Question

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

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

Answer 1

The correct option is C $p=3,q=1$ or $p=-3,q=-1$

Let the roots be $\alpha ,\beta$. Using sum and product of roots formula:
$\alpha +\beta =-(p+iq)$
$\alpha \beta =3i$
Given: ${\alpha }^2+{\beta }^2=8$
${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta$
$⟹8=(p+iq{)}^2-2(3i)$
$⟹8=p^2-q^2+(2pq-6)i$
Comparing real and imaginary parts on both sides:
$p^2-q^2=8$ $pq=3$
$⟹p^2-9/p^2=8$
$⟹p^4-8p^2-9=0$
$⟹(p^2-9)(p^2+1)=0$
As $p\in R$, $p^2+1\ne 0$
$implies{p}^2-9=0⟹p=\pm 3$
$\because q=3/p$:
If $p=3$, $q=1$
If $p=-3$, $q=-1$