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Algebra

Cool goIITian

Joined:	14 Mar 2008
Post:	36

15 Mar 2008 14:18:12 IST

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292

quite easy

None

if (a-b)² + (b-c)²+(c-a)² = (a + b - 2c)²+ (a + c - 2b)²+ (c+ b - 2a)² with a,b,c real and non zero then if a/b = k, what will b/c be?

1. 2k

2. k + k²

3. k² - k

4. 1

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Comments (4)

gokul subramanian

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Joined: 28 Jan 2008

Posts: 1451

15 Mar 2008 14:19:15 IST

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if u havnt noticed yet chk out ans 4 ur probability ques...

sandeep ramesh

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Posts: 1183

15 Mar 2008 14:21:19 IST

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solve and get a = b = c and hence k=1 and so b/c = k

Hari Shankar

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Joined: 28 Feb 2007

Posts: 2173

15 Mar 2008 14:34:24 IST

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(a-b)² + (c-b)² = (a+c-2b)² + 2(a-b)(b-c) etc.

Hence 2

(a+c-2b)² + 2(a-b)(b-c) =

(a+c-2b)²

(a+c-2b)²+ 2(a-b)(b-c) = 0

or 2

(a-b)²= 0

Hence a=b=c

sandeep ramesh

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15 Mar 2008 14:38:54 IST

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thank you sir for verifying my soln, i had a doubt abt that bcos he didnt specify a value for k n treated it as generic

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