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Radical Pair

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Radical Pair

by kv_ajay » Wed Dec 03, 2008 8:12 am
If (xy)^1/2 = xy what is value of x+y

(1) x = -1/2
(2) y is not equal to zero

OA is C

but explanation starts by squaring both sides of equation. I think we do not need do that. Just substituting right side from xy to (xy)^1/2 * (xy)^1/2 would be sufficient. Then we can say that (xy)^1/2 = 1 and y can not be 0. So A is sufficient.

Is there something i am doing wrong here..
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Re: Radical Pair

by vish150783 » Wed Dec 03, 2008 8:28 am
kv_ajay wrote:If (xy)^1/2 = xy what is value of x+y

(1) x = -1/2
(2) y is not equal to zero

OA is C

but explanation starts by squaring both sides of equation. I think we do not need do that. Just substituting right side from xy to (xy)^1/2 * (xy)^1/2 would be sufficient. Then we can say that (xy)^1/2 = 1 and y can not be 0. So A is sufficient.

Is there something i am doing wrong here..
Not sure but i think if u assume xy = 0 then you cant cancel zero on either side. But you could square zeros on both sides.
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Re: Radical Pair

by sudhir3127 » Wed Dec 03, 2008 10:12 am
kv_ajay wrote:If (xy)^1/2 = xy what is value of x+y

(1) x = -1/2
(2) y is not equal to zero

OA is C

but explanation starts by squaring both sides of equation. I think we do not need do that. Just substituting right side from xy to (xy)^1/2 * (xy)^1/2 would be sufficient. Then we can say that (xy)^1/2 = 1 and y can not be 0. So A is sufficient.

Is there something i am doing wrong here..
I go with C

Sqrt XY = XY

squarring on both sides
XY = xy*XY

dividing both sides by XY
XY/XY = XY*XY/XY

Now if either X or Y = 0 its will be in " indeterminate" form

Statement 1 gives u X

Statement Y says its non zero

solve the above equation

x= -1/2 Y = -2

u will get X+Y

hope this helps...
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