Question-91744

Question Number 91744 by ajfour last updated on 02/May/20

Commented by ajfour last updated on 02/May/20

I f t h e r e d c u b i c c u r v e h a s e q u a t i o n

y = x^{3} + {a x}^{2} + b x + c w h i l e t h e

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

o n e, s h i f t e d s u c h t h a t t w o r o o t s

a r e s t i l l c o m m o n, f i n d e q u a t i o n

o f t h e b l u e o n e i n t e r m s o f

a, b, c . x, y .

Answered by mr W last updated on 02/May/20

y = x^{3} + {a x}^{2} + b x + c

w i t h

p = \frac{3 b - a^{2}}{9}

q = \frac{2 a^{3} - 9 a b + 27 c}{54}

a n d

p^{3} + q^{2} < 0 (s u c h t h a t t h r e e r e a l r o o t s)

w e g e t t h e t h r e e r o o t s

α = 2 \sqrt{- p} \sin (\frac{1}{3} \sin^{- 1} \frac{q}{\sqrt{- p^{3}}} - \frac{2 π}{3}) - \frac{a}{3}

β = 2 \sqrt{- p} \sin (\frac{1}{3} \sin^{- 1} \frac{q}{\sqrt{- p^{3}}}) - \frac{a}{3}

γ = 2 \sqrt{- p} \sin (\frac{1}{3} \sin^{- 1} \frac{q}{\sqrt{- p^{3}}} + \frac{2 π}{3}) - \frac{a}{3}

t h e e q n . o f s h i f t e d c u b i c c u r v e i s

y - k = {(x - h)}^{3} + a {(x - h)}^{2} + b (x - h) + c

o r

y = {(x - h)}^{3} + a {(x - h)}^{2} + b (x - h) + c + k

s i n c e β a n d γ a r e a l s o i t s r o o t s,

{(β - h)}^{3} + a {(β - h)}^{2} + b (β - h) + c + k = 0 . . (i)

{(γ - h)}^{3} + a {(γ - h)}^{2} + b (γ - h) + c + k = 0 . . (i i)

(i) - (i i) :

3 h^{2} - [2 a + 3 (β + γ)] h + [β^{2} + γ^{2} + β γ + a (β + γ) + b] = 0

\Rightarrow h = \frac{2 a + 3 (β + γ) + \sqrt{4 a^{2} - 12 b - 3 {(β - γ)}^{2}}}{6}

\Rightarrow k = - {(β - h)}^{3} - a {(β - h)}^{2} - b (β - h) - c

Commented by mr W last updated on 02/May/20

Commented by mr W last updated on 03/May/20

s i r, p l e a s e a l s o h a v e a l o o k a t Q 91446 .

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