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(1) ab = 2ab --> \(ab=0\) --> since given that \(a>0\) then \(b=0\). Sufficient.

(2) |a + b| = |a − b| --> the distance between \(a\) and \(b\) is the same as distance between \(a\) and \({-b}\) --> again since given that \(a>0\) then \(b=0\) (if \(a\) is not zero then --0----a---- and \(b\) must be 0: --(b=0)----a----). Or since both parts of the expression are positive we can safely apply squaring: (\(a+b)^2=(a-b)^2\) --> \((a+b)^2-(a-b)^2=0\) --> \((a+b-a+b)(a+b+a-b)=0\) --> \(2b*2a=0\) --> \(ab=0\). The same info as above: since given that \(a>0\) then \(b=0\). Sufficient.

If a is positive, what is the value of b? [#permalink]

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07 Nov 2014, 22:47

Bunuel Thank you for all your posts. They've been helping me a lot.

I am a little confused here. I did not quite get the first sentence of your S2 explanation. I know that absolute value is the number's distance from 0, in which case a+b and a-b are equidistant from 0. But that is all I know.

Also, you squared the equation because...? is it because the signs could then be ignored?

what about two cases a+b=a-b or a+b = -a+b ? can anything be done with this?

Bunuel Thank you for all your posts. They've been helping me a lot.

I am a little confused here. I did not quite get the first sentence of your S2 explanation. I know that absolute value is the number's distance from 0, in which case a+b and a-b are equidistant from 0. But that is all I know.

Also, you squared the equation because...? is it because the signs could then be ignored?

what about two cases a+b=a-b or a+b = -a+b ? can anything be done with this?

Thanks a lot D

|a - b| is the distance between a and b. Similarly |a + b|, or which is the same as |a - (-b)| is the distance between a and -b.

As for squaring, it allows us to get rid of the modulus, so it simplifies the expression to manipulate with it further.

Re: If a is positive, what is the value of b? [#permalink]

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18 Aug 2016, 00:19

1

This post was BOOKMARKED

Bunuel wrote:

If a is positive, what is the value of b?

(1) ab = 2ab --> \(ab=0\) --> since given that \(a>0\) then \(b=0\). Sufficient.

(2) |a + b| = |a − b| --> the distance between \(a\) and \(b\) is the same as distance between \(a\) and \({-b}\) --> again since given that \(a>0\) then \(b=0\) (if \(a\) is not zero then --0----a---- and \(b\) must be 0: --(b=0)----a----). Or since both parts of the expression are positive we can safely apply squaring: (\(a+b)^2=(a-b)^2\) --> \((a+b)^2-(a-b)^2=0\) --> \((a+b-a+b)(a+b+a-b)=0\) --> \(2b*2a=0\) --> \(ab=0\). The same info as above: since given that \(a>0\) then \(b=0\). Sufficient.

Answer: D.

Hope it helps.

I did not get |a + b| = |a − b| --> the distance between \(a\) and \(b\) is the same as distance between \(a\) and \({-b}\) --> again since given that \(a>0\) then \(b=0\) (if \(a\) is not zero then --0----a---- and \(b\) must be 0: --(b=0)----a----). part of the explanation?
_________________

(1) ab = 2ab --> \(ab=0\) --> since given that \(a>0\) then \(b=0\). Sufficient.

(2) |a + b| = |a − b| --> the distance between \(a\) and \(b\) is the same as distance between \(a\) and \({-b}\) --> again since given that \(a>0\) then \(b=0\) (if \(a\) is not zero then --0----a---- and \(b\) must be 0: --(b=0)----a----). Or since both parts of the expression are positive we can safely apply squaring: (\(a+b)^2=(a-b)^2\) --> \((a+b)^2-(a-b)^2=0\) --> \((a+b-a+b)(a+b+a-b)=0\) --> \(2b*2a=0\) --> \(ab=0\). The same info as above: since given that \(a>0\) then \(b=0\). Sufficient.

Answer: D.

Hope it helps.

I did not get |a + b| = |a − b| --> the distance between \(a\) and \(b\) is the same as distance between \(a\) and \({-b}\) --> again since given that \(a>0\) then \(b=0\) (if \(a\) is not zero then --0----a---- and \(b\) must be 0: --(b=0)----a----). part of the explanation?

Before attempting questions you should make sure that your fundamentals are clear. You should understand what an absolute value is. After that start with easier questions on modulus and move to harder ones.

Re: If a is positive, what is the value of b? [#permalink]

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21 Sep 2017, 22:14

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