x-4-5x-2-20x-104-0-solve-for-x-

Question Number 65859 by ajfour last updated on 05/Aug/19

x^{4} + 5 x^{2} + 20 x + 104 = 0

s o l v e f o r x .

Answered by ajfour last updated on 05/Aug/19

a = 5, b = 20, c = 104

p^{6} + 2 {a p}^{4} + (a^{2} - 4 c) p^{2} - b^{2} = 0

p^{6} + 10 p^{4} - 391 p^{2} - 400 = 0

\Rightarrow p^{2} = - 25, - 1, 16

x^{2} - p x + \frac{1}{2} (p^{2} + a + \frac{b}{p}) = 0

\underline{F o r p^{2} = - 25} f i r s t l e t p = 5 i

x^{2} - 5 i x + \frac{1}{2} (- 25 + 5 - 4 i) = 0

x^{2} - 5 i x - 10 - 2 i = 0

x = \frac{5 i \pm \sqrt{- 25 + 40 + 8 i}}{2}

= \frac{5 i \pm \sqrt{15 + 8 i}}{2}

l e t {(r + m i)}^{2} = 15 + 8 i

\Rightarrow r^{2} - m^{2} + (2 m r) i = 15 + 8 i

\Rightarrow r^{2} - m^{2} = 15, m r = 4

\Rightarrow r^{2} - \frac{16}{r^{2}} = 15

\Rightarrow r^{4} - 15 r^{2} - 16 = 0

\Rightarrow r^{2} = 16, - 1 (b u t r^{2} ⩾ 0 \Rightarrow r^{2} \neq - 1)

\Rightarrow m = \pm 1

\Rightarrow r + m i = \pm (4 + i)

s o x = \frac{5 i \pm (4 + i)}{2}

\Rightarrow x_{1} = 2 + 3 i, x_{2} = - 2 + 2 i

a n d f o r p = - 5 i

x^{2} + 5 i x + \frac{1}{2} (- 25 + 5 + 4 i) = 0

x^{2} + 5 i x - 10 + 2 i = 0

x = \frac{- 5 i \pm \sqrt{- 25 + 40 - 8 i}}{2}

= \frac{- 5 i \pm \sqrt{15 - 8 i}}{2}

l e t {(r + m i)}^{2} = 15 - 8 i

\Rightarrow r^{2} - m^{2} = 15, m r = - 4

\Rightarrow r^{2} - \frac{16}{r^{2}} = 15

\Rightarrow r^{2} = 16, m = \pm 1

\Rightarrow r + m i = \pm (4 - i)

x = \frac{- 5 i \pm (4 - i)}{2}

x_{3} = 2 - 3 i, x_{4} = - 2 - 2 i

s i m i l a r l y f o r p = \pm 1

\underline{N o w f o r p = \pm 4}

x^{2} \mp 4 x + \frac{1}{2} (21 \pm 5) = 0

\Rightarrow x^{2} - 4 x + 13 = 0 & x^{2} + 4 x + 8 = 0

\Rightarrow x_{1}, x_{2} = 2 \pm 3 i & x_{3}, x_{4} = - 2 \pm 2 i

(f i n e f o r m u l a i d e v e l o p e d;

h a n d l e s a l l b i q u a d r a t i c s w i t h

r e a l c o e f f i c i e n t s r e a l w e l l,

i d o n t j u s t i m a g i n e!)

Commented by ajfour last updated on 05/Aug/19

I t s T i m e t o t r y q u i n t i c a g a i n .

Commented by mr W last updated on 05/Aug/19

{t h a t}^{'} s g r e a t!

Commented by TawaTawa last updated on 05/Aug/19

This is great sir . God bless you sir .

Commented by TawaTawa last updated on 05/Aug/19

Sir, what if we have : {mx}^{4} + {ax}^{3} + {bx}^{2} + cx + d = 0

we must divide through by m ? to make coefficient of x^{4} to be 1 ?

Commented by behi83417@gmail.com last updated on 05/Aug/19

nice method sir Ajfour . I love it .

thanks for sharing this knowldege .

Commented by ajfour last updated on 05/Aug/19

y e s w e s h o u l d d i v i d e i t b y m, t h e n .

t h a n k s b e h i s i r .

Commented by TawaTawa last updated on 05/Aug/19

Oh, great sir . Thanks so much . More knowledge and more

discovery

Commented by TawaTawa last updated on 05/Aug/19

Have not underdtand one thing sir .

How is p^{2} = - 25, - 1, 16

Commented by ajfour last updated on 05/Aug/19

s o l v i n g i t a s a c u b i c e q . i n p^{2} .

Commented by TawaTawa last updated on 05/Aug/19

God bless you sir .

Commented by ajfour last updated on 05/Aug/19

p^{6} + 10 p^{4} - 391 p^{2} - 400 = 0

l e t p^{2} = u

u^{3} + {A u}^{2} + B u + C = 0

l e t u = z - \frac{A}{3}

z^{3} - {A u}^{2} + \frac{A^{2} z}{3} - \frac{A^{3}}{27} +

{A u}^{2} - \frac{2 A^{2} z}{3} + \frac{A^{3}}{9} + B z - \frac{A B}{3} + C = 0

\Rightarrow

z^{3} + (B - \frac{A^{2}}{3}) z + (\frac{2 A^{3}}{27} - \frac{A B}{3} + C) = 0

c a l l i t z^{3} + P z + Q = 0

I f \frac{Q^{2}}{4} + \frac{P^{3}}{27} < 0

t h e r e a r e t h r e e s o l u t i o n s t o z,

c a l l e d t r i g o n o m e t r i c s o l u t i o n

t o a s t a n d a r d c u b i c; i n e v e r

c o u l d r e m e m b e r t h e r e s u l t;

s o i h a d j u s t u s e d c a l c u l a t o r

t o g e t p^{2} f r o m t h e e q . i n p^{2} .

Commented by TawaTawa last updated on 05/Aug/19

Wow, God bless you sir . But i will use factorization sir

Commented by TawaTawa last updated on 05/Aug/19

since i can know one factor by trial and error

Commented by TawaTawa last updated on 05/Aug/19

or i can use sir MrW formular posted on cubic equation

Commented by TawaTawa last updated on 05/Aug/19

Sir, in the equation, no term in x^{3} . it means the method cannot

work for {ax}^{4} + {bx}^{3} + {cx}^{2} + dx + e = 0 except {ax}^{4} + {bx}^{2} + cx + d = 0

just want to know sir .

Commented by mr W last updated on 05/Aug/19

e q n . w i t h x^{3} t e r m c a n b e t r a n s f o r m e d

t o a n e q n . w i t h o u t x^{3} t e r m u s i n g a s i m p l e

s u b s t i t u t i o n x = t - \frac{a}{4} w h e r e a i s t h e

c o e f f i c i e n t o f x^{3} t e r m . s e e Q 65830 .

Commented by TawaTawa last updated on 05/Aug/19

Just checked now sir . God bless you sir . i now understand all