In the quadratic equation x2+(p+i q) x+3 i=0, p q are real. If the sum of the squares of the roots is 8 then :

Question

In the quadratic equation x2+(p+i q) x+3 i=0, p 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=3, q=1 or p=-3, q=-1 (D) p=-3, q=1

Tardigrade Admin · Accepted Answer

x2+(p+i q) x+3 i=0 α+β=-(p+i q), α β=3 i α2+β2=(α+β)2-2 α β=[-(p+i q)]2-6 i =(p2-q2)+i(2 p q-6)=8 ⇒ p2-q2=8 and p q=3 ⇒ p=3, q=1 or p=-3, q=-1

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

$p = 3, q = - 1$

$p = - 3, q = - 1$

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

$p = - 3, q = 1$

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

Q. In the quadratic equation x2+(p+iq)x+3i=0,p&q are real. If the sum of the squares of the roots is 8 then :

p=3,q=−1

p=−3,q=−1

p=3,q=1 or p=−3,q=−1

p=−3,q=1