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# If rs = 1, what is the value of 2(r+s)^2 / 2(r-s)^2

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If rs = 1, what is the value of 2(r+s)^2 / 2(r-s)^2 [#permalink]

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22 Jan 2006, 23:53
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If rs = 1, what is the value of 2(r+s)^2 / 2(r-s)^2

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23 Jan 2006, 00:13
it looks the ques shud b

if rs=1, what is the val of 2^((r+s)^2)/2^((r-s)^2).

if it is the case, it is 16. ==> 2^2/2^-2 ==> 24 = 16.

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23 Jan 2006, 00:14
Bhai wrote:
If rs = 1, what is the value of 2(r+s)^2 / 2(r-s)^2

No solution for this one Bhai!Probably some typo in the question stem
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23 Jan 2006, 00:19
Please see this from GMATPrep. I wonder if this one is "not-score" question.
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Director
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23 Jan 2006, 00:30
Bhai wrote:
Please see this from GMATPrep. I wonder if this one is "not-score" question.

http://www.gmatclub.com/phpbb/viewtopic.php?t=25643
Agree with you Bhai it looks so simple but can't solve it
BTW maybe you have OE for this one?
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23 Jan 2006, 00:33
If x and y are integers, then the value is infinity. If no bounds are given, then cannot be solved.

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23 Jan 2006, 00:33
....and I thought I was

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23 Jan 2006, 23:34
the same question has already been discussed as shown by yurik !

the question stem is wrong!
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24 Jan 2006, 01:15
Bhai wrote:
....and I thought I was

I know. Someone asked the same question previously and I was ed too.
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24 Jan 2006, 06:08
2(r^2+2rs+s^2)/2(r^2-2rs+s^2)

We know rs =1, so:

2(r^2+2rs+s^2)/2(r^2-2rs+s^2)

= 2(r^2+2+s^2)/2(r^2-2+s^2)

If rs=1, then (r,s) = (1,1) or (-1,-1). For both cases, r^2 and s^2 = 1.

So :

= 2(1+2+1)/2(1-2+1)

= Indefinite

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24 Jan 2006, 06:16
ywilfred wrote:
2(r^2+2rs+s^2)/2(r^2-2rs+s^2)

We know rs =1, so:

2(r^2+2rs+s^2)/2(r^2-2rs+s^2)

= 2(r^2+2+s^2)/2(r^2-2+s^2)

If rs=1, then (r,s) = (1,1) or (-1,-1). For both cases, r^2 and s^2 = 1.

So :

= 2(1+2+1)/2(1-2+1)

= Indefinite

What if r and s are not integers? for example r=1/2 s= 2 or r=2 s =1/2
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24 Jan 2006, 12:57
ywilfred wrote:
2(r^2+2rs+s^2)/2(r^2-2rs+s^2)

We know rs =1, so:

2(r^2+2rs+s^2)/2(r^2-2rs+s^2)

= 2(r^2+2+s^2)/2(r^2-2+s^2)

If rs=1, then (r,s) = (1,1) or (-1,-1). For both cases, r^2 and s^2 = 1.

So :

= 2(1+2+1)/2(1-2+1)

= Indefinite

Wilfred, if I were you and could assume x,y are integers then its obvious they are either 1,1 or -1,-1 in that case, I would not have to do the 1/2 pg long calculations to figure out that the ans is indefinite! The Dr contains (x-y)^2, which is obviously zero and hence the expression infinite!!!!!

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24 Jan 2006, 12:57
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# If rs = 1, what is the value of 2(r+s)^2 / 2(r-s)^2

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