Let p, q, r, s ∈ R (the set of all real numbers) then the number of real roots of the equation (x2+p x-3 q)(x2-r x+q)(x2-s x+2 q)=0 text are

Question

Let p, q, r, s ∈ R (the set of all real numbers) then the number of real roots of the equation (x2+p x-3 q)(x2-r x+q)(x2-s x+2 q)=0 text are  (A) exactly two (B) atleast two (C) atmost four (D) exactly four

Tardigrade Admin · Accepted Answer

Clearly, the discriminants of the given quadratic equations are

D_{1} = p^{2} + 12 q, D_{2} = r^{2} - 4 q and D_{3} = s^{2} - 8 q

∴ D_{1} + D_{2} + D_{3} = p^{2} + r^{2} + s^{2} \geq 0 (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.

Q. Let p,q,r,s∈R (the set of all real numbers) then the number of real roots of the equation (x2+px−3q)(x2−rx+q)(x2−sx+2q)=0 are

exactly two

atleast two

atmost four

exactly four

Clearly, the discriminants of the given quadratic equations are D1​=p2+12q,D2​=r2−4q and D3​=s2−8q ∴D1​+D2​+D3​=p2+r2+s2≥0( As, p, q, r, s ∈R) ⇒ Atleast one of D1​,D2​,D3​ is non-negative. Hence, the equation has atleast two real roots.

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