x-4-17-18-x-2-40-3-x-1625-144-0-Find-roots-Two-are-real-and-two-are-complex-

Question Number 201941 by ajfour last updated on 15/Dec/23

x^{4} - \frac{17}{18} x^{2} + \frac{40}{3} x + \frac{1625}{144} = 0

F i n d r o o t s . (T w o a r e r e a l a n d t w o

a r e c o m p l e x) ★

Commented by Frix last updated on 17/Dec/23

I use the common method. x^4 +px^2 +qx+r=0 matching with (x^2 −ux−v)(x^2 +ux−w)=0 leads to u=(√(s−((2p)/3))) where s is a solution of s^3 −(((p^2 +12r)s)/3)−((2p^3 −72pr+27q^2 )/(27))=0 v=−(p/2)−(q/(2u))+(u^2 /2) w=−(p/2)+(q/(2u))+(u^2 /2) [if there′s no useable s better approximate] In the given case p=−((17)/(18)) q=((40)/3) r=((1625)/(144)) ⇒ s=((226)/(27)) u=3 v=−((25)/4) w=−((65)/(36)) ⇒ x=−((13)/6)∨x=−(5/6)∨=(3/2)±2i How does your new method work?

I use the common method .

x^{4} + {p x}^{2} + q x + r = 0

matching with

(x^{2} - u x - v) (x^{2} + u x - w) = 0

leads to

u = \sqrt{s - \frac{2 p}{3}}

where s is a solution of

s^{3} - \frac{(p^{2} + 12 r) s}{3} - \frac{2 p^{3} - 72 p r + 27 q^{2}}{27} = 0

v = - \frac{p}{2} - \frac{q}{2 u} + \frac{u^{2}}{2}

w = - \frac{p}{2} + \frac{q}{2 u} + \frac{u^{2}}{2}

[if {there}^{'} s no useable s better approximate]

In the given case

p = - \frac{17}{18} q = \frac{40}{3} r = \frac{1625}{144}

\Rightarrow

s = \frac{226}{27}

u = 3 v = - \frac{25}{4} w = - \frac{65}{36}

\Rightarrow

x = - \frac{13}{6} \lor x = - \frac{5}{6} \lor = \frac{3}{2} \pm 2 i

How does your new method work ?

Answered by ajfour last updated on 17/Dec/23

x^4 +bx^2 +cx+d=0 (2x^2 +px+h)(2x^2 −px+k)=3x^2 (x^2 +m) ⇒ p^2 {(p^2 +b)^2 −4d}=c^2 say p^2 =p_i ^2 (if three roots i=0,1,2) or p^2 =p_0 ^2 (if one root real) p^2 +m=−b x=((±p_0 )/2)±(√(−((c/(±2p_0 ))+(p^2 /4)+(b/2)))) All four roots obtained from any p=±p_i . This example p_0 =±3 (x∓(3/2))^2 =−[(1/(±6))(((40)/3))+(9/4)−((17)/(36))] (x∓(3/2))^2 =(4/9), −4 This method really never fails. because at least one p_i ^2 >0 always. As p^2 {(p^2 +b)^2 −4d}−c^2 =0 Its been 2 nights, i developed this.

x^{4} + {b x}^{2} + c x + d = 0

(2 x^{2} + p x + h) (2 x^{2} - p x + k) = 3 x^{2} (x^{2} + m)

\Rightarrow p^{2} {{(p^{2} + b)}^{2} - 4 d} = c^{2}

s a y p^{2} = p_{i}^{2} (i f t h r e e r o o t s i = 0, 1, 2)

o r p^{2} = p_{0}^{2} (i f o n e r o o t r e a l)

p^{2} + m = - b

x = \frac{\pm p_{0}}{2} \pm \sqrt{- (\frac{c}{\pm 2 p_{0}} + \frac{p^{2}}{4} + \frac{b}{2})}

A l l f o u r r o o t s o b t a i n e d f r o m a n y

p = \pm p_{i} .

T h i s e x a m p l e p_{0} = \pm 3

{(x \mp \frac{3}{2})}^{2} = - [\frac{1}{\pm 6} (\frac{40}{3}) + \frac{9}{4} - \frac{17}{36}]

{(x \mp \frac{3}{2})}^{2} = \frac{4}{9}, - 4

T h i s m e t h o d r e a l l y n e v e r f a i l s .

b e c a u s e a t l e a s t o n e p_{i}^{2} > 0 a l w a y s . A s

p^{2} {{(p^{2} + b)}^{2} - 4 d} - c^{2} = 0

I t s b e e n 2 n i g h t s, i d e v e l o p e d t h i s .

Commented by Frix last updated on 17/Dec/23

From my version: s^3 −(p^2 +12r)s/3−(2p^3 −72pr+27q^2)/27=0 u=√(s−2p/3) u^6 +2pu^4 +(p^2 −4r)u^2 −q^2 =0 From your version: p^2 ((p^2 +b)^2 −4d)−c^2 =0 p^6 +2bp^4 +(b^2 −4d)p^2 −c^2 =0 ...so it′s basically the same... ��

Answered by ajfour last updated on 17/Dec/23

Commented by ajfour last updated on 17/Dec/23

Shouldn't i be awarded for this?

Facebook Tweet Pin

x-4-17-18-x-2-40-3-x-1625-144-0-Find-roots-Two-are-real-and-two-are-complex-

Leave a Reply Cancel reply