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

|B+6|–|B–5|=0
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Re: old question [#permalink] New post 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|–|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

I hope this is helpful.
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Re: old question [#permalink] New post 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

I hope this is helpful.


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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 [#permalink] New post 08 Jul 2004, 07:53
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
  [#permalink] 08 Jul 2004, 07:53
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