Let α and β are the roots of equation ax2+bx+c=0 (a ≠ 0). If 1, α +β , α β are in arithmetic progression and α , 2, β are in harmonic progression, then the value of ((α )2 + (β )2 - 2 (α )2 (β )2/2 ((α )2 + (β )2)) is equal to

Question

Let α and β are the roots of equation ax2+bx+c=0 (a ≠ 0). If 1, α +β , α β are in arithmetic progression and α , 2, β are in harmonic progression, then the value of ((α )2 + (β )2 - 2 (α )2 (β )2/2 ((α )2 + (β )2)) is equal to (A) 0 (B) 0.5 (C) 1 (D) 1.5

Tardigrade Admin · Accepted Answer

1, α +β ,α β are in A.P. ⇒ 1,(- b/a),(c/a) are in A.P. ⇒ 1+(c/a)=(- 2 b/a) ⇒ a+c+2b=0 â¦ (1) (1/α ),(1/2),(1/β ) are in A.P. ⇒ (1/α )+(1/β )=1 ⇒ α +β =α β ⇒ (- b/a)=(c/a) ⇒ b+c=0 â¦ (2) From (1) (2) we get, a=-b=c ⇒ α ,β are roots of equation x2-x+1=0 Now, (α2+β2-2 α2 β2/2(α2+β2))=(1/2)-((α β)2/(α+β)2-2 α β) =(1/2)-((1)2/(1)2 - 2 (1))=(1/2)+1=1.5

$0$

$0.5$

$1$

$1.5$

Q. Let α and β are the roots of equation ax2+bx+c=0(a=0). If 1,α+β,αβ are in arithmetic progression and α,2,β are in harmonic progression, then the value of 2((α)2+(β)2)(α)2+(β)2−2(α)2(β)2​ is equal to

0

0.5

1

1.5

$0$

$0.5$

$1$

$1.5$