R = 1 + 2xy + x^2y^2, what is the value of xy?

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R=1+2xy+(xy)^2, what is the value of x∗y ?

(1) R = 0
(2) x > 0

==> transforming the original condition and the question by variable approach method, we have, R=(1+xy)^2, 1+xy=R, -R, then xy=R-1, xy=-R-1. if we know R we can find xy
In case of 1), if R=0 then xy=-1, a unique answer. Therefore it is sufficient. Thus A is the answer.



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\(R = 1 + 2*x*y + x^2*y^2\), what is the value of \(x*y\)?

(1) R = 0
(2) x > 0

\(R = 1 + 2xy + (x^2)(y^2)\)
\(R = 1 + (xy)^2\)
\(xy = \sqrt{R}-1\)

Statement 1: R=0
Surely it gives xy =-1
Sufficient

Statement 2: x>0
Irrelevant . Not Sufficient

Answer A

Please what happened to the 2xy in the question? I'm confused

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The answer is indeed A

R=1+2xy+x^2y^2 => (1+xy)^2

Statement I

R=0, therefore (1+xy)^2=0
xy=-1

Statement II
x>0

It does not gives us any specific value of x or y or R

Not Sufficient.



Hence A.

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Hi, in the solutions that have been given before on this question. Please can someone help me with how we go from R=1+2xy+x^2y^2 => (1+xy)^2? Thanks - I don't get this final step/the work done.