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Re: if a>0, b>0, c>0, is a(b-c)=0? I. b-c = c-b II. b/c [#permalink]

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30 Jun 2013, 07:37

fozzzy wrote:

For statement 2

we get \(b^2=c^2\) then \(b^2 - C^2 = 0\) >> \((b-c) (b+c) = 0\)

Why is this manipulation incorrect?

That is not incorrect.

You get \((b-c) (b+c) = 0\) so one (or both) terms equal zero. However we know that both b and c are positive, so (b+c) is positive as well, so b-c=0 Hence \(a(b-c)=0\) => sufficient _________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: if a>0, b>0, c>0, is a(b-c)=0? I. b-c = c-b II. b/c [#permalink]

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30 Jun 2013, 07:52

Zarrolou wrote:

That is not incorrect.

You get \((b-c) (b+c) = 0\) so one (or both) terms equal zero. However we know that both b and c are positive, so (b+c) is positive as well, so b-c=0 Hence \(a(b-c)=0\) => sufficient

So basically B+C is just redundant information once we solve till that point? Hence that case is ignored and we focus on the second case? _________________

Click +1 Kudos if my post helped...

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Re: if a>0, b>0, c>0, is a(b-c)=0? I. b-c = c-b II. b/c [#permalink]

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30 Jun 2013, 07:56

1

This post received KUDOS

fozzzy wrote:

Zarrolou wrote:

That is not incorrect.

You get \((b-c) (b+c) = 0\) so one (or both) terms equal zero. However we know that both b and c are positive, so (b+c) is positive as well, so b-c=0 Hence \(a(b-c)=0\) => sufficient

So basically B+C is just redundant information once we solve till that point? Hence that case is ignored and we focus on the second case?

I do not know what you mean by "redundant information", but yes: we know that \(b+c\) cannot be zero, so the other term (\(b-c\)) must be zero.

All we get from \(b^2-c^2=0\) in this problem is that \(b-c=0\) _________________

It is beyond a doubt that all our knowledge that begins with experience.

Is \(a(b - c) = 0\)? --> is \(a=0\) or \(b-c=0\)? Since given that \(a > 0\), then the questions basically asks whether \(b-c=0\).

(1) b - c = c - b --> \(2b-2c=0\) --> \(b-c=0\). Sufficient.

(2) b/c = c/b --> \(b^2=c^2\) --> \((b-c)(b+c)=0\) --> \(b+c=0\) or \(b-c=0\) but since \(b\) and \(c\) are positive, then \(b+c>0\). Therefore \(b-c=0\). Sufficient.

Re: If a > 0, b > 0 and c > 0, is a(b - c) = 0? [#permalink]

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25 Sep 2014, 00:30

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