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

Options

A. `a^(2) - (bara)^(2) = 4b`
B. `a^(2) -(bara)^(2) = 2b`
C. `b^(2) - (bara)^(2) = 2a`
D. `b^(2) - (barb)^(2) = 2a`

Answer

A

Explanation

Let `alpha` be the real root and `ibeta` be the imaginary root of the given equation. Then
`alpha+ibeta = -a`
`rArr alpha-ibeta =-bar a`
So, `2alpha=-(a+bar a) and 2ibeta=-(a-bar a)`
Multiplying these, we get
`4ialphabeta=a^(2)-(bara)^(2)`
`therefore " " 4b=a^(2)-(bara)^(2)`