If the roots of $x^{3} + p x^{2} + q x + r = 0$ are in H.P., then

2 q^{3} - 27 r^{2} = 9 p q r

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

2 q^{3} + 27 r^{2} = 9 p q r

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

2 q^{3} + 27 r^{2} = - 9 p q r

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

2 q^{3} - 27 r^{2} = - 9 p q r

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

Open in App

Solution

The correct option is B $2 q^{3} + 27 r^{2} = 9 p q r$
Let $α, β, γ$ be the roots of $x^{3} + p x^{2} + q x + r = 0$
$α + β + γ = - p α β + β γ + γ α = q α β γ = - r$
Mean of H.P is
$β = \frac{3 α β γ}{α β + β γ + γ α} = \frac{- 3 r}{q}$
Substituting this in equation we get
${(\frac{- 3 r}{q})}^{3} + p {(\frac{- 3 r}{q})}^{2} + q (\frac{- 3 r}{q}) - r = 0 \Rightarrow 2 q^{3} + 27 r^{2} = 9 p q r$

Suggest Corrections

Similar questions

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

2 q^{3} = r (3 p q - r)

9 p q r ({2 p}^{2} q - {3 q}^{2} r)

1 - p^{3}

2 p^{3} - 3 p q^{2} - 2 q^{3}

2 \sqrt{2} q^{4} + 4 \sqrt{2} q^{3} + q^{2} by (- 2 \sqrt{2} q^{2})

Join BYJU'S Learning Program