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Question Number 10742 by okhema last updated on 24/Feb/17

l e t t h e r o o t s o f t h e e q u a t i o n 2 x^{3} - 5 x^{2} + 4 x + 6 = 0

b e α, β a n d γ .

i) s t a t e t h e v a l u e s o f α + β + γ, α β + α γ + β γ a n d α β γ .

i i) h e n c e o r o t h e r w i s e d e t e r m i n e a n e q u a t i o n w i t h i n t e g e r c o e f f i c i e n t s w h i c h a s r o o t s \frac{1}{α^{2}}, \frac{1}{β^{2}}, a n d \frac{1}{γ^{2}}

Answered by sandy_suhendra last updated on 24/Feb/17

(i) a = 2, b = - 5, c = 4 and d = 6

α + β + γ = \frac{- b}{a} = \frac{5}{2}

α β + α γ + β γ = \frac{c}{a} = 2

α β γ = \frac{- d}{a} = - 3

Answered by sandy_suhendra last updated on 24/Feb/17

ii)

let x_{1} = \frac{1}{α^{2}}, x_{2} = \frac{1}{β^{2}}, x_{3} = \frac{1}{γ^{2}}

x_{1} + x_{2} + x_{3} = \frac{1}{α^{2}} + \frac{1}{β^{2}} + \frac{1}{γ^{2}}

= \frac{β^{2} γ^{2} + α^{2} γ^{2} + α^{2} β^{2}}{{(α β γ)}^{2_{}}}

= \frac{{(α β + α γ + β γ)}^{2} - 2 α β γ (α + β + γ)}{{(α β γ)}^{2}}

= \frac{2^{2} - 2 (- 3) (2.5)}{{(- 3)}^{2}}

= \frac{19}{9} = \frac{- b}{a}

x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = \frac{1}{α^{2} β^{2}} + \frac{1}{α^{2} γ^{2}} + \frac{1}{β^{2} γ^{2}}

= \frac{γ^{2} + β^{2} + α^{2}}{α^{2} β^{2} γ^{2}}

= \frac{{(α + β + γ)}^{2} - 2 (α β + α γ + β γ)}{{(α β γ)}^{2}}

= \frac{{(2.5)}^{2} - 2 (2)}{{(- 3)}^{2}}

= \frac{1}{4} = \frac{c}{a}

x_{1} x_{2} x_{3} = \frac{1}{α^{2}} \times \frac{1}{β^{2}} \times \frac{1}{γ^{2}} = \frac{1}{{(α β γ)}^{2}}

= \frac{1}{{(- 3)}^{2}} = \frac{1}{9} = \frac{- d}{a}

the new equation :

x^{3} - (\frac{- b}{a}) x^{2} + (\frac{c}{a}) x - (\frac{- d}{a}) = 0

x^{3} - \frac{19}{9} x^{2} + \frac{1}{4} x - \frac{1}{9} = 0 \dots . (\times 36)

{36 x}^{3} - {76 x}^{2} + 9 x - 4 = 0