What is the value of (a+b)^2? 1) ab=0 2) (a-b)^2=36 : DS Archive
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# What is the value of (a+b)^2? 1) ab=0 2) (a-b)^2=36

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What is the value of (a+b)^2? 1) ab=0 2) (a-b)^2=36 [#permalink]

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24 May 2006, 14:16
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What is the value of (a+b)^2?

1) ab=0
2) (a-b)^2=36
Senior Manager
Joined: 17 Aug 2005
Posts: 392
Location: Boston, MA
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Kudos [?]: 93 [0], given: 0

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24 May 2006, 16:23
for 2) can't you take the SQRT of both sides to get (a-b)=+or- 6
So if 6 is + then,
a-b=6
a+b=-6

if 6 is - then,
a-b=-6
a+b=6

either way (a+b)^2 = 36

so B is sufficient? what incorrect assumption am i making??
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24 May 2006, 16:46
St1:
ab = 0 --> either a or b or both are 0. But we do not know which. So we can't derive a value for (a+b)^2

St2:
(a-b)^2 - 36 = 0
(a-b+36)(a-b-36) = 0

a-b+36=0 --> a+b = 36
a-b-36=0 --> a+b = -36

Insuffcient.

Using St1 and St2:
(a-b)^2 = a^2-2ab+b^2 = a^2+b^2 = 36

The question is asking or (a+b)^2 = a^2 + 2ab + b^2. We know ab = 0, so we need to find a^2+b^2 which we have derived to be 36.

Sufficient.

Ans C
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24 May 2006, 16:54
buckkitty wrote:
for 2) can't you take the SQRT of both sides to get (a-b)=+or- 6
So if 6 is + then,
a-b=6
a+b=-6

if 6 is - then,
a-b=-6
a+b=6
either way (a+b)^2 = 36

so B is sufficient? what incorrect assumption am i making??

a-b = 6 cannot be turned into a+b = -6.
a-b=-6 also cannot be turned into a+b=6.
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24 May 2006, 18:40
Thanks professor. Not sure how I came up with THAT arithmetic It is frustrating how the simple stuff screws me up the longer I study.

Good job everyone, OA is C
24 May 2006, 18:40
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