Rigorously over one month′s time, I developed a formula for general cubic. x^3 +ax^2 +bx+c=0 let x=((pt+q)/(t+1)) pq=m, p+q=s ________________________ m^2 {(a^2 +b)^2 −6a(ab−c)} +m{2(b^2 +ac)(a^2 +b)− 3(ab−c)(ab+3c)} +(b^2 +ac)^2 −6bc(ab−c)=0 ________________________ s=−(2/3){((m(a^2 +b)+b^2 +ac)/(ab−c))} +{(8/(27))[((m(a^2 +b)+b^2 +ac)/(ab−c))]^3 −8[((m^3 +bm^2 +acm+c^2 )/(ab−c))]}^(1/3) p,q = (s/2)±(√((s^2 /4)−m)) t=−(((3pq^2 +2apq+ap^2 +2bp+bq+3c))/((p^3 +ap^2 +bp+c))) x=((pt+q)/(t+1)) . (Please help checking..) (edited a digit 1 in place of 4)


Question and Answers Forum

All Questions Topic List
Others Questions
Previous in All Question Next in All Question
Previous in Others Next in Others
Question Number 79015 by ajfour last updated on 22/Jan/20
$Rigorously over one month′s time, I developed a formula for general cubic. x^3 +ax^2 +bx+c=0 let x=((pt+q)/(t+1)) pq=m, p+q=s ________________________ m^2 {(a^2 +b)^2 −6a(ab−c)} +m{2(b^2 +ac)(a^2 +b)− 3(ab−c)(ab+3c)} +(b^2 +ac)^2 −6bc(ab−c)=0 ________________________ s=−(2/3){((m(a^2 +b)+b^2 +ac)/(ab−c))} +{(8/(27))[((m(a^2 +b)+b^2 +ac)/(ab−c))]^3 −8[((m^3 +bm^2 +acm+c^2 )/(ab−c))]}^(1/3) p,q = (s/2)±(√((s^2 /4)−m)) t=−(((3pq^2 +2apq+ap^2 +2bp+bq+3c))/((p^3 +ap^2 +bp+c))) x=((pt+q)/(t+1)) . (Please help checking..) (edited a digit 1 in place of 4)$
$R i g o r o u s l y o v e r o n e {m o n t h}^{'} s$ $t i m e, I d e v e l o p e d a f o r m u l a f o r$ $g e n e r a l c u b i c .$ $x^{3} + {a x}^{2} + b x + c = 0$ $l e t x = \frac{p t + q}{t + 1}$ $p q = m, p + q = s$ $________________________$ $m^{2} {{(a^{2} + b)}^{2} - 6 a (a b - c)}$ $+ m {2 (b^{2} + a c) (a^{2} + b) -$ $3 (a b - c) (a b + 3 c)}$ $+ {(b^{2} + a c)}^{2} - 6 b c (a b - c) = 0$ $________________________$ $s = - \frac{2}{3} {\frac{m (a^{2} + b) + b^{2} + a c}{a b - c}}$ $+ {\frac{8}{27} {[\frac{m (a^{2} + b) + b^{2} + a c}{a b - c}]}^{3}$ ${- 8 [\frac{m^{3} + {b m}^{2} + a c m + c^{2}}{a b - c}]}}^{1 / 3}$ $p, q = \frac{s}{2} \pm \sqrt{\frac{s^{2}}{4} - m}$ $t = - \frac{(3 {p q}^{2} + 2 a p q + {a p}^{2} + 2 b p + b q + 3 c)}{(p^{3} + {a p}^{2} + b p + c)}$ $x = \frac{p t + q}{t + 1} .$ $(P l e a s e h e l p c h e c k i n g . .)$ $(e d i t e d a d i g i t 1 i n p l a c e o f 4)$
Commented by mr W last updated on 22/Jan/20

$g r e a t s i r! y o u d i d i t!$ $j u s t o n e q u e s t i o n b e f o r e f u r t h e r$ $s t u d y i n g :$ $w e g e t g e n e r a l l y t w o v a l u e s f o r m a n d$ $s, f o r e a c h p a i r o f m a n d s w e g e t$ $a p a i r o f p a n d q a n d o n e v a l u e o f t,$ $a n d f i n a l l y a r o o t x . t o t a l l y w e g e t$ $t w o r o o t s . b u t g e n e r a l l y w e h a v e t h r e e$ $r o o t s . w h a t d i d i u n d e r s t a n d w r o n g l y ?$
Commented by MJS last updated on 22/Jan/20

${doesn}^{'} t work for$ $x^{3} + 3 x^{2} - 13 x - 15 = 0$ $with x_{1} = - 5, x_{2} = - 1, x_{3} = 3$ $your method gives depending on the choice$ $of p and q [t is not symmetric in p, q; so we$ $have to decide p = \frac{s}{2} + . . . or p = \frac{s}{2} - . . .]$ $x \approx . 172 \pm . 763 or x \approx - 1.43 \pm 2.53$ $btw . in this case m \notin R$
Commented by ajfour last updated on 22/Jan/20

$i t s n o t a t a l l g o o d! b u t j u s t n o w$ $i h a d b e e n m o d i f y i n g t h e$ $C a r d a n o f o r m u l a, f i r s t l e t m e$ $c o n f i r m a n d i s h a l l p o s t t h e n . .$
Commented by TawaTawa last updated on 23/Jan/20

$Weldone sir$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum