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-

Question Number 79015 by ajfour last updated on 22/Jan/20

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}

\boldsymbol p q = \boldsymbol m, \boldsymbol p + \boldsymbol q = \boldsymbol s

________________________

\boldsymbol m^{2} {{(\boldsymbol a^{2} + \boldsymbol b)}^{2} - 6 \boldsymbol a (\boldsymbol a b - \boldsymbol c)}

+ \boldsymbol m {2 (\boldsymbol b^{2} + \boldsymbol a c) (\boldsymbol a^{2} + \boldsymbol b) -

3 (\boldsymbol a b - \boldsymbol c) (\boldsymbol a b + 3 \boldsymbol c)}

+ {(\boldsymbol b^{2} + \boldsymbol a c)}^{2} - 6 \boldsymbol b c (\boldsymbol a b - \boldsymbol c) = 0

________________________

\boldsymbol s = - \frac{2}{3} {\frac{\boldsymbol m (\boldsymbol a^{2} + \boldsymbol b) + \boldsymbol b^{2} + \boldsymbol a c}{\boldsymbol a b - \boldsymbol c}}

+ {\frac{8}{27} {[\frac{\boldsymbol m (\boldsymbol a^{2} + \boldsymbol b) + \boldsymbol b^{2} + \boldsymbol a c}{\boldsymbol a b - \boldsymbol c}]}^{3}

{- 8 [\frac{\boldsymbol m^{3} + \boldsymbol {b m}^{2} + \boldsymbol a c m + \boldsymbol c^{2}}{\boldsymbol a b - \boldsymbol c}]}}^{1 / 3}

\boldsymbol p, \boldsymbol q = \frac{\boldsymbol s}{2} \pm \sqrt{\frac{\boldsymbol s^{2}}{4} - \boldsymbol m}

\boldsymbol t = - \frac{(3 \boldsymbol {p q}^{2} + 2 \boldsymbol a p q + \boldsymbol {a p}^{2} + 2 \boldsymbol b p + \boldsymbol b q + 3 \boldsymbol c)}{(\boldsymbol p^{3} + \boldsymbol {a p}^{2} + \boldsymbol b p + \boldsymbol c)}

\boldsymbol x = \frac{\boldsymbol p t + \boldsymbol q}{\boldsymbol 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} + \dots or p = \frac{s}{2} - \dots]

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

Weldone sir