If p, q, r, s ∈ R, then equation (x2+p x+3 q)(-x2+r x+q)(-x2+S x-2 q)=0 has

Question

If p, q, r, s ∈ R, then equation (x2+p x+3 q)(-x2+r x+q)(-x2+S x-2 q)=0 has (A) 6 real roots. (B) at least two real roots. (C) 2 real and 4 imaginary roots. (D) 4 real and 2 imaginary roots.

Tardigrade Admin · Accepted Answer

D1=p2-12 q D2=r2+4 q D3=s2-8 q Case-I: q <0 ⇒ D 1>0, D 3>0 But D 2 not defined. Case-II: q =0 ⇒ D 1>0, D 2>0, D 3>02 Case-III: q >0 ⇒ D 2>0 but D 1 and D 3 not defined. Hence, at least two real roots.

Q. If p,q,r,s∈R, then equation (x2+px+3q)(−x2+rx+q)(−x2+Sx−2q)=0 has

6 real roots.

at least two real roots.

2 real and 4 imaginary roots.

4 real and 2 imaginary roots.

D1​=p2−12q D2​=r2+4q D3​=s2−8q Case-I : q<0⇒D1​>0,D3​>0 But D2​ not defined. Case-II : q=0⇒D1​>0,D2​>0,D3​>02 Case-III : q>0⇒D2​>0 but D1​ and D3​ not defined. Hence, at least two real roots.

Q. If $p, q, r, s \in R$ , then equation $(x^{2} + p x + 3 q) (- x^{2} + r x + q) (- x^{2} + S x - 2 q) = 0$ has