Question

If the equation $z^2+(p+iq)z+r+is=0$, wherre $p,q,r$ and $s$ are real and non-zero roots, then

• A. $pqr=r^2+p^{2}s$
• B. $prs=q^2+r^{2}p$
• C. $qrs=p^2+s^{2}q$
• D. $pqs=s^2+q^{2}r$

Answer 1

Answer: (D)

$pqs=s^2+q^{2}r$

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

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

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

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

On equating real and imaginary parts, we get

${\alpha }^2+p\alpha +r=0$ ..... (ii)

and $q\alpha +s=0\Rightarrow \alpha =\frac{-s}{q}$

On putting the value of $\alpha$ in Eq. (ii), we get

${\left(\frac{-s}{q}\right)}^2+\left(\frac{-s}{q}\right)p+r=0$

$\Rightarrow s^2-pqs+q^{2}r=0$

$\Rightarrow pqs=s^2+q^{2}r$.