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# if xy=1, what is the value of 2(x+y)^2/2(x-y)^2? A. 2 B. 4

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Senior Manager
Joined: 17 Aug 2005
Posts: 388
Location: Boston, MA
if xy=1, what is the value of 2(x+y)^2/2(x-y)^2? A. 2 B. 4 [#permalink]

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16 Feb 2007, 17:07
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if xy=1, what is the value of 2(x+y)^2/2(x-y)^2?
A. 2
B. 4
C. 8
D. 16
E. 32
Intern
Joined: 17 Dec 2006
Posts: 22

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16 Feb 2007, 17:41
buckkitty wrote:
if xy=1, what is the value of 2(x+y)^2/2(x-y)^2?
A. 2
B. 4
C. 8
D. 16
E. 32

Are you sure you copied the question right? 1s cancel off right away, doesnt look right
SVP
Joined: 08 Nov 2006
Posts: 1554
Location: Ann Arbor
Schools: Ross '10

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16 Feb 2007, 19:01
buckkitty wrote:
if xy=1, what is the value of 2(x+y)^2/2(x-y)^2?
A. 2
B. 4
C. 8
D. 16
E. 32

I think you mean 2^((x+y)^2)/2^((x-y)^2)

If this is the case, then

A=(x+y)^2 = x^2+y^2+2xy
B=(x-y)^2 = x^2+y^2-2xy

2^A/2^B=2^(A-B)

Substitute A and B and you get 2^(4xy)

If xy=1, then the expression becomes 2^4=16.

D.
Senior Manager
Joined: 12 Mar 2006
Posts: 365
Schools: Kellogg School of Management

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16 Feb 2007, 21:58
buckkitty wrote:
if xy=1, what is the value of 2(x+y)^2/2(x-y)^2?
A. 2
B. 4
C. 8
D. 16
E. 32

I think you mean 2^((x+y)^2)/2^((x-y)^2)

If this is the case, then

A=(x+y)^2 = x^2+y^2+2xy
B=(x-y)^2 = x^2+y^2-2xy

2^A/2^B=2^(A-B)

Substitute A and B and you get 2^(4xy)

If xy=1, then the expression becomes 2^4=16.

D.

this makes sense... the question as it is will not give a numerical solution...
Senior Manager
Joined: 17 Aug 2005
Posts: 388
Location: Boston, MA

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17 Feb 2007, 11:53

as ncprasad correctly pointed out, the question should be as follows:

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

A. 2
B. 4
C. 8
D. 16
E. 32
17 Feb 2007, 11:53
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