Question

Let $p, q, r, s \in R$ (the set of all real numbers) then the number of real roots of the equation
$\left(x^2+p x-3 q\right)\left(x^2-r x+q\right)\left(x^2-s x+2 q\right)=0 \text { are }$

• A. exactly two
• B. atleast two
• C. atmost four
• D. exactly four

Answer 1

Answer: (B)

Clearly, the discriminants of the given quadratic equations are
$D_1=p^2+12 q, D_2=r^2-4 q \text { and } D_3=s^2-8 q $
$\therefore D_1+D_2+D_3=p^2+r^2+s^2 \geq 0(\text { As, p, q, r, s } \in R)$
$\Rightarrow $ Atleast one of $D _1, D _2, D _3$ is non-negative.
Hence, the equation has atleast two real roots.