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roots

by smallsorrow » Tue Nov 11, 2008 1:38 am
I guessed correctly... I know what the essense of this problem is (4+1-x -> when does the result get negative - but mathematically ...??
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by vishubn » Tue Nov 11, 2008 2:07 am
Plug in the values !!

A) u get 0 which is a real number
B)root2 is 1.41 ... so 2-1.41 is 0.61
root of 0.61 is < 1
so when 1 is subtracted ! it is again some numebr which is a real number

C) root 3 is 1.73 !! so u go in the same line as B) and u get a real number
D)u get 0 as the answer
E)root 5 is >2 .. so 2- root 5 is negative which is not real number

OA E

VIshu
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by stop@800 » Tue Nov 11, 2008 3:10 am
for results to be imaginary

x can be -ve
but not given so move one step out

2 - sqrt(x) has to be < 1

so if sqrt(x) is > 2 we are done
which is possible with x=5

so we have solved it at 1 level.
in some cases we may be required to move out....
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IMO E

by iamcste » Tue Nov 11, 2008 8:05 am
For X to be not real, sqrt x > 2 ( Based on the equation)

Square it ob both sides

x >4

Only one option E

Dont do intensive calculations !
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