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Question Number 129320 by ajfour last updated on 14/Jan/21

Commented by gaferrafie last updated on 14/Jan/21
سلام چطور میشه تو خروجی آدرس سایت را حذف کرد
	Answered by ajfour last updated on 14/Jan/21
	$y=(x−p)(x^3 −x−c) =x^4 −px^3 −x^2 +(p−c)x+cp (dy/dx)=4x^3 −3px^2 −2x+p−c (d^2 y/dx^2 )=12x^2 −6px−2 y=p−kx^2 let at x=α, (d^2 y/dx^2 )=0 and y=0 ⇒ 12x^2 −6kx^3 −2=0 ⇒ 6x^2 −3k(x+c)−1=0 ⇒ 6x^2 −3kx−(3ck+1)=0 ..(i) 6(x+c)−3kx^2 −(3ck+1)x=0 ⇒ 3kx^2 +(3ck−5)x−6c=0 ..(ii) ⇒ 2(ii)−k(i) gives (6ck−10+3k^2 )x=12c−k(3ck+1) ⇒ x=((12c−k(3ck+1))/(3k^2 +6ck−10)) 6{12c−k(3ck+1)}^2 −3k{12c−k(3ck+1)}(3k^2 +6ck−10) −(3ck+1)(3k^2 +6ck−10)^2 =0 ......$
	$y = (x - p) (x^{3} - x - c)$ $= x^{4} - {p x}^{3} - x^{2} + (p - c) x + c p$ $\frac{d y}{d x} = 4 x^{3} - 3 {p x}^{2} - 2 x + p - c$ $\frac{d^{2} y}{{d x}^{2}} = 12 x^{2} - 6 p x - 2$ $y = p - {k x}^{2}$ $l e t a t x = α, \frac{d^{2} y}{{d x}^{2}} = 0 a n d y = 0$ $\Rightarrow 12 x^{2} - 6 {k x}^{3} - 2 = 0$ $\Rightarrow 6 x^{2} - 3 k (x + c) - 1 = 0$ $\Rightarrow 6 x^{2} - 3 k x - (3 c k + 1) = 0 . . (i)$ $6 (x + c) - 3 {k x}^{2} - (3 c k + 1) x = 0$ $\Rightarrow 3 {k x}^{2} + (3 c k - 5) x - 6 c = 0 . . (i i)$ $\Rightarrow 2 (i i) - k (i) g i v e s$ $(6 c k - 10 + 3 k^{2}) x = 12 c - k (3 c k + 1)$ $\Rightarrow x = \frac{12 c - k (3 c k + 1)}{3 k^{2} + 6 c k - 10}$ $6 {12 c - k (3 c k + 1)}^{2}$ $- 3 k {12 c - k (3 c k + 1)} (3 k^{2} + 6 c k - 10)$ $- (3 c k + 1) {(3 k^{2} + 6 c k - 10)}^{2} = 0$ $. . . . . .$
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