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by bhumika.k.shah » Sun Jan 17, 2010 12:48 am
If x is not equal to 0 ,then (x + 1/x)^2 - (x - 1/x)^2=?

A. 1/x
B. x
C. 1
D. 2
E. 4

OA E

Can i plug in #s?
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by ace_gre » Sun Jan 17, 2010 12:53 am
Hi, I would use the identities (a+b)^2 = a^2 + b^2 + 2ab and (a-b)^2 = a^2 + b^2 - 2ab
Here a = x , b = 1/x

(x + 1/x)^2 - (x - 1/x)^2 ==> 4 ab = 4 *x * 1/x = 4
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by linkinpark » Sun Jan 17, 2010 1:17 am
bhumika.k.shah wrote:If x is not equal to 0 ,then (x + 1/x)^2 - (x - 1/x)^2=?

A. 1/x
B. x
C. 1
D. 2
E. 4

OA E

Can i plug in #s?
in my view yes you could use numbers in this case because it involves only one variable x, if there were 2 variables then you might have to consider of scenarios like one variable is +ve and other -ve.
530->480->580
when posting a question don't post OA(even masked) before some discussion.
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