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Question Number 114720 by Algoritm last updated on 20/Sep/20

	Answered by 1549442205PVT last updated on 20/Sep/20
	$x^4 −2x^3 +4x−2⇔(x^2 −x+1)^2 −3x^2 +6x−3=0 ⇔(x^2 −x+1)^2 −3(x−1)^2 =0 ⇔[x^2 +((√3)−1)x+1−(√3)][x^2 −(1+(√3))x+1+(√3)]=0 i)⇔x^2 +((√3)−1)x+1−(√3)=0 Δ=((√3)−1)^2 +4((√3)−1)=2(√3) x=((1−(√3)±(√(2(√3))))/2) ii)x^2 −(1+(√3))x+1+(√3)=0 Δ=(1+(√3))^2 −4(1+(√3))=−2(√3)<0 ⇒has no roots Thus,the given equation has two roots x∈{((1−(√3)+(√(2(√3))))/2),((1−(√3)−(√(2(√3))))/2)}$
	$x^{4} - {2 x}^{3} + 4 x - 2 \Leftrightarrow {(x^{2} - x + 1)}^{2} - {3 x}^{2} + 6 x - 3 = 0$ $\Leftrightarrow {(x^{2} - x + 1)}^{2} - 3 {(x - 1)}^{2} = 0$ $\Leftrightarrow [x^{2} + (\sqrt{3} - 1) x + 1 - \sqrt{3}] [x^{2} - (1 + \sqrt{3}) x + 1 + \sqrt{3}] = 0$ $i) \Leftrightarrow x^{2} + (\sqrt{3} - 1) x + 1 - \sqrt{3} = 0$ $Δ = {(\sqrt{3} - 1)}^{2} + 4 (\sqrt{3} - 1) = 2 \sqrt{3}$ $x = \frac{1 - \sqrt{3} \pm \sqrt{2 \sqrt{3}}}{2}$ $ii) x^{2} - (1 + \sqrt{3}) x + 1 + \sqrt{3} = 0$ $Δ = {(1 + \sqrt{3})}^{2} - 4 (1 + \sqrt{3}) = - 2 \sqrt{3} < 0$ $\Rightarrow has no roots$ $Thus, the given equation has two$ $roots x \in {\frac{1 - \sqrt{3} + \sqrt{2 \sqrt{3}}}{2}, \frac{1 - \sqrt{3} - \sqrt{2 \sqrt{3}}}{2}}$
	Answered by MJS_new last updated on 20/Sep/20

	$x^{4} - 2 x^{3} + 4 x - 2 = 0$ $(x^{2} - (1 - \sqrt{3}) x + 1 - \sqrt{3}) (x^{2} - (1 + \sqrt{3}) x + 1 + \sqrt{3}) = 0$ $x_{1, 2} = \frac{1 - \sqrt{3} \pm \sqrt{2 \sqrt{3}}}{2}$ $x_{3, 4} = \frac{1 + \sqrt{3}}{2} \pm \frac{\sqrt{2 \sqrt{3}}}{2} i$
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