If $α, β, γ$ be the roots of $a x^{3} + b x^{2} + c x + d = 0$ , then find the value of :
$\sum α^{2}$
$\sum \frac{1}{α}$
$\sum α^{2} (β + γ)$

Open in App

Solution

$a x^{3} + b x^{2} + c x + d = 0$
$α + β + γ = \frac{- b}{a}, α β γ = \frac{- d}{a}, α β + β γ + γ α = \frac{c}{a}$

(i) $\sum α^{2} = α^{2} + β^{2} + γ^{2} = (α + β + γ)^{2} - 2 (α β + β γ + γ α)$
$= {(\frac{- b}{a})}^{2} - 2 (\frac{c}{a})$
$\sum α^{2} \Rightarrow \frac{b^{2} - 2 a c}{a^{2}}$

(ii) $\sum \frac{1}{α} = \frac{1}{α} + \frac{1}{β} + \frac{1}{γ} = \frac{β γ + γ α + α β)}{α β γ}$
$\sum \frac{1}{α} = \frac{\frac{c}{a}}{- \frac{d}{c}} = - \frac{c}{d}$

(iii) $\sum α^{2} (β + γ) \Rightarrow α^{2} (β + γ) + β^{2} (γ + α) + γ^{2} (α + β)$

$\Rightarrow α^{2} β + α^{2} γ + β^{2} γ + β^{2} α + γ^{2} α + γ^{2} β$

$\Rightarrow α β (α + β) + β γ (β + γ) + γ α (γ + α)$

$\Rightarrow α β γ [\frac{(α + β}{γ} + \frac{(β + γ}{α} + \frac{γ + α}{β}]$

$\Rightarrow α β γ [\frac{α + β + γ}{γ} - \frac{γ}{γ} + \frac{α + β + γ}{α} - \frac{α}{α} + \frac{α + β + γ}{β} - \frac{β}{β}]$

$\Rightarrow α β γ (α + β + γ) [\frac{1}{α} + \frac{1}{β} + \frac{1}{γ} - 3]$

$\sum x^{2} (β + γ) \Rightarrow \frac{- d}{a} [\frac{b c}{q d} - 3] \Rightarrow \frac{3 a d - b c}{a^{2}}$

Suggest Corrections

Similar questions

α, β, γ

a x^{3} + b x^{2} + c x + d = 0

∣ ∣ ∣ ∣ ∣ \begin{matrix} α & β & γ β & γ & α γ & α & β \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0

α \neq β \neq γ

α + β - γ, γ + α - β, β + γ - α

α, β, γ

a x^{3} + b x^{2} + c x + d = 0

α + β + γ = \frac{- b}{a}

α, β, γ

P (x) = a x^{3} + b x^{2} + c x + d (a \neq 0)

\frac{1}{α} + \frac{1}{β} + \frac{1}{γ} =

If $α, β$ and $γ$ are the zeroes of the polynomial $f (x) = a x^{3} + b x^{2} + c x + d$ , then $\frac{1}{α} + \frac{1}{β} + \frac{1}{γ}$ is _____.

α, β, γ

a x^{3} + b x^{2} + c x + d

α β γ =

Join BYJU'S Learning Program