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Is |xy| > x^2y^2 ?

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Is |xy| > x^2y^2 ?

by massi2884 » Fri May 18, 2012 8:09 am
Is |xy| > x^2y^2 ?

1) 0<x^2>1/4
2) 0<y^2<1/9

OA C
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by neelgandham » Fri May 18, 2012 8:35 am
Is |xy| > x^2y^2 ?
1) 0<x^2<1/4
Statement 1 is insufficient to answer the question as the value of(or the limits of) y isn't provided.
2) 0<y^2<1/9
Statement 2 is insufficient to answer the question as the value of(or the limits of) x isn't provided.
Statement 1 + 2
0<(x^2)*(y^2)<(1/9)*(1/4)
0<(x^2)*(y^2)<(1/36)
let xy be A, then
0<A^2<1/36
-1/6<A<1/6
Since -1<xy<1, |xy| > x^2y^2
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by greenwich » Tue Oct 16, 2012 8:11 am
The OE provided states that |xy|>x^2y^2===>(x^2)(y^2)>(x^4)(y^4)===>1>x^2y^2

Can anyone explain how |xy|>x^2y^2===>(x^2)(y^2)>(x^4)(y^4)?
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by Whitney Garner » Wed Oct 17, 2012 10:14 am
greenwich wrote:The OE provided states that |xy|>x^2y^2===>(x^2)(y^2)>(x^4)(y^4)===>1>x^2y^2

Can anyone explain how |xy|>x^2y^2===>(x^2)(y^2)>(x^4)(y^4)?
Hi greenwich!

This actually uses a little-known definition of absolute value:

|x| = sqrt(x^2).

So if we redefine the original expression using the following:

Image

and then square both sides - you get the equation the OE provides! And YES, you can always use this alternate definition of absolute value, just make sure that the square root is on the outside and the squared in on the inside (you CANNOT interchange them since you don't know the signs on the variables and might be accidentally taking the square root of the negative BEFORE you square it!!).

Hope this helps!!
:)
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Math is a lot like love - a simple idea that can easily get complicated :)
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