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# old question

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Senior Manager
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old question [#permalink]  03 Jul 2004, 23:32
I saw this question in old posts..but i was wondering how the solution was arrived.

|B+6|â€“|Bâ€“5|=0
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Re: old question [#permalink]  04 Jul 2004, 04:22
crackgmat750 wrote:
I saw this question in old posts..but i was wondering how the solution was arrived.

|B+6|â€“|Bâ€“5|=0

For every equation of type |a|=|b|, we can write its equivalent representation: a^2=b^2

|B+6|â€“|Bâ€“5|=0
|B+6|^2=|B-5|^2
b^2+12b+36=b^2-10b+25
22b=-11
b=-1/2

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Re: old question [#permalink]  08 Jul 2004, 05:11
SmashingGrace wrote:
crackgmat750 wrote:
I saw this question in old posts..but i was wondering how the solution was arrived.

|B+6|–|B–5|=0

For every equation of type |a|=|b|, we can write its equivalent representation: a^2=b^2

|B+6|–|B–5|=0
|B+6|^2=|B-5|^2
b^2+12b+36=b^2-10b+25
22b=-11
b=-1/2

One note, however. If you square any equation, you may get extraneous roots. Therefore, once roots are identified, each one has to be checked by plugging into the initial equation..
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Probably ,this solution is easier

|B+6|=|B-5|
B+6 = -(B-5) or B+6 = B-5
B+6 = -B+5 or 6=-5 not possible
B=-1/2
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# old question

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