Find the roots of $x^{3} - 9 x^{2} + 24 x - 16 = 0$

1, 4, 4

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

1, 2, 6

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

2, 2, 8

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

3, 1, 9

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

Open in App

Solution

The correct option is A $1, 4, 4$
The given equation can be factorise as:
$x^{3} - 9 x^{2} + 24 x - 16 = (x - 1) (x^{2} - 8 x + 16) = (x - 1) (x - 4)^{2}$
Hence roots of given equation are $1, 4, 4$

Suggest Corrections

Similar questions

x^{3} - 9 x^{2} + 24 x + k = 0

(2, 4)

k

k

9 x^{2} - 24 x + k = 0

9 x^{2}

k = 16

{9 x}^{2} + 24 x + 16

zeroes of $9 x^{2} - 24 x + 16$

Join BYJU'S Learning Program