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Question Number 160837 by mnjuly1970 last updated on 07/Dec/21

Commented by cortano last updated on 07/Dec/21
$⇔ x^2 +x^4 −2x^2 +1=x(x^2 +2x+1) ⇔x^4 −x^2 +1 = x^3 +2x^2 +x ⇔x^4 −x^3 −3x^2 −x+1 = 0 ⇔(x^2 +ax+1)(x^2 +bx+1)=x^4 −x^3 −3x^2 −x+1 ⇒x^4 +(a+b)x^3 +(ab+2)x^2 +(a+b)x+1= x^4 −x^3 −3x^2 −x+1 ⇒ { ((a+b=−1)),((ab+2=−3)) :}→ { ((b=−1−a)),((a(−1−a)=−5)) :} ⇒a^2 +a−5=0 ⇒a = ((−1±(√(21)))/2) ⇒ { ((a=((−1+(√(21)))/2) ; b=((−1−(√(21)))/2))),((a=((−1−(√(21)))/2) ; b=((−1+(√(21)))/2))) :} ⇒ { ((x^2 +((((√(21))−1)/2))x+1=0)),((x^2 −(((1+(√(21)))/2))x+1=0)) :} ⇒ { ((Δ_1 =((((√(21))−1)/2))^2 −4<0 (reject))),((Δ_2 =((((√(21))+1)/2))^2 −4>0 (accept))) :} so ⇒2x^2 −((√(21))+1)x+2=0 { (α),(β) :} { ((α+β=(((√(21))+1)/2))),((αβ=1)) :} ⇒(((α−1)(β−1))/((α+1)(β+1))) = ((αβ−(α+β)+1)/(αβ+(α+β)+1)) ⇒ ((2−((((√(21))+1)/2)))/(2+((((√(21))+1)/2)))) = ((3−(√(21)))/(5+(√(21)))) = (((3−(√(21)))(5−(√(21))))/4) = ((36−8(√(21)))/4)= 9−2(√(21))$
$\Leftrightarrow x^{2} + x^{4} - {2 x}^{2} + 1 = x (x^{2} + 2 x + 1)$ $\Leftrightarrow x^{4} - x^{2} + 1 = x^{3} + {2 x}^{2} + x$ $\Leftrightarrow x^{4} - x^{3} - {3 x}^{2} - x + 1 = 0$ $\Leftrightarrow (x^{2} + ax + 1) (x^{2} + bx + 1) = x^{4} - x^{3} - {3 x}^{2} - x + 1$ $\Rightarrow x^{4} + (a + b) x^{3} + (ab + 2) x^{2} + (a + b) x + 1 =$ $x^{4} - x^{3} - {3 x}^{2} - x + 1$ $\Rightarrow {\begin{cases} a + b = - 1 \\ ab + 2 = - 3 \end{cases} \to {\begin{cases} b = - 1 - a \\ a (- 1 - a) = - 5 \end{cases}$ $\Rightarrow a^{2} + a - 5 = 0 \Rightarrow a = \frac{- 1 \pm \sqrt{21}}{2}$ $\Rightarrow {\begin{cases} a = \frac{- 1 + \sqrt{21}}{2}; b = \frac{- 1 - \sqrt{21}}{2} \\ a = \frac{- 1 - \sqrt{21}}{2}; b = \frac{- 1 + \sqrt{21}}{2} \end{cases}$ $\Rightarrow {\begin{cases} x^{2} + (\frac{\sqrt{21} - 1}{2}) x + 1 = 0 \\ x^{2} - (\frac{1 + \sqrt{21}}{2}) x + 1 = 0 \end{cases}$ $\Rightarrow {\begin{cases} Δ_{1} = {(\frac{\sqrt{21} - 1}{2})}^{2} - 4 < 0 (reject) \\ Δ_{2} = {(\frac{\sqrt{21} + 1}{2})}^{2} - 4 > 0 (accept) \end{cases}$ $so \Rightarrow {2 x}^{2} - (\sqrt{21} + 1) x + 2 = 0 {\begin{cases} α \\ β \end{cases}$ ${\begin{cases} α + β = \frac{\sqrt{21} + 1}{2} \\ α β = 1 \end{cases}$ $\Rightarrow \frac{(α - 1) (β - 1)}{(α + 1) (β + 1)} = \frac{α β - (α + β) + 1}{α β + (α + β) + 1}$ $\Rightarrow \frac{2 - (\frac{\sqrt{21} + 1}{2})}{2 + (\frac{\sqrt{21} + 1}{2})} = \frac{3 - \sqrt{21}}{5 + \sqrt{21}} = \frac{(3 - \sqrt{21}) (5 - \sqrt{21})}{4}$ $= \frac{36 - 8 \sqrt{21}}{4} = 9 - 2 \sqrt{21}$
Commented by mnjuly1970 last updated on 07/Dec/21

$t h a n k s a l o t m a s t e r . . .$
Commented by Tawa11 last updated on 07/Dec/21

$Great sir .$
Commented by HongKing last updated on 08/Dec/21

$x^{2} + {(x^{2} - 1)}^{2} = x \cdot {(x + 1)}^{2}$ ${(x^{2} - 1)}^{2} + x^{2} - x \cdot {(x + 1)}^{2} = 0$ $x^{4} - x^{3} - {3 x}^{2} - x + 1 = 0$ $x^{2} \cdot (x^{2} - x - 3 - \frac{1}{x} + \frac{1}{x^{2}}) = 0$ $(i) x^{2} - \frac{1}{2} \cdot (1 + \sqrt{21}) \cdot x + 1 = 0$ $(i i) x^{2} + \frac{1}{2} \cdot (\sqrt{21} - 1) \cdot x + 1 = 0 (no)$ $(i) \to roots α and β$ $\frac{(α - 1) \cdot (β - 1)}{(α + 1) \cdot (β + 1)} = \frac{3 - \sqrt{21}}{5 + \sqrt{21}} = 9 - 2 \sqrt{21} ∙$
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