If $α, β, γ$ are the roots of $4 x^{3} - 7 x^{2} + 2 x - 6 = 0$ , then the equation whose roots are $\frac{α}{2}$ , $\frac{β}{2}$ , $\frac{γ}{2}$ , is:

32 x^{3} - 28 x^{2} + 4 x + 6 = 0

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

32 x^{3} - 28 x^{2} + 4 x - 6 = 0

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

32 x^{3} - 28 x^{2} - 4 x - 6 = 0

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

32 x^{3} - 28 x^{2} - 4 x + 6 = 0

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

Open in App

Solution

The correct option is B $32 x^{3} - 28 x^{2} + 4 x - 6 = 0$
$4 x^{3} - 7 x^{2} + 2 x - 6 = 0$
$α + β + γ = \frac{7}{4}$
$α β + β γ + γ α = \frac{1}{2}$
$α β γ = \frac{3}{2}$
For a new equation,
The roots are $\frac{α}{2}, \frac{β}{2}, \frac{γ}{2}$
The sum of roots: $\frac{α + β + γ}{2} = \frac{7}{8}$
The product of roots: $\frac{α β γ}{8} = \frac{3}{16}$
The product of the roots taken two at a time : $\frac{α β + β γ + γ α}{4} = \frac{1}{8}$
$∴$ the equation is: $x^{3} - \frac{7}{8} x^{2} + \frac{x}{8} - \frac{3}{16} = 0$
$\Rightarrow 32 x^{3} - 28 x^{2} + 4 x - 6 = 0$

Suggest Corrections

Similar questions

x^{6} - 18 x^{4} + 16 x^{3} + 28 x^{2} - 32 x + 8 = 0

\sqrt{6} - 2

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} + 4 x + 1 = 0

(α + β)^{- 1} + (β + γ)^{- 1} + (γ + α)^{- 1}

α, β, γ

x^{3} - 7 x + 6 = 0

(α - β)^{2}, (β - γ)^{2}, (γ - α)^{2}

α, β, γ

x^{3} - 3 x^{2} + 4 x - 7 = 0

(α + 2) (β + 2) (γ + 2) =

Join BYJU'S Learning Program