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Question Number 162533 by mnjuly1970 last updated on 30/Dec/21

	Answered by Ar Brandon last updated on 30/Dec/21

	$x^{2} + 2 x - 2 = 0$ $α + β = - 2$ $α β = - 2$ $\frac{α^{4}}{β} + \frac{β^{4}}{α} = \frac{α^{5} + β^{5}}{α β} = \frac{- 152}{- 2} = 76$ ${(α + β)}^{5} = α^{5} + 5 α^{4} β + 10 α^{3} β^{2} + 10 α^{2} β^{3} + 5 α β^{4} + β^{5}$ $α^{5} + β^{5} = {(α + β)}^{5} - 5 α β (α^{3} + β^{3}) - 10 α^{2} β^{2} (α + β)$ $= {(α + β)}^{5} - 5 α β (α + β) [{(α + β)}^{2} - 3 α β] - 10 α^{2} β^{2} (α + β)$ $= {(- 2)}^{5} - 5 (- 2) (- 2) [{(- 2)}^{2} - 3 (- 2)] - 10 {(- 2)}^{2} (- 2)$ $= - 32 - 20 (10) + 80 = 80 - 32 - 200 = - 152$ $1.1$ $1.2.1$ $1.3.3.1$ $1.4.6.4.1$ $1.5.10.10.5.1$
	Commented by mnjuly1970 last updated on 30/Dec/21

	$v e r y n i c e . t h a n k y o u s o m u c h$ $s i r b r a n d o n$
	Answered by Rasheed.Sindhi last updated on 30/Dec/21

	$x^{2} + 2 x - 2 = 0; a & b a r e r o o t s$ $a + b = - 2,$ $a b = - 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A$ $∙ (a + b) (a + b) = - 2 (- 2)$ $a^{2} + b^{2} = 4 - 2 a b = 4 - 2 (- 2) = 8$ $∙ (a^{2} + b^{2}) (a + b) = 8 (- 2) = - 16$ $a^{3} + b^{3} + a b (a + b) = - 16$ $a^{3} + b^{3} = - 16 - (- 2) (- 2) = - 20$ $∙ (a^{3} + b^{3}) (a + b) = - 20 (- 2) = 40$ $a^{4} + b^{4} + a b (a^{2} + b^{2}) = 40$ $a^{4} + b^{4} + (- 2) (8) = 40$ $a^{4} + b^{4} = 40 + 16 = 56$ $∙ (a^{4} + b^{4}) (a + b) = 56 (- 2) = - 112$ $a^{5} + b^{5} + a b (a^{3} + b^{3}) = - 112$ $a^{5} + b^{5} + (- 2) (- 20) = - 112$ $a^{5} + b^{5} = - 112 - 40 = - 152 . . . . . . . B$ $∙ A / B :$ $\frac{a^{5} + b^{5}}{a b} = \frac{a^{4}}{b} + \frac{b^{4}}{a} = \frac{- 152}{- 2} = 76$
	Commented by JDamian last updated on 30/Dec/21
	Mr. Rasheed: -16-(-2)(-2) = -20
	Commented by Rasheed.Sindhi last updated on 30/Dec/21

	$TH α n X S i r! I^{'} v e c o r r e c t e d!$
	Answered by Rasheed.Sindhi last updated on 30/Dec/21

	$x^{2} + 2 x - 2 = 0$ $a + b = - 2 \land a b = - 2$ $p_{n} = a^{n} + b^{n}$ $p_{1} = e_{1} = - 2$ $p_{2} = e_{1} p_{1} - 2 e_{2} = {(- 2)}^{2} - 2 (- 2) = 8$ $p_{3} = e_{1} p_{2} - e_{2} p_{1}$ $= (- 2) (8) - (- 2) (- 2) = - 20$ $p_{4} = e_{1} p_{3} - e_{2} p_{2}$ $= (- 2) (- 20) - (- 2) (8) = 56$ $p_{5} = e_{1} p_{4} - e_{2} p_{3}$ $= (- 2) (56) - (- 2) (- 20) = - 152$ $\frac{a^{4}}{b} + \frac{b^{4}}{a} = \frac{a^{5} + b^{5}}{a b} = \frac{p_{5}}{e_{2}} = \frac{- 152}{- 2} = 76$ $N o t e : T h e a b o v e m e t h o d I^{'} v e l e a r n t$ $f r o m m r W s i r .$
	Commented by Rasheed.Sindhi last updated on 30/Dec/21

	$S i r m r W, p l c h e c k m y a b o v e a n s w e r .$ $I s i t c o r r e c t ?$
	Commented by mr W last updated on 30/Dec/21
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	Commented by mnjuly1970 last updated on 30/Dec/21

