:: α , β and γ are roots of the following equation . Find the value of ” F ” : Equation : x^( 3) −2x −1=0 F := α^( 5) + β^( 5) + γ^( 5)


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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

	$t h a n k y o u s o m u c h . . .$
	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

	$x^{3} = 2 x + 1 \Rightarrow x^{4} = 2 x^{2} + x \Rightarrow x^{5} = 2 x^{3} + x^{2} =$ $2 (2 x + 1) + x^{2} = x^{2} + 4 x + 2$ $α + β + γ = 0 \Rightarrow Σ α^{2} = {(Σ α)}^{2} - 2 (Σ α β) =$ $= 0 - 2 (- 2) = 4$ $\Rightarrow Σ α^{5} = Σ α^{2} + 4 Σ α + 3 = 4 + 4 (0) + 6 = 10 . ◼$
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