and-are-roots-of-the-following-equation-Find-the-value-of-F-Equation-x-3-2x-1-0-F-5-5-

Question Number 209187 by mnjuly1970 last updated on 03/Jul/24

:: α, β a n d γ a r 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^{″} F^{″} :

E q u a t i o n : x^{3} - 2 x - 1 = 0

F := α^{5} + β^{5} + γ^{5}

Answered by A5T last updated on 03/Jul/24

α + β + γ = 0; α β + β γ + γ α = - 2; α β γ = 1

x^{3} = 2 x + 1 \Rightarrow x^{5} = 2 x^{3} + x^{2} = 2 (2 x + 1) + x^{2} = x^{2} + 4 x + 2

\Rightarrow α^{5} + β^{5} + γ^{5} = α^{2} + β^{2} + γ^{2} + 4 (α + β + γ) + 6

= {(α + β + γ)}^{2} - 2 (α β + β γ + γ α) + 4 (0) + 6 = 10

Commented by mnjuly1970 last updated on 03/Jul/24

t h a n k y o u s o m u c h \dots

Answered by lepuissantcedricjunior last updated on 05/Jul/24

x^{3} - 2 x - 1 = (x + 1) (x^{2} - x - 1)

= (x + 1) (x - \frac{1 - \sqrt{5}}{2}) (x - \frac{1 + \sqrt{5}}{2})

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

F : {(- 1)}^{5} + {(\frac{1 - \sqrt{5}}{2})}^{5} + {(\frac{1 + \sqrt{5}}{2})}^{5}

= - 1 + (\frac{1 - \sqrt{5}}{2}) {(\frac{6 - 2 \sqrt{5}}{4})}^{2} + (\frac{1 + \sqrt{5}}{2}) {(\frac{6 + 2 \sqrt{5}}{4})}^{2}

= - 1 + (\frac{1 - \sqrt{5}}{2}) (\frac{56 - 24 \sqrt{5}}{16}) + (\frac{1 + \sqrt{5}}{2}) (\frac{56 + 24 \sqrt{5}}{16})

= - 1 + (\frac{56 + 120 - 80 \sqrt{5}}{32}) + (\frac{56 + 120 + 80 \sqrt{5}}{32})

= - 1 + \frac{176 \times 2}{32} = - 1 + \frac{352}{32}

= - 1 + 11 = 10

=> F : α^{5} + β^{5} + γ^{5} = 10

Commented by Spillover last updated on 06/Jul/24

g o o d

Answered by behi834171 last updated on 09/Jul/24

x^{3} = 2 x + 1 \Rightarrow x^{4} = 2 x^{2} + x \Rightarrow x^{5} = 2 x^{3} + x^{2} =

2 (2 x + 1) + x^{2} = x^{2} + 4 x + 2

α + β + γ = 0 \Rightarrow Σ α^{2} = {(Σ α)}^{2} - 2 (Σ α β) =

= 0 - 2 (- 2) = 4

\Rightarrow Σ α^{5} = Σ α^{2} + 4 Σ α + 3 = 4 + 4 (0) + 6 = 10 .