32x^3 −48x^2 −22x−3=0


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Question Number 63602 by pavithra last updated on 06/Jul/19

$32 x^{3} - 48 x^{2} - 22 x - 3 = 0$
	Answered by MJS last updated on 06/Jul/19

	$x^{3} - \frac{3}{2} x^{2} - \frac{11}{16} x - \frac{3}{32} = 0$ $x = z + \frac{1}{2}$ $z^{3} - \frac{23}{16} z - \frac{11}{16} = 0$ $D = \frac{1}{27} {(- \frac{23}{16})}^{3} + \frac{1}{4} {(- \frac{11}{16})}^{2} > 0$ $u = \sqrt[3]{\frac{11}{32} + \frac{\sqrt{2703}}{576}} v = \sqrt[3]{\frac{11}{32} - \frac{\sqrt{2703}}{576}}$ $x_{1} = \frac{1}{2} + u + v$ $x_{2} = \frac{1}{2} + (- \frac{1}{2} - \frac{\sqrt{3}}{2} i) u + (- \frac{1}{2} + \frac{\sqrt{3}}{2} i) v$ $x_{3} = \frac{1}{2} + (- \frac{1}{2} + \frac{\sqrt{3}}{2} i) u + (- \frac{1}{2} - \frac{\sqrt{3}}{2} i) v$ $x_{1} \approx 1.89000138$ $x_{2} \approx - . 195000688 - . 107600511 i$ $x_{3} \approx - . 195000688 + . 107600511 i$
	Commented by pavithra last updated on 07/Jul/19

	$t h a n k y o u$
	Answered by ajfour last updated on 07/Jul/19

	$l e t x = \frac{a t + b}{t + 1}$ $32 (a^{3} t^{3} + 3 a^{2} {b t}^{2} + 3 {a b}^{2} t + b^{3})$ $- 48 (a^{2} t^{2} + 2 a b t + b^{2}) (t + 1)$ $- 22 (a t + b) (t^{2} + 2 t + 1)$ $- 3 (t^{3} + 3 t^{2} + 3 t + 1) = 0$ $\Rightarrow$ $(32 a^{3} - 48 a^{2} - 22 a - 3) t^{3}$ $+ (96 a^{2} b - 48 a^{2} - 96 a b - 44 a - 22 b - 9) t^{2}$ $+ (96 {a b}^{2} - 96 a b - 48 b^{2} - 22 a - 44 b - 9) t$ $+ 32 b^{3} - 48 b^{2} - 22 b - 3 = 0$ $l e t c o e f f i c i e n t s o f t^{2} a n d t b e z e r o;$ $\Rightarrow$ $96 a^{2} b - 48 a^{2} - 96 a b - 44 a - 22 b - 9 = 0$ $96 {a b}^{2} - 96 a b - 48 b^{2} - 22 a - 44 b - 9 = 0$ $s u b t r a c t i n g, k n o w i n g a \neq b$ $48 a b - 24 (a + b) - 11 = 0 . . . (i)$ $a d d i n g$ $48 a b (a + b) - 24 {(a + b)}^{2} - 48 a b$ $- 33 (a + b) - 9 = 0$ $u s i n g (i)$ $24 {(a + b)}^{2} + 11 (a + b) - 24 {(a + b)}^{2}$ $- 24 (a + b) - 11 - 33 (a + b) - 9 = 0$ $\Rightarrow a + b = - \frac{10}{23}$ $48 a b = - \frac{240}{23} + 11 = \frac{13}{23}$ $\Rightarrow a b = \frac{13}{23 \times 48}$ $a, b a r e r o o t s o f e q .$ $x^{2} + \frac{10 x}{23} + \frac{13}{23 \times 48} = 0$ $a, b = - \frac{5}{23} \pm \sqrt{{(\frac{5}{23})}^{2} - \frac{13}{23 \times 16}}$ $a, b = \frac{- 5 \pm \sqrt{101}}{23}$ $t^{3} = \frac{32 b^{3} - 48 b^{2} - 22 b - 3}{32 a^{3} - 48 a^{2} - 22 a - 3}$ $. . . . .$
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