If α, β and γ are three consecutive terms of a non-constant G.P. such that the equations α x2 + 2β x + γ = 0 and x2 + x - 1 = 0 have a common root, then α(β + γ) is equal to :

Question

If α, β and &gamma; are three consecutive terms of a non-constant G.P. such that the equations α x2 + 2β x + &gamma; = 0 and x2 + x - 1 = 0 have a common root, then α(β + &gamma;) is equal to : (A) β &gamma; (B) 0 (C) α &gamma; (D) α β

Tardigrade Admin · Accepted Answer

α x2 + 2β x + &gamma; = 0 Let β = α t, &gamma; = α t2 ∴ α x2 + 2tx + t2 = 0 ⇒ x2 + 2tx + t2 = 0 ⇒ (x + t)2 = 0 ⇒ x = -t it must be root of equation x2 + x - 1 = 0 ∴ t2 - t - 1 = 0 Now α (β + &gamma;) = α2(t + t2) Option 1 β &gamma; = α t . α t2 = α2 t3 = a2 (t2 + t) (from equation 1)

Q. If $α, β$ and $γ$ are three consecutive terms of a non-constant G.P. such that the equations $α x^{2} + 2 β x + γ = 0$ and $x^{2} + x - 1 = 0$ have a common root, then $α (β + γ)$ is equal to :

$β γ$

$0$

$α γ$

$α β$

Q. If α,β and γ are three consecutive terms of a non-constant G.P. such that the equations αx2+2βx+γ=0 and x2+x−1=0 have a common root, then α(β+γ) is equal to :

βγ

0

αγ

αβ

αx2+2βx+γ=0 Letβ=αt,γ=αt2 ∴αx2+2tx+t2=0 ⇒x2+2tx+t2=0 ⇒(x+t)2=0 ⇒x=−t it must be root of equation x2+ x - 1 = 0 ∴t2−t−1=0 Now α(β+γ)=α2(t+t2) Option 1 βγ=αt.αt2=α2t3=a2(t2+t) (from equation 1)

Q. If $α, β$ and $γ$ are three consecutive terms of a non-constant G.P. such that the equations $α x^{2} + 2 β x + γ = 0$ and $x^{2} + x - 1 = 0$ have a common root, then $α (β + γ)$ is equal to :

$β γ$

$0$

$α γ$

$α β$