Q-are-the-roots-of-the-following-equation-find-the-value-of-Eq-n-x-3-2x-2-x-2-0-E-

Question Number 199109 by mnjuly1970 last updated on 28/Oct/23

Q : α, β, γ a r e t h e r o o t s o f t h e f o l l o w i n g

e q u a t i o n . f i n d t h e v a l u e o f :

{E q}^{n} : x^{3} - 2 x^{2} + x + 2 = 0

E = \frac{α}{β + γ} + \frac{β}{α + γ} + \frac{γ}{α + β}

Commented by cortano12 last updated on 28/Oct/23

x^{3} - {2 x}^{2} + x + 2 = 0 {\begin{cases} α \\ β \\ γ \end{cases}

then {4 x}^{3} - {5 x}^{2} + 4 x - 1 = 0 {\begin{cases} \frac{1}{2 - α} \\ \frac{1}{2 - β} \\ \frac{1}{2 - γ} \end{cases}

by {Vieta}^{'} s

\frac{1}{2 - α} + \frac{1}{2 - β} + \frac{1}{2 - γ} = \frac{5}{4}

Answered by cortano12 last updated on 28/Oct/23

E = 2 (\frac{1}{2 - α} + \frac{1}{2 - β} + \frac{1}{2 - γ}) - 3

E = 2 (\frac{5}{4}) - 3 = \frac{5}{2} - 3 = - \frac{1}{2}

Answered by Rasheed.Sindhi last updated on 28/Oct/23

α, β, γ a r e r o o t s o f x^{3} - 2 x^{2} + x + 2 = 0

E = \frac{α}{β + γ} + \frac{β}{α + γ} + \frac{γ}{α + β} = ?

α + β + γ = - (- 2) = 2

α β + β γ + γ α = 1

α β γ = - 2

E + 3 = \frac{α}{β + γ} + 1 + \frac{β}{α + γ} + 1 + \frac{γ}{α + β} + 1

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

= (α + β + γ) (\frac{1}{β + γ} + \frac{1}{α + γ} + \frac{1}{α + β})

= (α + β + γ) (\frac{1}{α + β + γ - α} + \frac{1}{α + β + γ - β} + \frac{1}{α + β + γ - γ})

= (2) (\frac{1}{2 - α} + \frac{1}{2 - β} + \frac{1}{2 - γ})

\frac{E + 3}{2} = \frac{(2 - α) (2 - β) + (2 - β) (2 - γ) + (2 - γ) (2 - α)}{(2 - α) (2 - β) (2 - γ)}

= \frac{- 4 (α + β + γ) + (α β + β γ + γ α) + 12}{8 - α β γ - 4 (α + β + γ) + 2 (α β + β γ + γ α)}

= \frac{- 4 (2) + (1) + 12}{8 - (- 2) - 4 (2) + 2 (1)} = \frac{5}{4}

E = \frac{5}{4} \times 2 - 3 = - \frac{1}{2} ✓

Commented by mnjuly1970 last updated on 30/Oct/23

t h a n k s a l o t s i r ?