If one root of $x^{3} - 12 x^{2} + k x - 18 = 0$ is thrice the sum fo the remaining two roots then $k =$

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

-29

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

Open in App

Solution

The correct option is A 29
let roots are $a, b, 3 (a + b)$
sum of roots $- (- 12)$
$4 (a + b) = 12$
$a + b = 3$
product $= - (- 18)$
$3 a b (a + b) = + 18$
$a b = + 2$
sum of two at $= k$
$a b + 3 a (a + b) + 3 b (a + b) = k$
$k = 2 + 9 a + 9 b$
$k = 9 \times 3 + 2$
$k = 29$

Suggest Corrections

Similar questions

x^{3} - 3 x^{2} + k x + 48 = 0

k =

x^{3} - 12 x^{2} + 39 x + k = 0

k =

f (x) = x^{3} - 12 x^{2} + 34 x - k

f (x) = 0

1, 3, - 4

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

k =

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

Join BYJU'S Learning Program