Question

In the quadratic equation $x^2+(p+iq)x+3i=0$ , p & q are real. If the sum of the squares of the roots is 8 then,values of $p$ and $q$ are

• A. $p=2,q=2$
• B. $p=-3,q=4$
• C. $p=3,q=1$ or $p=-3,q=-1$
• D. $p=-2,q=-2$

Answer 1

Answer: (C)

$p=3,q=1$ or $p=-3,q=-1$

$x^2+(p+iq)x+3i=0$
$\alpha +\beta =-(p-iq),\alpha \beta =3i$
${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta =[-(p+iq){]}^2-6i$
$=(p^2-q^2)+i(2pq-6)=8$
$\Rightarrow p^2-q^2=8$ and $pq=3$
$\Rightarrow p=3,q=1$ or $p=-3,q=-1$