Consider a quadratic equation `az^(2)bz+c=0,` where a,b and c are complex numbers.
The condition that the equation has one purely real root, is

Options

A. `(overline(b)c-boverline(a))/(aoverline(a)-coverline(c))=(aoverline(a)+coverline(c))/(overline(c)b+aoverline(b))`
B. `(overline(b)(c)+boverline(a))/(aoverline(a)+coverline(c))=(aoverline(a)+boverline(a))/(aoverline(a)+coverline(c))=(aoverline(a)+coverline(c))/(overline(c)b+aoverline(b))`
C. `(overline(b)c-boverline(a))(overline(c)b-aoverline(b))=(aoverline(a)-coverline(c))^(2)`
D. `(overline(b)c+boverline(a))/(aoverline(a)+coverline(c))=(aoverline(a)+coverline(c))/(overline(c)b-aoverline(b))`

Answer

C

Explanation

Let `alpha` be one root of equation then `overline(alpha)=-alpha`
Given equation is `az^(2)+bz+c=0`
`becauseaalpha^(2)+balpha+c=0`
Taking conjugate we get `becauseoverline(alpha)overline(alpha)^(2)+overline(b).overline(c)=0`
`overline(a).alpha^(2)-overline(b).apha+overline(c)=0` and `overline(a)z^(2)-overline(b)z+overline(c)=0`
So `(overline(a))/(a)=(overline(b))/(b)=(overline(c))/(c)`
if `|alpha|=1` then `overline(alpha)=(1)/(alpha)` as `aalpha^(2)+balpha+c=0` ..(i)
`impliesoverline(a)overline(alpha)^(2)+overline(b)overlinealpha)+overline(c)=0`
`implies(overline(a))/(alpha^(2))+(overline(b))/(alpha)+overline(c)=0`
So `overline(c).alpha^(2)+overline(b).alpha+overline(a)=0` .(ii)
Now appying condition for one common root for (1) and (@) we get `(aoverline(b)-boverline(c))(boverline(a)-coverline(b))=(aoverline(a)-coverline(c))^(2)`