Two roots of the cubic $x^{3} + 3 x^{2} + k x - 12 = 0$ are real and unequal but have the same absolute value. The value of $k$ is

4

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

- 4

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

6

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

- 9

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

Open in App

Solution

The correct option is B $- 4$
Let the roots are $r, - r$ and $t$
Hence, $r - r + t = - 3$ $\Rightarrow$ $t = - 3$
Now product of the roots is $(r) (- r) (t) = 12$
$- r^{2} (- 3) = 12 \Rightarrow r^{2} = 4$
Hence roots are $2, - 2$ and $- 3$
Now, sum of the product taken two at a time is $r (- r) + (- r) t + t r = k$
$- 4 - 6 + 6 = k$ $\Rightarrow$ $k = - 4$

Suggest Corrections

Similar questions

P (x) = x^{3} - 3 x^{2} + k x + 4

A P,

| k |

Determine the value of k so that the two roots of the equation x² − kx + 36 = 0 are equal.

4 x^{2}

k = \pm 12

1, 3, - 4

x^{3} + k x + 12 = 0

k =

k

k x^{2} - 2 \sqrt{5} x + 4 = 0

Join BYJU'S Learning Program