((x^(11) +x)/(x^7 +x^5 )) = ((205)/(16)) solve for x (undecic equation )


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Question Number 175972 by Linton last updated on 10/Sep/22

$\frac{x^{11} + x}{x^{7} + x^{5}} = \frac{205}{16}$ $s o l v e f o r x (u n d e c i c e q u a t i o n)$
	Answered by BaliramKumar last updated on 10/Sep/22

	$l e t f (x) = \frac{x^{11} + x}{x^{7} + x^{5}} = f (- x) = f (\frac{1}{x}) = f (\frac{- 1}{x})$ $\frac{(x^{10} + 1)}{x^{4} (x^{2} + 1)} = \frac{205}{16} \times \frac{5}{5} = \frac{1025}{16 \times 5} = \frac{1024 + 1}{16 (4 + 1)} = \frac{2^{10} + 1}{2^{4} (2^{2} + 1)}$ $\frac{(x^{10} + 1)}{x^{4} (x^{2} + 1)} = \frac{2^{10} + 1}{2^{4} (2^{2} + 1)}$ $x = 2$ $x = \pm 2 o r x = \pm \frac{1}{2}$
	Answered by Rasheed.Sindhi last updated on 11/Sep/22
	$((x^(11) +x)/(x^7 +x^5 )) = ((205)/(16)) (( ((x^(11) +x)/x^6 ) )/((x^7 +x^5 )/x^6 ))=((205)/(16)) ((x^5 +(1/x^5 ))/(x+(1/x)))=((205)/(16)) x+(1/x)=u (x+(1/x))^3 =u^3 x^3 +(1/x^3 )+3(x+(1/x))=u^3 x^3 +(1/x^3 )=u^3 −3u..............(i) (x+(1/x))^2 =u^2 x^2 +(1/x^2 )=u^2 −2................(ii) (i)×(ii): x^5 +(1/x^5 )+x+(1/x)=u^5 −3u^3 −2u^3 +6u x^5 +(1/x^5 )=u^5 −5u^3 +5u ((u^5 −5u^3 +5u)/u)=((205)/(16)) u^4 −5u^2 +5−((205)/(16))=0 16u^4 −80u^2 −125=0 u^2 =((80±(√(6400+8000)))/(32))=((80±120)/(32))=((25)/4),−(5/4) u=±(5/2),±((i(√5))/2) x+(1/x)=±(5/2) ∣ x+(1/x)=±((i(√5))/2)^((A)) 2x^2 +2=±5x { ((2x^2 +2=5x⇒2x^2 −5x+2=0^• )),((2x^2 +2=−5x⇒2x^2 +5x+2=0^(••) )) :} ^• 2x^2 −5x+2=0 x=((5±(√(25−16)))/4) =((5±3)/4)=2,(1/2) ^(••) 2x^2 +5x+2=0 x=((−5±(√(25−16)))/4) =((−5±3)/4)=−(1/2),−2 ^((A)) x+(1/x)=±((i(√5))/2)⇒ { ((2x^2 −i(√5) x+2=0)),((2x^2 +i(√5) x+2=0)) :} 2x^2 −i(√5) x+2=0 ∣ 2x^2 +i(√5) x+2=0 x=((i(√5) ±(√(−5−16)))/4) ∣ x=((−i(√5) ±(√(−5−16)))/4) x=((i(√5) ±i(√(21)))/4) ∣ x=((−i(√5) ±i(√(21)))/4) x=((i((√5) ±(√(21)) ))/4) ∣ x=((i(−(√5) ±(√(21)) ))/4) These also may be valid solutions.$
	$\frac{x^{11} + x}{x^{7} + x^{5}} = \frac{205}{16}$ $\frac{\frac{x^{11} + x}{x^{6}}}{\frac{x^{7} + x^{5}}{x^{6}}} = \frac{205}{16}$ $\frac{x^{5} + \frac{1}{x^{5}}}{x + \frac{1}{x}} = \frac{205}{16}$ $x + \frac{1}{x} = u$ ${(x + \frac{1}{x})}^{3} = u^{3}$ $x^{3} + \frac{1}{x^{3}} + 3 (x + \frac{1}{x}) = u^{3}$ $x^{3} + \frac{1}{x^{3}} = u^{3} - 3 u . . . . . . . . . . . . . . (i)$ ${(x + \frac{1}{x})}^{2} = u^{2}$ $x^{2} + \frac{1}{x^{2}} = u^{2} - 