	$t h x a l o t s i r$
	Answered by bobhans last updated on 30/Dec/21
	$x^2 +2x−2=0 ⇒ { ((α^2 =2−2α)),((β^2 =2−2β)) :} → { ((α^3 =2α−2α^2 )),((β^3 =2β−2β^2 )) :} { ((α^2 +β^2 =4−2(−2)=8)),((α^3 +β^3 =2(−2)−2(8)=−20)) :} ⇔ (α^2 +β^2 )(α^3 +β^3 )=−160 ⇔ α^5 +α^2 β^3 +α^3 β^2 +β^5 =−160 ⇔α^5 +β^5 +(αβ)^2 (α+β)=−160 ⇔α^5 +β^5 +4(−2)=−160 ⇒α^5 +β^5 =−152 ((α^5 +β^5 )/(αβ)) = ((−152)/(−2)) = 76$
	$x^{2} + 2 x - 2 = 0 \Rightarrow {\begin{cases} α^{2} = 2 - 2 α \\ β^{2} = 2 - 2 β \end{cases} \to {\begin{cases} α^{3} = 2 α - 2 α^{2} \\ β^{3} = 2 β - 2 β^{2} \end{cases}$ ${\begin{cases} α^{2} + β^{2} = 4 - 2 (- 2) = 8 \\ α^{3} + β^{3} = 2 (- 2) - 2 (8) = - 20 \end{cases}$ $\Leftrightarrow (α^{2} + β^{2}) (α^{3} + β^{3}) = - 160$ $\Leftrightarrow α^{5} + α^{2} β^{3} + α^{3} β^{2} + β^{5} = - 160$ $\Leftrightarrow α^{5} + β^{5} + {(α β)}^{2} (α + β) = - 160$ $\Leftrightarrow α^{5} + β^{5} + 4 (- 2) = - 160 \Rightarrow α^{5} + β^{5} = - 152$ $\frac{α^{5} + β^{5}}{α β} = \frac{- 152}{- 2} = 76$
	Commented by mnjuly1970 last updated on 30/Dec/21

	$g r a t e f u l . . . n i c e s o l u t i o n s i r$
	Answered by mindispower last updated on 30/Dec/21

	$x^{2} = 2 - 2 x$ $x^{3} = 6 x - 4, x^{4} = - 4 x + 6 (2 - 2 x) = - 16 x + 12$ $\frac{- 16 a + 12}{β} + \frac{12 - 16 β}{α}$ $= \frac{12 (α + β) - 16 (α^{2} + β^{2})}{α β} = \frac{12 (α + β) - 16 ({(α + β)}^{2} - 2 α β)}{α β}$ $= \frac{12 (- 2) - 16 (4 + 4)}{- 2} = 12 + 64 = 76$
	Commented by Rasheed.Sindhi last updated on 31/Dec/21

	$S i r, d i d y o u i n t e g r a t e t h e f i r s t l i n e ?$
	Commented by tounghoungko last updated on 31/Dec/21

	$x^{2} = 2 - 2 x$ $x^{3} = 2 x - 2 x^{2} = 2 x - 2 (2 - 2 x)$ $x^{3} = 2 x - 4 + 4 x = 6 x - 4$
	Commented by Rasheed.Sindhi last updated on 31/Dec/21

	$T han k s sir! I f e l t a l r e a d y m y b l u n d e r a n d$ $d e l e t e d m y c o m m e n t! T h a n k s f o r$ $d e t a i l e d e x p l n a t i o n . I u n d e r s t o o d$ $f r o m i t a l o t!$
	Commented by mr W last updated on 31/Dec/21

	$t a k e c a r e s i r! w h a t l o o k s l i k e t h e$ $s a m e {d o e s n}^{'} t m e a n t h e s a m e!$ $h e r e x i s j u s t a s y m b o l f o r a f i x e d$ $v a l u e (c o n s t a n t) . {d o n}^{'} t c o n f u s e i t$ $w i t h a v a r i a b l e i n a f u n c t i o n w h i c h$ $i s a l s o d e n o t e d a s x . y o u c a n n o t d o$ $o p e r a t i o n l i k e d e r i v a t i o n o r$ $i n t e g r a t i o n w i t h a f i x e d v a l u e x$ $i n a n e q u a t i o n a s w i t h a v a r i a b l e x$ $i n a f u n c t i o n!$ $s o y o u c a n n o t t r a n s f e r f o l l o w i n g$ $e q u a t i o n x^{3} + 5 x^{2} - 3 = 0 b y d e r i v a t i n g$ $i t i n t o e q u a t i o n 3 x^{2} + 10 x = 0!$
	Commented by Rasheed.Sindhi last updated on 31/Dec/21

	$T han k s also sir tounghoungko!$
	Commented by mindispower last updated on 31/Dec/21

	$H e l l o s i r$ $x^{2} = a x + b$ $x . x^{2} = x^{3} = {a x}^{2} + b x = a (a x + b) + b x = a b + x (a^{2} + a b)$ $i d i d t h i s 3 T i m e s f o r r e d u c e d g r e e s$ $i r e p l a c e x^{2} . b y a x + b$ $h a v e a n i c e D a y$
	Commented by Rasheed.Sindhi last updated on 31/Dec/21

	$T han k s sir mindispower!$ $N i c e t o m e e t y o u!$ 🌹🌷🥀
	Commented by mindispower last updated on 31/Dec/21

	$y o u a r e w e l c o m n i c e t o m e e t Y o u s i r$
	Commented by Rasheed.Sindhi last updated on 31/Dec/21
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