If $α, β, γ$ are the roots of $x^{3} + 3 x^{2} + 2 = 0$ then find the equation whose roots are $\frac{α}{β + γ}, \frac{β}{γ + α}, \frac{γ}{α + β}$

2 x^{3} + 33 x^{2} - 6 x - 2 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 x^{3} + 33 x^{2} + 6 x - 2 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 x^{3} + 33 x^{2} - 6 x + 2 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 x^{3} + 33 x^{2} + 6 x + 2 = 0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $2 x^{3} + 33 x^{2} + 6 x + 2 = 0$
Given $α + β + γ = - 3, α β + β γ + γ α = 0, α β γ = - 2$
Let
$y = \frac{α}{β + γ} = \frac{α}{(α + β + γ) - α} = \frac{α}{- 3 - α}$
$\Rightarrow α = \frac{- 3 y}{y + 1}$
which is a root of the given equation.
$∴ {(\frac{- 3 y}{y + 1})}^{3} + 3 {(\frac{- 3 y}{y + 1})}^{2} + 2 = 0$
$\Rightarrow - 27 y^{3} + 27 y^{2} (y + 1) + 2 (y + 1)^{3} = 0$
$\Rightarrow - 27 y^{3} + 27 y^{3} + 27 y^{2} + 2 y^{3} + 6 y^{2} + 6 y = 0$
$\Rightarrow 2 y^{3} + 33 y^{2} + 6 y + 2 = 0$
$∴$ The required equation is $2 x^{3} + 33 x^{2} + 6 x + 2 = 0$

Suggest Corrections

Similar questions

α, β, γ

x^{3} + 3 x^{2} + 2 = 0

\frac{α}{β + γ}, \frac{β}{γ + α}, \frac{γ}{α + β}

α, β, γ

x^{3} - {p x}^{2} + q x - r = 0

(β + γ - α), (γ + α - β), (α + β - γ)

α, β, γ

x^{3} + 2 x - 3 = 0

\frac{α}{β} + \frac{β}{α}, \frac{β}{γ} + \frac{γ}{β}, \frac{γ}{α} + \frac{α}{γ}

x =

If α, β, γ are the roots of the equation $x^{3} + 4 x + 1 = 0$ ,then $(α + β)^{- 1} + (β + γ)^{- 1} + (γ + α)^{- 1} =$

α, β, γ

x^{3} + x^{2} + 2 x + 3 = 0

f (x) = 0

α + β, β + γ, γ + α

f (x) =

Join BYJU'S Learning Program