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testing numbers.

by resilient » Wed Mar 05, 2008 5:48 pm
DO you know of an easier way to do this? I was picking numbers and it got pretty messy. Easier question though.

Does a=b?

1. a/b=b/a

2. (a-b)^2=a^2-b^2



qa is C, I chose E
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by hemanth28 » Wed Mar 05, 2008 7:49 pm
from (1)we have a^2-b^2 = 0
=>a=b (or) a=-b so you cant conclude if a=b.

from (2) we have
(a-b)^2=a^2-b^2
=>(a-b)(-2b)=0
=>either a=b or b=0

if both have to be true a=b or b=0(in which case again a=b)
hence [C]
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by xilef » Thu Mar 06, 2008 11:57 am
hemanth28,

I was wondering how you got:
(a-b)^2=a^2-b^2
=>(a-b)(-2b)=0

The way I solved it

(a-b)^2=a^2-b^2 =>
a^2-2ab+b^2=a^2-b^2 =>
ab=b^2
a=b

or

(a-b)(a-b)=(a-b)(a+b) =>
(a-b)=(a+b)
b=0
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