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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
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Re: If (xy)^)(1/2) = xy , what is the value of x + y? [#permalink]
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06 Sep 2009, 08:54
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.
Re: If (xy)^)(1/2) = xy , what is the value of x + y? [#permalink]
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06 Sep 2009, 10:38
1
This post received KUDOS
dhushan wrote:
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.
Re: If (xy)^)(1/2) = xy , what is the value of x + y? [#permalink]
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06 Sep 2009, 13:42
yezz wrote:
dhushan wrote:
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.
Re: If (xy)^)(1/2) = xy , what is the value of x + y? [#permalink]
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28 Sep 2009, 04:11
dhushan wrote:
yezz wrote:
dhushan wrote:
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 (xy)^)(1/2) = xy , what is the value of x + y? [#permalink]
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09 Mar 2016, 18:52
Bunuel wrote:
If \(\sqrt{xy} = xy\) what is the value of x + y?
\(\sqrt{xy} = xy\) --> \(xy=x^2y^2\) --> \(x^2y^2-xy=0\) --> \(xy(xy-1)=0\) --> 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.
Answer: C.
On combining both the statements, we still wouldn't know the value of x. What if x is 0?
Last edited by atirajak on 09 Mar 2016, 22:08, edited 1 time in total.
\(\sqrt{xy} = xy\) --> \(xy=x^2y^2\) --> \(x^2y^2-xy=0\) --> \(xy(xy-1)=0\) --> 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.
Answer: C.
What if x is 0? On combining both the statements, we still wouldn't know the value of x.
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.
Re: If (xy)^)(1/2) = xy , what is the value of x + y? [#permalink]
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21 Sep 2017, 17:48
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61% (01:09) correct
