Question

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

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

Answer 1

The correct option is A $pqs=s^2+q^{2}r$

Putting $z=x+iy$, we get

${(x+iy)}^2+(p+iq)(x+iy)+r+is=0$

$\Rightarrow (x^2-y^2+px-qy+r)+i(2xy+py+qx+s)=0$

$\Rightarrow x^2-y^2+px-qy+r=0$ ...(1)

and $2xy+py+qx+s=0$ ...(2)

If the roots are real, then $y=0$

Therefore (1) gives $x^2+px+r=0$ ....(3)

and (2) gives $qx+s=0$ ...(4)

From (4), $x=-\frac{s}{q}$

Putting in (3), we get $\frac{s^2}{q^2}-\frac{ps}{q}+r=0$

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

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