The-two-roots-of-an-equation-x-3-9x-2-14x-24-0-are-in-the-ratio-3-2-Find-the-roots-

Question Number 20372 by Tinkutara last updated on 26/Aug/17

The two roots of an equation x^{3} - 9 x^{2}

+ 14 x + 24 = 0 are in the ratio 3 : 2 .

Find the roots .

Answered by $@ty@m last updated on 26/Aug/17

α + 3 β + 2 β = 9 \Rightarrow α + 5 β = 9 - - (1)

3 α β + 6 β^{2} + 2 α β = 14 \Rightarrow 6 β^{2} + 5 α β = 14 - - (2)

6 α β^{2} = - 24 - - (3)

E l i m i n a t i n g α f r o m (1) a n d (2),

6 β^{2} + 5 (9 - 5 β) β = 14

19 β^{2} - 45 β + 14 = 0

β = \frac{45 \pm \sqrt{2025 - 1064}}{38}

β = \frac{45 \pm 31}{38}

β = \frac{76}{38}, \frac{14}{38}

β = 2, \frac{7}{19}

\Rightarrow α = - 1, \frac{136}{19}

t h e f r a c t i o n a l v a l u e s o f α a n d

β d o n o t s a t i s f y e q u a t i o n (3)

∴ α = - 1, β = 2

∴ R o o t s o f t h e g i v e n e q u a t i o n a r e

- 1, 6 a n d 4

Commented by Tinkutara last updated on 26/Aug/17

Thank you very much Sir!

Answered by ajfour last updated on 26/Aug/17

l e t r o o t s b e 3 r, 2 r, γ

5 r + γ = 9; 6 r^{2} + 5 γ r = 14;

6 r^{2} γ = - 24 \Rightarrow r^{2} γ = - 4

5 r^{3} + r^{2} γ = 9 r^{2}

\Rightarrow 5 r^{3} - 9 r^{2} - 4 = 0 \dots (i)

f r o m f i r s t t w o e q u a t i o n s, w e g e t

\Rightarrow 5 r^{2} + (\frac{14 - 6 r^{2}}{5}) = 9 r

19 r^{2} - 45 r + 14 = 0

19 r^{2} - 38 r - 7 r + 14 = 0

(19 r - 7) (r - 2) = 0

r = 2 s a t i s f i e s (i)

i f r = \frac{7}{19}, t h e n

5 {(\frac{7}{19})}^{3} - 9 {(\frac{7}{19})}^{2} - 4 = \frac{{(7)}^{2}}{{(19)}^{3}} [35 - 171]

\neq 0 . S o r \neq \frac{7}{19}

r = 2, α = 3 r = 6, β = 2 r = 4

γ = 9 + 5 r = 9 - 10 = - 1

r o o t s a r e 6, 4, - 1 .

Commented by Tinkutara last updated on 26/Aug/17

Thank you very much Sir!

Commented by $@ty@m last updated on 26/Aug/17

w h y r i s a n i n t e g e r ?

U n l e s s y o u j u s t i f y i t,

3 \times \frac{7}{19} a n d 2 \times \frac{7}{19} a r e a l s o p o s s i b l e r o o t s .

Commented by ajfour last updated on 26/Aug/17

y e s, y o u a r e r i g h t .

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