Find-all-solutions-x-to-the-equation-x-3-bx-2-cx-d-0-where-b-c-d-are-constants-from-C-

Question Number 3548 by Yozzii last updated on 15/Dec/15

F i n d a l l s o l u t i o n s x t o t h e e q u a t i o n

x^{3} + {b x}^{2} + c x + d = 0 w h e r e b, c, d a r e

c o n s t a n t s f r o m C .

Answered by RasheedSindhi last updated on 15/Dec/15

L e t b = b_{1} + b_{2} i

c = c_{1} + c_{2} i

d = d_{1} + d_{2} i

w h e r e b_{1}, b_{2}, c_{1}, c_{2}, d_{1}, d_{2} a r e a l l r e a l s

x^{3} + b_{1} x^{2} + c_{1} x + d_{1} = 0 \land b_{2} x^{2} + c_{2} x + d_{2} = 0

Commented by prakash jain last updated on 15/Dec/15

By substituting :

(x^{3} + b_{1} x^{2} + c_{1} x + d_{1}) + i (b_{2} x^{2} + c_{2} x + d_{2}) = 0

It cannot be inferred that

(b_{2} x^{2} + c_{2} x + d_{2}) = 0

unless it is given that x \in R .

If x \in C, then (b_{2} x^{2} + c_{2} x + d_{2}) = u + v \sqrt{- 1}

Commented by RasheedSindhi last updated on 15/Dec/15

TH α n k S f o r g u i d a n c e!

Answered by prakash jain last updated on 15/Dec/15

x = y - \frac{b}{3}

{(y - \frac{b}{3})}^{3} + b {(y - \frac{b}{3})}^{2} + c (y - \frac{b}{3}) + d = 0

y^{3} - {b y}^{2} + y \frac{b^{2}}{3} - \frac{b^{3}}{27} + {b y}^{2} - 2 b^{3} \frac{y}{3} + \frac{b^{3}}{9} + c y - \frac{b c}{3} + d = 0

y^{3} + y (c - \frac{b^{3}}{3}) + (d - \frac{b c}{3} + \frac{2 b^{3}}{27}) = 0

p = c - \frac{b^{3}}{3}, q = d - \frac{b c}{3} + \frac{2 b^{3}}{27}

y^{3} + p y + q = 0

subtitute y = u - v

{(u - v)}^{3} + p (u - v) + q = 0

(q - (v^{3} - u^{3})) + (u - v) (p - 3 u v) = 0

u - v is a solution if q = v^{3} - u^{3}, p = 3 u v

p = 3 u v \Rightarrow v = \frac{p}{3 u}

q = v^{3} - u^{3} \Rightarrow q = \frac{p^{3}}{27 u^{3}} - u^{3} \dots . (A)

(A) is quadratic in u^{3} \Rightarrow u c a n b e f o u n d

t h e n w e f i n d v .

s o w e k n o w u - v i s s o l u t i o n .

o n c e w e k n o w o n e s o l u t i o n o f c u b i c t h e n

o t h e r 2 c a n b e f o u n d a f t e r f a c t o r i z i n g .

There is also a cubic formula which is derived

based on the above steps . But it is very hard

even to type . The steps are easy to remember .

Commented by Yozzii last updated on 15/Dec/15

T h a n k s a l o t . I^{'} m w o n d e r i n g h o w

o n e c r e a t e s s u c h m e t h o d s b e c a u s e

I^{'} m i n t e r e s t e d n o w i n s o l v i n g

a g o o d f e w o f t h e s e p o l y n o m i a l s,

p e r h a p s u p t o d e g r e e 10 (b e i n g

t e r r i b l y a m b i t i o u s) .

Commented by prakash jain last updated on 15/Dec/15

Abel Ruffini Theorem : There is no general

algebraic solution for polynomial of degree

5 or higher with arbitrary coefficients .

So you cannot find a formula .

Formula upto degree 4 are already available .

Commented by Yozzii last updated on 15/Dec/15

I s e e . {I t}^{'} d b e g o o d t o l e a r n t h e p r o o f s

f o r d e g r e e 4 .

Commented by Yozzii last updated on 15/Dec/15

A f t e r d e g r e e 4 {w e}^{'} d r e l y o n

n u m e r i c a l m e t h o d s o n l y t o s o l v e

p o l y n o m i a l s ?

Commented by prakash jain last updated on 15/Dec/15

For a general equation with arbitrary coefficients

only numeric solution . However there are

equation of degree 5 which can be solved

if coefficients meet certain criteria .

So for your further study you may want to

work on solvable equation and extending

the result wherever you can .