find the roots of 8x^3 −4x−1 =0


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Question Number 39026 by maxmathsup by imad last updated on 01/Jul/18

$f i n d t h e r o o t s o f 8 x^{3} - 4 x - 1 = 0$
	Answered by behi83417@gmail.com last updated on 02/Jul/18

	$x_{1} = - \frac{1}{2}$ $8 x^{3} - 4 x - 1 = (2 x + 1) (4 x^{2} + a x + b)$ $x = 0 \Rightarrow - 1 = (1) (b) \Rightarrow b = - 1$ $x = 1 \Rightarrow 3 = (3) (4 + a - 1) \Rightarrow a = - 2$ $\Rightarrow 4 x^{2} - 2 x - 1 = 0 \Rightarrow x = \frac{2 \pm \sqrt{4 + 16}}{8} = \frac{1}{4} (1 \pm \sqrt{5})$ $\Rightarrow x = - \frac{1}{2}, \frac{1}{4} (1 \pm \sqrt{5}) . ◼$
	Answered by tanmay.chaudhury50@gmail.com last updated on 02/Jul/18

	$8 x^{3} - 4 x - (2 - 1) = 0$ $8 x^{3} - 4 x - 2 + 1 = 0$ $8 x^{3} + 1 - 2 (2 x + 1) = 0$ $(2 x + 1) (4 x^{2} - 2 x . 1 + 1^{2}) - 2 (2 x + 1) = 0$ $(2 x + 1) (4 x^{2} - 2 x + 1 - 2) = 0$ $(2 x + 1) (4 x^{2} - 2 x - 1) = 0$ $x = - \frac{1}{2}$ $4 x^{2} - 2 x - 1 = 0$ $x = \frac{2 \pm \sqrt{4 + 16}}{8}$ $= \frac{2 \pm 2 \sqrt{5}}{8} = \frac{1 \pm \sqrt{5}}{4}$
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