Let cos (α-β)+ cos (β-γ)+ cos (α-γ)=3, where α, β, γ ∈[0, π]. If α, β, γ are roots of x 3+ px 2+ qx + r =0 then which of the following must be true?

Question

Let cos (α-β)+ cos (β-&gamma;)+ cos (α-&gamma;)=3, where α, β, &gamma; ∈[0, π]. If α, β, &gamma; are roots of x 3+ px 2+ qx + r =0 then which of the following must be true? (A) pq =9 r 2 (B) pqr =9 (C) pq =9 r (D) q=9 r p

Tardigrade Admin · Accepted Answer

cos (α-β)+ cos (β-&gamma;)+ cos (α-&gamma;)=3 ⇒ α-β=0=β-&gamma;=α-&gamma; ⇒ α=β=&gamma; For cubic equation, x 3+ px 2+ qx + r ⇒ 3 α=- p ....(1) ⇒ 3 α2= q ....(2) ⇒ α3=- r ....(3) text From (3), α ⋅ α2=- r ⇒ (- p /3) ⋅ ( q /3)=- r ⇒ pq =9 r

Q. Let $cos (α - β) + cos (β - γ) + cos (α - γ) = 3$ , where $α, β, γ \in [0, π]$ . If $α, β, γ$ are roots of $x^{3} + p x^{2} + q x + r = 0$ then which of the following must be true?

$pq = 9 r^{2}$

$pq r = 9$

$pq = 9 r$

$q = 9 r p$

Q. Let cos(α−β)+cos(β−γ)+cos(α−γ)=3, where α,β,γ∈[0,π]. If α,β,γ are roots of x3+px2+qx+r=0 then which of the following must be true?

pq=9r2

pqr=9

pq=9r

q=9rp

cos(α−β)+cos(β−γ)+cos(α−γ)=3 ⇒α−β=0=β−γ=α−γ ⇒α=β=γ For cubic equation, x3+px2+qx+r ⇒3α=−p ....(1) ⇒3α2=q....(2) ⇒α3=−r....(3) From (3), α⋅α2=−r⇒3−p​⋅3q​=−r ⇒pq=9r

Q. Let $cos (α - β) + cos (β - γ) + cos (α - γ) = 3$ , where $α, β, γ \in [0, π]$ . If $α, β, γ$ are roots of $x^{3} + p x^{2} + q x + r = 0$ then which of the following must be true?

$pq = 9 r^{2}$

$pq r = 9$

$pq = 9 r$

$q = 9 r p$