Correct Answer - 18
`A= (1+ 2i)x^(3) - 2(3 + i) x^(2) + (5-4i)x + 2a^(2) = 0`
Let the real root of equations be `alpha`. Then
`(1+2i) alpha^(3) -2(3+i) alpha^(2) + (5-4i) alpha + 2alpha^(2)= 0`
Equating imaginary part zero, we get
`2apha^(3) - 2alpha^(2) - 4alpha = 0`
` rArr or alpha(alpha^(2) - alpha -2) = 0`
`rArr alpha = 0 or alpha = - 1,2`
Now equating real part zero, we have
`alpha^(3) - 6alpha^(2) + 5alpha+ 2alpha^(2) = 0` "
`alpha = 0 rArr a =0`
` alpha= -1 rArr a = pm sqrt(6)`
`alpha = 2 rArr alpha = pm sqrt(3)`
`rArr sum alpha^(2) = (0)^(2) + (+sqrt(6))^(2) + (-sqrt(6))^(2) + (+sqrt(3))^(2)+ (-sqrt(3))^(2)`
`= 18`