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PS: x^2

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Senior Manager
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PS: x^2 [#permalink]

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New post 20 Aug 2008, 00:23
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A
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D
E

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Re: PS: x^2 [#permalink]

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New post 20 Aug 2008, 00:32
\(x^2y^2+3xy-18=0\)
\(xy(xy+6)-3(xy+6) = 0\)
xy=3 or xy=-6

both positive so xy cant be -6

\(x^2y^2=9\)

\(x^2 = 9/y^2\)

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Re: PS: x^2 [#permalink]

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New post 20 Aug 2008, 00:36
Answer: 9/(y^2).

x^2y^2 =18-3xy

One other method is to substitute the answer choices.

Try the easier ones first. When x^2=9/(y^2) is substituted in the given equation, we get 9=9 and hence the answer choice D.
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Re: PS: x^2 [#permalink]

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New post 20 Aug 2008, 00:46
It seemed easier to solve the quadratic equation for xy. You get (xy-3)(xy+6)=0. Then xy = 3; -6. But both are positive so only one answer left.
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Re: PS: x^2 [#permalink]

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New post 20 Aug 2008, 15:17
Nerdboy wrote:
It seemed easier to solve the quadratic equation for xy. You get (xy-3)(xy+6)=0. Then xy = 3; -6. But both are positive so only one answer left.


Agree quadratic is faster than plugging in.

x & y = positive is key to answering 9/y^2

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Re: PS: x^2 [#permalink]

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New post 21 Aug 2008, 12:50
Why not E? that would work as well.

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Re: PS: x^2 [#permalink]

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New post 21 Aug 2008, 12:55
sarzan wrote:
Why not E? that would work as well.


x,y must be positive. -6^2 yields 36 but cannot use b/c does not follow rules

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Re: PS: x^2 [#permalink]

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New post 21 Aug 2008, 12:56
As mentioned before, the two solutions are:

xy = -6, and xy = 3

If we sub in xy = 3, we get X^2 = 9/y^2

But, if we sub in xy = -6, we get X^2 = 36/y^2, which is also in the answer choices.

So what am I doing wrong?

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Re: PS: x^2   [#permalink] 21 Aug 2008, 12:56
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