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If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y

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Re: If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y  [#permalink]

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New post 27 Jun 2017, 04:48
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GMATinsight wrote:
Bunuel wrote:
If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

(1) |xy| is NOT a square of an integer
(2) Point (x, y) is above x-axis



SOLUTION (OE) IS HERE.



√(x^2-y^2) = (x+y)
Only If y=0

Each statement says that y#0
Hence sufficient

Answer option D


There is another possibility, when x = -y. For example, x = 1 and y = -1.
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Re: If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y  [#permalink]

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New post 27 Jun 2017, 05:05
Bunuel wrote:
GMATinsight wrote:
Bunuel wrote:
If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

(1) |xy| is NOT a square of an integer
(2) Point (x, y) is above x-axis



SOLUTION (OE) IS HERE.



√(x^2-y^2) = (x+y)
Only If y=0

Each statement says that y#0
Hence sufficient

Answer option D


There is another possibility, when x = -y. For example, x = 1 and y = -1.


Good point... :) +1Kudos
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If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y  [#permalink]

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New post 27 Jun 2017, 05:09
Bunuel wrote:
If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

(1) |xy| is NOT a square of an integer
(2) Point (x, y) is above x-axis



SOLUTION (OE) IS HERE.


\(\sqrt{x^2 - y^2} = x + y\)
-> x^2-y^2 = (x+y)^2
-> (x+y)(x-y ) = (x+y)^2
(x+y )(y) = 0
-> Either y =0 or x = -y which we require .............................................................DS

Option 1 : |xy| is NOT a square of an integer
so x =/= y,-y,0 .. Since these values will make the value |xy| as a square of an integer.

Hence SUFFICIENT.


Option 2 : Point (x, y) is above x-axis, So y>0
Also y=/= 0,

Now since -x<=y <= x....
Since y>0, y =/=-x

Hence SUFFICIENT....
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Re: If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y  [#permalink]

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New post 28 Jun 2017, 13:51
egghead2017 wrote:
Squaring both sides we get 2y(x+y) = 0 , so the question becomes "Is 2y(x+y) =0 ?" or y =0 or x=-y ?

Statement 1:

|xy| != perfect square ..
means x!=y or x!=-y , but Y could be zero, or non-zero. Not Sufficient

Statement 2 :

Point (x,y) is above X- axis.

Therefore, y>0 and x = anything.

Point could be (-3,3) or ( -4,3)

So, NOT Sufficient .


Answer E


Correct me if i am wrong.



2y(x+y) =0 ... can't we expand it further? we can. The final equation will be x+y=0
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Re: If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y  [#permalink]

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New post 28 Jun 2017, 13:55
seeker14 wrote:
egghead2017 wrote:
Squaring both sides we get 2y(x+y) = 0 , so the question becomes "Is 2y(x+y) =0 ?" or y =0 or x=-y ?

Statement 1:

|xy| != perfect square ..
means x!=y or x!=-y , but Y could be zero, or non-zero. Not Sufficient

Statement 2 :

Point (x,y) is above X- axis.

Therefore, y>0 and x = anything.

Point could be (-3,3) or ( -4,3)

So, NOT Sufficient .


Answer E


Correct me if i am wrong.



2y(x+y) =0 ... can't we expand it further? we can. The final equation will be x+y=0


That's not correct. \(y(x+y)=0\) means \(y=0\) or \(x=-y\).

I suggest to read solution HERE.
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Re: If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y  [#permalink]

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New post 09 Mar 2019, 01:07
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Re: If x and y are integers and -x < y < x, does (x^2 - y^2)^(1/2) = x + y   [#permalink] 09 Mar 2019, 01:07

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