Question

Given that the equation $z^2+(p+iq)z+r+is$=0 where p,q,r,s are real and non-zero has a real root, then

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

Given that $z^2$+(p+iq)z+r+is=0 .......(i)

Let z=$\alpha$ (where $\alpha$ is real) be a root of (i), then

${\alpha }^2$+(p+iq)$\alpha$+r+is=0

or ${\alpha }^2$+p$\alpha$+r+i(q$\alpha$+s)=0

Equating real and imaginary parts,we have

${\alpha }^2$+p$\alpha$+r=0 and q$\alpha$+s=0

Eliminating $\alpha$, we get ${\left(\frac{-s}{q}\right)}^2$+p$\left(\frac{-s}{q}\right)$+r=0

or $s^2$-pqs+$q^2$r=0 or pqs=$s^2$+$q^2$r