Solve the equation x^4 −2x^3 −5x^2 +10x−3=0


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Question Number 141672 by iloveisrael last updated on 22/May/21

$S o l v e t h e e q u a t i o n$ $x^{4} - 2 x^{3} - 5 x^{2} + 10 x - 3 = 0$
	Answered by MJS_new last updated on 22/May/21

	$(x^{2} + x - 3) (x^{2} - 3 x + 1) = 0$ $x = - \frac{1}{2} \pm \frac{\sqrt{13}}{2} \lor x = \frac{3}{2} \pm \frac{\sqrt{5}}{2}$
	Commented by MJS_new last updated on 22/May/21
	$x^4 −2x^3 −5x^2 +10x−3=0 let x=t+(1/2) t^4 −((13)/2)t^2 +4t+(9/(16))=0 (t^2 −at−b)(t^2 +at−c)=0 t^4 −(a^2 +b+c)t^2 +a(c−b)t+bc=0 now match the constants { ((a^2 +b+c−((13)/2)=0)),((a(c−b)−4=0)),((bc−(9/(16))=0)) :} { ((b=−((2a^3 −13a+8)/(4a)))),((c=−((2a^3 −13a−8)/(4a)))),((((a^6 −13a^4 +40a^2 −16)/(4a^2 ))=0)) :} a^6 −13a^4 +40a^2 −16=0 select a “nice” solution of this a=2 ⇒ b=(1/4)∧c=(9/4) no insert backwards and it′s finished$
	$x^{4} - 2 x^{3} - 5 x^{2} + 10 x - 3 = 0$ $let x = t + \frac{1}{2}$ $t^{4} - \frac{13}{2} t^{2} + 4 t + \frac{9}{16} = 0$ $(t^{2} - a t - b) (t^{2} + a t - c) = 0$ $t^{4} - (a^{2} + b + c) t^{2} + a (c - b) t + b c = 0$ $now match the constants$ ${\begin{cases} a^{2} + b + c - \frac{13}{2} = 0 \\ a (c - b) - 4 = 0 \\ b c - \frac{9}{16} = 0 \end{cases}$ ${\begin{cases} b = - \frac{2 a^{3} - 13 a + 8}{4 a} \\ c = - \frac{2 a^{3} - 13 a - 8}{4 a} \\ \frac{a^{6} - 13 a^{4} + 40 a^{2} - 16}{4 a^{2}} = 0 \end{cases}$ $a^{6} - 13 a^{4} + 40 a^{2} - 16 = 0$ $select a ‘ ‘ {nice}^{″} solution of this$ $a = 2$ $\Rightarrow b = \frac{1}{4} \land c = \frac{9}{4}$ $no insert backwards and {it}^{'} s finished$
	Commented by iloveisrael last updated on 22/May/21

	$y e s . i t s a m e b y {F e r r a r i}^{'} s m e t h o d$
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