solve 6x^4 −25x^3 +12x^2 −25x+6=0


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Question Number 114653 by bemath last updated on 20/Sep/20

$s o l v e 6 x^{4} - 25 x^{3} + 12 x^{2} - 25 x + 6 = 0$
	Answered by bobhans last updated on 20/Sep/20
	$⇔6(x^4 +1)−25(x^3 +x)+12x^2 =0 assume x ≠ 0 6(x^2 +(1/x^2 ))−25(x+(1/x))+12=0 let x+(1/x) = b ⇔ 6(b^2 −2)−25b+12=0 6b^2 −25b = 0 ⇒ { ((b=0⇒x+(1/x)=0; x^2 +1=0; x=±i)),((b=((25)/6)⇒x+(1/x)=((25)/6);6x^2 −25x+6=0)) :} x = ((25 ± (√(625−144)))/(12)) = ((25 ± (√(481)))/(12))$
	$\Leftrightarrow 6 (x^{4} + 1) - 25 (x^{3} + x) + 12 x^{2} = 0$ $a s s u m e x \neq 0$ $6 (x^{2} + \frac{1}{x^{2}}) - 25 (x + \frac{1}{x}) + 12 = 0$ $l e t x + \frac{1}{x} = b$ $\Leftrightarrow 6 (b^{2} - 2) - 25 b + 12 = 0$ $6 b^{2} - 25 b = 0$ $\Rightarrow {\begin{cases} b = 0 \Rightarrow x + \frac{1}{x} = 0; x^{2} + 1 = 0; x = \pm i \\ b = \frac{25}{6} \Rightarrow x + \frac{1}{x} = \frac{25}{6}; 6 x^{2} - 25 x + 6 = 0 \end{cases}$ $x = \frac{25 \pm \sqrt{625 - 144}}{12} = \frac{25 \pm \sqrt{481}}{12}$
	Answered by MJS_new last updated on 20/Sep/20

	$x^{4} - \frac{25}{6} x^{3} + 2 x^{2} - \frac{25}{6} x + 1 = 0$ $(x^{2} + 1) (x^{2} - \frac{25}{6} x + 1) = 0$ $x_{1, 2} = \frac{25 \pm \sqrt{481}}{12}$ $x_{3, 4} = \pm i$
	Commented by bobhans last updated on 21/Sep/20

	$g a v e k u d o s s i r M J S - n e w$
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