2 . . . . . . . . . . . . . . . . (i i)$ $(i) \times (i i) :$ $x^{5} + \frac{1}{x^{5}} + x + \frac{1}{x} = u^{5} - 3 u^{3} - 2 u^{3} + 6 u$ $x^{5} + \frac{1}{x^{5}} = u^{5} - 5 u^{3} + 5 u$ $\frac{u^{5} - 5 u^{3} + 5 u}{u} = \frac{205}{16}$ $u^{4} - 5 u^{2} + 5 - \frac{205}{16} = 0$ $16 u^{4} - 80 u^{2} - 125 = 0$ $u^{2} = \frac{80 \pm \sqrt{6400 + 8000}}{32} = \frac{80 \pm 120}{32} = \frac{25}{4}, - \frac{5}{4}$ $u = \pm \frac{5}{2}, \pm \frac{i \sqrt{5}}{2}$ $x + \frac{1}{x} = \pm \frac{5}{2} ∣ x + \frac{1}{x} = \pm \frac{i \sqrt{5}}{2}^{(A)}$ $2 x^{2} + 2 = \pm 5 x$ ${\begin{cases} 2 x^{2} + 2 = 5 x \Rightarrow 2 x^{2} - 5 x + 2 = 0^{∙} \\ 2 x^{2} + 2 = - 5 x \Rightarrow 2 x^{2} + 5 x + 2 = 0^{∙ ∙} \end{cases}$ $^{∙} 2 x^{2} - 5 x + 2 = 0$ $x = \frac{5 \pm \sqrt{25 - 16}}{4}$ $= \frac{5 \pm 3}{4} = 2, \frac{1}{2}$ $^{∙ ∙} 2 x^{2} + 5 x + 2 = 0$ $x = \frac{- 5 \pm \sqrt{25 - 16}}{4}$ $= \frac{- 5 \pm 3}{4} = - \frac{1}{2}, - 2$ $^{(A)} x + \frac{1}{x} = \pm \frac{i \sqrt{5}}{2} \Rightarrow {\begin{cases} 2 x^{2} - i \sqrt{5} x + 2 = 0 \\ 2 x^{2} + i \sqrt{5} x + 2 = 0 \end{cases}$ $2 x^{2} - i \sqrt{5} x + 2 = 0 ∣ 2 x^{2} + i \sqrt{5} x + 2 = 0$ $x = \frac{i \sqrt{5} \pm \sqrt{- 5 - 16}}{4} ∣ x = \frac{- i \sqrt{5} \pm \sqrt{- 5 - 16}}{4}$ $x = \frac{i \sqrt{5} \pm i \sqrt{21}}{4} ∣ x = \frac{- i \sqrt{5} \pm i \sqrt{21}}{4}$ $x = \frac{i (\sqrt{5} \pm \sqrt{21})}{4} ∣ x = \frac{i (- \sqrt{5} \pm \sqrt{21})}{4}$ $T h e s e a l s o m a y b e v a l i d s o l u t i o n s .$
	Commented by BaliramKumar last updated on 11/Sep/22

	$x = \frac{\pm 5 \pm 3}{4}$ $x = \frac{5 + 3}{4} = 2$ $x = \frac{- 5 - 3}{4} = - 2$ $x = \frac{5 - 3}{4} = \frac{1}{2}$ $x = \frac{- 5 + 3}{4} = - \frac{1}{2}$
	Commented by Rasheed.Sindhi last updated on 11/Sep/22

	$T h a n k y o u B a l i r a m, I^{'} v e a d d e d$ $t h e s o l u t i o n s : x = \frac{1}{2}, x = - \frac{1}{2} n o w .$ $P l e a s e r e v i s i t m y a n s w e r .$
	Commented by BaliramKumar last updated on 11/Sep/22

	$I c h e c k t h e a n s w e r s b y H i p e r s c i e n t i f i c c a l c u l a t o r A p p ✓$
	Commented by Rasheed.Sindhi last updated on 11/Sep/22
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	Commented by Tawa11 last updated on 15/Sep/22

	$Great sir$
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