24x^3 −26x^2 +9x−1=0(solve)


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Question Number 31677 by gyugfeet last updated on 12/Mar/18

$24 x^{3} - 26 x^{2} + 9 x - 1 = 0 (s o l v e)$
	Answered by Joel578 last updated on 12/Mar/18

	$Trial and error$ $x^{3} - \frac{13}{12} x^{2} + \frac{3}{8} x - \frac{1}{24} = 0$ $x = \frac{1}{2} \to \frac{1}{8} - \frac{13}{12 . 4} + \frac{3}{8 . 2} - \frac{1}{24} = 0 (T r u e)$ $x = - \frac{1}{2} \to - \frac{1}{8} - \frac{13}{12 . 4} - \frac{3}{8 . 2} - \frac{1}{24} \neq 0$ $x = \frac{1}{3} \to \frac{1}{27} - \frac{13}{12 . 9} + \frac{3}{8 . 3} - \frac{1}{24} = 0 (T r u e)$ $x = - \frac{1}{3} \to - \frac{1}{27} - \frac{13}{12 . 9} - \frac{3}{8 . 3} - \frac{1}{24} \neq 0$ $x = \frac{1}{4} \to \frac{1}{64} - \frac{13}{12 . 16} + \frac{3}{8 . 4} - \frac{1}{24} = 0 (T r u e)$ $24 x^{3} - 26 x^{2} + 9 x - 1 = (2 x - 1) (3 x - 1) (4 x - 1) = 0$ $x = \frac{1}{2} \lor x = \frac{1}{3} \lor x = \frac{1}{4}$
	Commented by MJS last updated on 12/Mar/18

	$you could solve the quadratic$ $equation after you found one$ $root$ $(x^{3} - \frac{13}{12} x^{2} + \frac{3}{8} x - \frac{1}{24}) / (x - \frac{1}{2}) =$ $= x^{2} - \frac{7}{12} x + \frac{1}{12}$ $x = \frac{7}{24} \pm \sqrt{\frac{49}{576} - \frac{1}{12}} = \frac{7}{24} \pm \frac{1}{24}$ $x_{2} = \frac{1}{4}; x_{3} = \frac{1}{3}$
	Commented by Joel578 last updated on 12/Mar/18

	$t h a n k s f o r s u g g e s t i o n$
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