106
Asish Mahapatra
·2010-04-03 06:02:59
Q1. the mistake ur doing is like this
x2+1=0
=> x(x2+1) = 0
=> x=0
multiplying by zero wont be valid here, so ur getting an absurd answer.
Put z=0 in the original equation it doesnt satisfy
1
The Enlightened One - jsg
·2010-04-03 10:29:57
thanks yaar ashish..u saved my one day ! [1]
1
Bicchuram Aveek
·2010-04-03 10:52:28
ANS 1 :
Z3 + iZ - 1 = 0 ..... (1)
IF Z HAS REAL ROOTS THEN Z = Z' (Z BAR) FOR IMAGINARY PART TO BE 0.
SO TAKING CONJUGATE OF THE ENTIRE EXPRESSION :
(Z')3 + i' Z' - 1' = 0
AS Z = Z'
Z3 - i Z - 1 = 0 ..... (2)
SOLVING EQUATIONS (1) AND (2) :
Z3- 2 i = Z-1+1 = 1- 2 i
SO WE GET Z3 = 1 AND Z =0 WHICH IS HIGHLY CONTRADICTORY .... SO THIS EQUATION HAS NO REAL ROOTS.
1
Bicchuram Aveek
·2010-04-03 11:01:06
AND AS FAR AS THE 2ND QUESTION IS CONCERNED....THE ANSWER GIVEN IS TOTALLY WRONG.
IT SHOULD BE (B) AS YOU HAVE GOT ... NOT [a] .