If two roots of $x^{3} - 9 x^{2} + 14 x + 24 = 0$ are in the ratio 3 : 2 then the roots are

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

6, 4, –1

$\frac{1}{2}, \frac{1}{3}, \frac{49}{6}$

3, 2, 4

6, 4, 2

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} L e t 3 α, 2 α, β a r e r o o t s o f x^{3} - 9 x^{2} + 14 x + 24 = 0 \\ ∴ 3 α + 2 α + β = 9 \Rightarrow 5 α + β = 9 - - (1) \\ ∴ 3 α (2 α) + 2 α (β) + 3 α (β) = 14 \\ \Rightarrow 6 α^{2} + 5 α β = 14 - - (2) \\ ∴ 3 α (2 α) β = - 24 \Rightarrow 6 α^{2} β = - 24 \Rightarrow α^{2} β = - 4 - - (3) \\ f r o m (1) β = 9 - 5 α p u t i n (2) \\ 6 α^{2} + 5 α (9 - 5 α) = 14 \\ \Rightarrow 6 α^{2} + 45 α - 25 α^{2} = 14 \\ \Rightarrow - 19 α^{2} + 45 α = 14 \\ \Rightarrow 19 α^{2} - 45 α + 14 = 0 \\ \Rightarrow 19 α^{2} - 38 α - 7 α + 14 = 0 \\ \Rightarrow 19 α (α - 2) - 7 (α - 2) = 0 \\ \Rightarrow (α - 2) (19 α - 7) = 0 \\ \Rightarrow α = 2, α = \frac{7}{19} \\ I f α = 2 t h e n β = 9 - 10 = - 1 \\ ∴ T h e r o o t s a r e 6, 4, - 1 \end{array}$