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The equation having $4-3j$ and $4+3j $ as roots is

\[\begin{array}{1 1}(1)x^{2}+8x+25=0&(2)x^{2}+8x-25=0\\(3)x^{2}-8x+25=0&(4)x^{2}-8x-25=0\end{array}\]

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The given roots are :
$\alpha =4 -3 i, \beta =4+3 i$
$\alpha+ \beta =4-3 i +4 +3i=8$
$\alpha \beta =(4-3i) (4+3i) =4^2-(3i)^2$
The required quadratic equation is
$x^2$- (sum of the roots)x +products of the roots=0
Hence 3 is the correct answer
answered May 15, 2014 by meena.p

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