Question

If $p,q,r,s\in R,$ then the equation $(x^2+px-3q)(x^2-rx+q)(x^2-sx+2q)=0$ has

• A. exactly six real roots
• B. at least three real roots
• C. at least four real roots
• D. at least two real roots

Answer 1

Answer: (D)

at least two real roots

Discriminant of $(x^2+px-3q)=0$ is $D_1=p^2+12q$
Discriminant of $(x^2-rx+q)=0$ is $D_2=r^2-4q$
And discriminant of $(x^2-sx+2q)=0$ is $D_3=s^2-8q$
Now $D_1+D_2+D_3=p^2+r^2+s^2\ge 0\forall p,q,r,s\in R$
Thus at least one of $D_1,D_2,D_3$ is +ve
Hence equation $(x^2+px-3q)(x^2-rx+q)(x^2-sx+2q)=0$ will have atleast two real roots