Problem Solving

If X*Y = 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 hope that the question is clear, thans you...

There need to be a mistake

X=2, Y=1/2 => X*Y=1

(5/2)^2)/(3/2)^2=(5/3)^2=25/9

and several other results...

CIAO1982 wrote:If X*Y = 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 hope that the question is clear, thans you...

Definitely a typo.

The fraction reduces to:

(x^2 + 2xy + y^2)/(x^2 - 2xy + y^2)

(x^2 + 2 + y^2)/(x^2 - 2 + y^2)

Which has an infinite number of solutions.

Stuart,
U r correct; ther is a typo. The actual question if I am not
mistaken is 2 ^ (x+y)^2 / 2 ^ (x-y)^2
given xy=1




CIAO1982,

Here's the solution:

2 ^ (x+y)^2 / 2 ^ (x-y)^2

{
((x+y)^2 = x^2+y^2+2xy
(x-y)^2 = x^2+y^2-2xy)
}

= 2 ^ (x^2+y^2+2xy) / 2^ (x^2+y^2- 2xy)

(a^m/a^n = a ^ (m-n) where a=2 m = (x+y)^2 n = (x-y)^2)

= 2 ^ (x^2+y^2+2xy - ( x^2+y^2-2xy)
= 2 ^ (x^2+y^2+2xy-x^2-y^2+2xy)
= 2^4xy
= 2 ^4 (since xy=1)
= 16




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OA: whatever the oa is

Regards,
Cramya