Question

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

• A. $p=3$, $q=-1$
• B. $p=-3$, $q=-1$
• C. $p=\pm 3$, $q=\pm 1$
• D. $p=-3$, $q=1$

Answer 1

The correct option is C $p=\pm 3$, $q=\pm 1$

Let roots be $\alpha ,\beta$

$\alpha +\beta =-(p+iq)$

$\alpha \beta =3i$

${\alpha }^2+{\beta }^2=8$

$(\alpha +\beta {)}^2-2\alpha \beta =8$

$(p+iq{)}^2-6i=8$

$p^2-q^2+2pqi-6i=8$

$(p^2-q^2)+i(2pq-6)=8$

$p^2-q^2=8$, $2pq-6=0$

$p^2-q^2=8$, $pq=3$

$q=\frac{3}{p}$

$\Rightarrow p^2-\frac{q}{p^2}=8$

$p^4-8p^2-q=0$

$p^4-9p^2+p^2-9=0$

$p^2(p^2-9)+1(p^2-9)=0$

$p=\pm 3$

$q=\pm 1$.