find x x^4 −x^3 −19x^2 +93x−128=0


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Question Number 183746 by ali009 last updated on 29/Dec/22

$f i n d x$ $x^{4} - x^{3} - 19 x^{2} + 93 x - 128 = 0$
Commented by Frix last updated on 29/Dec/22

$No useable exact solution, you can only$ $approximate .$
Commented by Michaelfaraday last updated on 29/Dec/22

$t h i s q u e s t i o n h a s n o r e a l r o o t b u t y o u$ $c a n u s e N R I M .$
Commented by mr W last updated on 30/Dec/22

$h o w c a n y o u s a y t h a t i t h a s n o r e a l$ $r o o t ?$ $f (- \infty) = + \infty$ $f (+ \infty) = + \infty$ $f (0) = - 128 < 0$ $\Rightarrow i t h a s a t l e a s t t w o r e a l r o o t s!$ $o n e i s p o s i t i v e, o n e i s n e g a t i v e .$
Commented by Michaelfaraday last updated on 30/Dec/22

$o k a y s i r b u t s i r s h o w u s t h e r e a l r o o t$ $t h a t t h e e q u a t i o n h a v e .$
Commented by MJS_new last updated on 30/Dec/22

$x_{1} \approx - 5.76397627403$ $x_{2} \approx 2.26841847295$ $x_{3, 4} \approx 2.24777890054 \pm 2 . 17648395669 i$
	Answered by aleks041103 last updated on 29/Dec/22

	$f (x) = x^{4} - x^{3} - 19 x^{2} + 93 x - 128$ $l e t f (x) = (x^{2} + a x + b) (x^{2} + c x + d)$ $\Rightarrow a + c = - 1$ $b + d + a c = - 19$ $a d + b c = 93$ $b d = - 128$ $\Rightarrow b + d - a (1 + a) = - 19$ $a d - b (1 + a) = 93$ $b d = - 128$ $\Rightarrow b^{2} - 128 - a b (1 + a) = - 19 b$ $- 128 a - b^{2} (1 + a) = 93 b$ $\Rightarrow b^{2} + 19 b - 128 - a b - a^{2} b = 0$ $b^{2} + {a b}^{2} + 93 b + 128 a = 0$ $b^{2} + (19 - a - a^{2}) b - 128 = 0$ $(1 + a) b^{2} + 93 b + 128 a = 0$ $a = 0 ?$ $b^{2} + 19 b - 128 = 0$ $b^{2} + 93 b = 0 \Rightarrow b = 0, - 93, b u t b d = - 128 \Rightarrow b \neq 0$ $\Rightarrow a \neq 0$ $a = - 1 ?$ $b^{2} + 19 b - 128 = 0$ $93 b - 128 = 0 \Rightarrow b = 128 / 93$ $\Rightarrow b^{2} - 74 b = 0, b \neq 0 \Rightarrow b = 74$ $\Rightarrow a \neq - 1$ $\Rightarrow {a b}^{2} + a (19 - a - a^{2}) b - 128 a = 0$ $(1 + a) b^{2} + 93 b + 128 a = 0$ $\Rightarrow (1 + 2 a) b^{2} + b (93 + 19 a - a^{2} - a^{3}) = 0$ $\Rightarrow b = \frac{a^{3} + a^{2} - 19 a - 93}{1 + 2 a}$ $a l s o$ $b^{2} + (19 - a - a^{2}) b - 128 = 0$ $(1 + a) b^{2} + 93 b + 128 a = 0$ $(1 + a) b^{2} + (1 + a) (19 - a - a^{2}) b - 128 - 128 a = 0$ $(1 + a) b^{2} + 93 b + 128 a = 0$ $\Rightarrow (19 - a - a^{2} + 19 a - a^{2} - a^{3} - 93) b - 128 (1 + 2 a) = 0$ $(- 74 + 18 a - 2 a^{2} - a^{3}) b = 128 (1 + 2 a)$ $\Rightarrow b = \frac{128 (1 + 2 a)}{- 74 + 18 a - 2 a^{2} - a^{3}}$ $\Rightarrow 128 {(1 + 2 a)}^{2} = (- 74 + 18 a - 2 a^{2} - a^{3}) (a^{3} + a^{2} - 19 a - 93)$ $128 + 512 a + 512 a^{2} = . . .$ $. . .$ $. . .$
	Commented by Michaelfaraday last updated on 29/Dec/22

	$s i r i d o n t u n d e r s t a n d y o u r s t e p ?$
	Commented by MJS_new last updated on 29/Dec/22

	${It}^{'} s not possible this way .$ $x^{4} + {a x}^{3} + {b x}^{2} + c x + d = 0$ $1 . let x = t - \frac{a}{4} to get$ $t^{4} + {p t}^{2} + q t + r = 0$ $2 . now we might find 2 square factors$ $(t^{2} - α t + β) (t^{2} + α t + γ) = 0$ $by matching the constants .$ $solve 2 of the 3 equations for β and γ, then$ $insert in the 3^{rd} equation to get$ ${(α^{2})}^{3} + A {(α^{2})}^{2} + B α^{2} + C = 0$ $only if this has got ‘ ‘ {nice}^{″} solutions we get$ $square factors we can work with .$
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