Help with simple maths problem

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Topic: Help with simple maths problem
PostThu Jan 11, 2007 8:26 am
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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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b00ndgl
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PostThu Jan 11, 2007 2:19 pm
If (x-y)^2 = (x-y)(x-y)=0, then we can immediately say that (x-y)=0. This equation consists of the multiplication of two terms that are the same, so we need only consider one of the terms (since they are both the same).
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chandooed
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PostThu Jan 11, 2007 2:26 pm
(x-y)^2=0;
apply sqrt on both sides of the '=' sign
sqrt [ (x-y)^2 ] = sqrt (0); ( sqrt of (x-y)^2 is (x-y) & sqrt of 0 is 0)
so, (x-y) = 0.

Hope this helps.
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wildrebel
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PostThu Jan 11, 2007 3:37 pm
In simpler terms,

if (x-y)^2 = 0,

that means (x-y)(x-y)=0

Simple logic- if ZERO is the product of two numbers, one or both of those numbers HAS to be ZERO.
For eg. (1)(0)=0, (0)(534)=0 or (0)(0)=0

In this case, since you're squaring (x-y), and the answer is 0, that means x-y HAS TO BE 0.

Other than the number 0, no other number can mathematically square to 0.

Hope that was easier to understand than lines of equations.
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