now we can write eq as:- [square_root]xy=xy...... xy=(xy)^2.....ie (xy)^2-xy=0.....or xy(xy-1)=0.... so xy=0 or xy=1 i)x=-1/2..... substituting this value in xy we get (-1/2)y=0 ....so y=0... also (-1/2)y=0....y=-2.... not sufficient.... ii)y not equal to 0.... not sufficient... combining the two.... y=-2... sufficient

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.

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?

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.

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?

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.

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?

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 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...

Re: If sqrt(xy)=xy , what is the value of x + y? (1) x = -1/2 [#permalink]
13 May 2014, 09:09

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