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Help with simple maths problem

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Help with simple maths problem [#permalink] New post 11 Jan 2007, 07:26
Help needed with the following:

(x-y)^2 = 0.

I understand that this translates to (x-y)(x-y) = 0. But kaplan says that this is the same as (x-y) = 0, why is this.

Thanks
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Re: Help with simple maths problem [#permalink] New post 11 Jan 2007, 07:31
GMAT? wrote:
Help needed with the following:

(x-y)^2 = 0.

I understand that this translates to (x-y)(x-y) = 0. But kaplan says that this is the same as (x-y) = 0, why is this.

Thanks


can we give any other value to (x - y), and ensure that (x-y)(x-y) = 0 ? :lol:
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Re: Help with simple maths problem [#permalink] New post 11 Jan 2007, 07:47
GMAT? wrote:
(x-y)^2 = 0.

I understand that this translates to (x-y)(x-y) = 0.


Actually, (x-y)^2 = x^2 - 2xy + y^2
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 [#permalink] New post 11 Jan 2007, 08:13
This is what I thought it translated to as well:

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

The whole question is:

Does x = y

(1) x^2 - y^2 = 0
(2) (x-y)^2 = 0

I know that the answer if B.

With (1): x^2 = y^2 but that does not mean x = y as 2^2 and -2^2 both = 16.

With (2): (x-y)^2 = 0. So x^2 - 2xy + y^2 = 0, hence x^2 -2xy = -y^2 hence x=y.

Just was unsure how kaplan got that (x-y)(x-y) = 0 is the same as (x-y) = 0.

Thanks
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 [#permalink] New post 11 Jan 2007, 11:44
In order to have (x-y)(x-y) equal 0, one of the two things you are multiplying must be equal to 0.

So either (x-y) or (x-y) has to be equal to 0.

And if one of them is, then the other one has to be as well.

Try working the logic using X * X = 0 (substitute X for x-y)

Hope that helps
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  [#permalink] 11 Jan 2007, 11:44
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