If `a`, `b` are complex numbers and one of the roots of the equation `x^(2)+ax+b=0` is purely real whereas the other is purely imaginery, and `a^(2)-bara^(2)=kb`, then `k` is

Options

A. `2`
B. `4`
C. `6`
D. `8`

Answer

B

Explanation

`(b)` Let us consider `alpha` as the real and `ibeta` as the imaginergy root. Then
`alpha+ibeta=-a`
`implies alpha-ibeta=-bara`
`implies2alpha=-(a+bara)` and `2ibeta=-(a-bara)`
`implies 4ialphabeta=a^(2)-bara^(2)`
`implies a^(2)-bara^(2)=4b`