Find-the-real-root-of-x-2-1-x-c-

Question Number 26235 by ajfour last updated on 22/Dec/17

F i n d t h e r e a l r o o t o f

x^{2} + \frac{1}{x} = c .

Answered by mrW1 last updated on 22/Dec/17

x \neq 0

x^{3} - c x + 1 = 0

Δ = {(\frac{1}{2})}^{2} + {(- \frac{c}{3})}^{3} = \frac{1}{4} - \frac{c^{3}}{27}

Δ = 0 \Rightarrow c = \frac{3}{^{3} \sqrt{4}}

c a s e 1 : i f c = \frac{3}{^{3} \sqrt{4}}, t w o r e a l s o l u t i o n s .

x_{1} = 2^{3} \sqrt{- \frac{1}{2}} = - \frac{2}{^{3} \sqrt{2}} \approx - 1.587

x_{2} = -^{3} \sqrt{- \frac{1}{2}} = \frac{1}{^{3} \sqrt{2}} \approx 0.794

c a s e 2 : i f c < \frac{3}{^{3} \sqrt{4}}, o n e r e a l s o l u t i o n .

x =^{3} \sqrt{- \frac{1}{2} + \sqrt{Δ}} -^{3} \sqrt{\frac{1}{2} + \sqrt{Δ}}

e . g . c = 1 \Rightarrow Δ = \frac{1}{4} - \frac{1}{27} = \frac{23}{108}

x =^{3} \sqrt{- \frac{1}{2} + \sqrt{\frac{23}{108}}} -^{3} \sqrt{\frac{1}{2} + \sqrt{\frac{23}{108}}} \approx - 1.324

c a s e 3 : i f c > \frac{3}{^{3} \sqrt{4}}, t h r e e r e a l s o l u t i o n s .

e . g . c = 3,

r = \sqrt{\frac{c^{3}}{27}} = 1

\cos \emptyset = - \frac{1}{2 r} = - \frac{1}{2}

\Rightarrow \emptyset = \frac{2 π}{3}

x_{1} = 2^{3} \sqrt{r} \cos \frac{\emptyset}{3} = 2 \cos \frac{2 π}{9} \approx 1.532

x_{2} = 2^{3} \sqrt{r} \cos \frac{\emptyset + 2 π}{3} = 2 \cos \frac{8 π}{9} \approx - 1.879

x_{3} = 2^{3} \sqrt{r} \cos \frac{\emptyset + 4 π}{3} = 2 \cos \frac{14 π}{9} \approx 0.347

Commented by ajfour last updated on 23/Dec/17

p l e a s e s i r, i d o n t f o l l o w t h i s .

H e l p m e d e r i v e f o r m u l a f o r t h i s .

h o w d o y o u o b t a i n Δ a s a f u n c t i o n

o f c ? p l e a s e s e e Q . 26251

s o m e c o n s r r u c t i o n m e t h o d

s h o u l d b e t h e r e . M o r e o v e r a l l

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

t o t h i s c u b i c e q u a t i o n, i t h i n k . .

Commented by mrW1 last updated on 23/Dec/17

Y e s, e v e r y c u b i c e q u a t i o n c a n b e

r e d u c e d t o t h e f o r m

x^{3} + p x + q = 0 \dots (i)

l e t v = \frac{p}{3 u}, u^{3} - v^{3} = - q

a n d x = u - v

(i) w i l l b e c o m e

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

o r

{(u^{3})}^{2} + q (u^{3}) - {(\frac{p}{3})}^{3} = 0 . . (i i)

\Rightarrow u^{3} = \frac{- q \pm \sqrt{q^{2} + 4 {(\frac{p}{3})}^{3}}}{2}

= - \frac{q}{2} \pm \sqrt{{(\frac{q}{2})}^{2} + {(\frac{p}{3})}^{3}}

h o w t h e s o l u t i o n s w i l l b e d e p e n d s o n

Δ = {(\frac{q}{2})}^{2} + {(\frac{p}{3})}^{3}

o n e c a s e i s e . g . :

\Rightarrow u =^{3} \sqrt{- \frac{q}{2} \pm \sqrt{{(\frac{q}{2})}^{2} + {(\frac{p}{3})}^{3}}}

v^{3} = u^{3} + q = - \frac{q}{2} \pm \sqrt{{(\frac{q}{2})}^{2} + {(\frac{p}{3})}^{3}} + q

v^{3} = \frac{q}{2} \pm \sqrt{{(\frac{q}{2})}^{2} + {(\frac{p}{3})}^{3}}

\Rightarrow v =^{3} \sqrt{\frac{q}{2} \pm \sqrt{{(\frac{q}{2})}^{2} + {(\frac{p}{3})}^{3}}}

x = u - v

\Rightarrow x =^{3} \sqrt{- \frac{q}{2} \pm \sqrt{{(\frac{q}{2})}^{2} + {(\frac{p}{3})}^{3}}} -^{3} \sqrt{\frac{q}{2} \pm \sqrt{{(\frac{q}{2})}^{2} + {(\frac{p}{3})}^{3}}}

Commented by ajfour last updated on 23/Dec/17

g r e a t f o r t u n e s i r . t h a n k y o u

v e r y s i n c e r e l y!

Commented by ajfour last updated on 10/Jun/18

m y f a v o u r i t e p o s t .

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