If α,βα,β be the roots of ax2+bx+c=0,c≠0ax2+bx+c=0,c≠0 α+β=−baα+β=−ba and αβ=caαβ=ca, then
Q: a(α^3+β^3)+b(α^2+β^2)+c(α+β)
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Step-by-step explanation:
α+β,α
2
+β
2
,α
3
+β
3
are in G.P.
∴(α
2
+β
2
)
2
=(α+β)(α
3
+β
3
)
∴α
4
+β
4
+2α
2
β
2
=α
4
+β
4
+α
3
β+αβ
3
∴α
2
β
2
−α
3
β−αβ
3
+α
2
β
2
=0
∴α
2
β(β−α)−αβ
2
(β−α)=0
∴(α
2
β−αβ
2
)(β−α)=0
∴αβ(α−β)(β−α)=0
∴αβ(α−β)
2
=0
∴α=0 or β=0 or α−β=0
Case (1) α=0 or β=0,
⟹x=0 is a solution of given equation
∴a(0)
2
+b(0)+c=0
∴c=0
Case (2) α−β=0
⟹α=β
So, the equation has equal roots.
∴Δ=b
2
−4ac=0
∴c=0 or Δ=0
∴cΔ=0
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