If $1, 3, - 4$ are the roots of $x^{3} + k x + 12 = 0$ , then $k =$

11

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

- 11

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

13

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

- 13

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

Open in App

Solution

The correct option is C $- 13$
Given $1, 3, - 4$ are roots of $x^{3} + k x - 12 = 0$
$S_{2} = 1 \cdot 3 + 1 \cdot (- 4) + 3 \cdot (- 4) = k \Rightarrow k = - 13$

Suggest Corrections

Similar questions

α, β, γ

x^{3} - 3 x + 11 = 0

(α + β), (β + γ)

(γ + α)

⎡ ⎢ ⎣ \begin{matrix} cos θ - sin θ 0 \end{matrix} \begin{matrix} sin θ cos θ 0 \end{matrix} \begin{matrix} 001 \end{matrix} ⎤ ⎥ ⎦

a_{11} A_{11} + a_{12} A_{12} + a_{13} A_{13} =

A_{11}, A_{12} A_{13}

a_{11}, a_{12}, a_{13}

If the roots of the equation $4 x^{3} - 12 x^{2} + 11 x + k = 0$ are in A.P. Then $k =$

3 x^{2} + (k - 1) x + 9 = 0,

Join BYJU'S Learning Program