If $α$ and $β$ be two real roots of the equation $x^{3} + p x^{2} + q x + r = 0$ satisfying the relation $α β + 1 = 0$ , then prove that $r^{2} + p r + q + 1 = 0$ .

Open in App

Solution

$α β + 1 = 0 α β = - 1$
$α β γ = - r$
$(+ 1) γ = - r$
$γ = r$
$∴ r^{3} + p r^{2} + q r + r = 0$
$r (r^{2} + p r + q + 1) = 0$
$∴ r^{2} + p r + q + 1 = 0$
Hence, proved.

Suggest Corrections

Similar questions

α, β, γ

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

α β + 1 = 0

α, β, γ

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

(α - \frac{1}{β γ}) (β - \frac{1}{γ α}) (γ - \frac{1}{α β})

α, β, γ

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

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

x^{3} + a x^{2} + b x + c = 0, c \neq 0,

α, β, γ

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

α (β + γ), β (γ + α), γ (α + β)

If α,β,γ are the roots of $x 3$ +p $x^{2}$ +qx+r=0 , then (α- $\frac{1}{β γ}$ ) (β – $\frac{1}{γ α}$ ) (γ – $\frac{1}{α β}$ ) =

Join BYJU'S Learning Program