In the quadratic equation ax 2+ bx + c =0, if Δ= b 2-4 ac and α+β, α2+β2, α3+β3 are in G.P. where α, β are the roots of a x2+b x+c=0, then

Question

In the quadratic equation ax 2+ bx + c =0, if Δ= b 2-4 ac and α+β, α2+β2, α3+β3 are in G.P. where α, β are the roots of a x2+b x+c=0, then (A) Δ ≠ 0 (B) b Δ=0 (C) c Δ=0 (D) Δ=0

Tardigrade Admin · Accepted Answer

(a2+b2)2=(a+b)(a3+b3) α4+β4+2 α2 β2=α4+β4+α β(β2+α2) α β(α2+β2-2 α β)=0 α β[(α+β)2-4 α β]=0 (c/a)((b2/a2)-(4 c/a))=0 c(b2-4 a c)=0 c Δ=0

Q. In the quadratic equation $a x^{2} + b x + c = 0$ , if $Δ = b^{2} - 4 a c$ and $α + β, α^{2} + β^{2}, α^{3} + β^{3}$ are in G.P. where $α, β$ are the roots of $a x^{2} + b x + c = 0$ , then

$Δ \neq = 0$

$b Δ = 0$

$c Δ = 0$

$Δ = 0$

$(a^{2} + b^{2})^{2} = (a + b) (a^{3} + b^{3})$
$α^{4} + β^{4} + 2 α^{2} β^{2} = α^{4} + β^{4} + α β (β^{2} + α^{2})$
$α β (α^{2} + β^{2} - 2 α β) = 0$
$α β [(α + β)^{2} - 4 α β] = 0$
$\frac{c}{a} (\frac{b ^{2}}{a ^{2}} - \frac{4 c}{a}) = 0$
$c (b^{2} - 4 a c) = 0$
$c Δ = 0$

Q. In the quadratic equation ax2+bx+c=0, if Δ=b2−4ac and α+β,α2+β2,α3+β3 are in G.P. where α,β are the roots of ax2+bx+c=0, then

Δ=0

bΔ=0

cΔ=0

Δ=0