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Question Number 192780 by ajfour last updated on 27/May/23

Commented by ajfour last updated on 27/May/23
Find m in terms of c, hence p.
	Answered by ajfour last updated on 27/May/23
	$y=x^3 −x−c 0=p^3 −p−c y=mx y=−x^2 +hx+1−h c=−p^2 +hp+1−h p^3 =p+c p+c=h(hp+1−h−c)+(1−h−c)p ⇒ {h^2 −(h+c)}p=h^2 −(1−c)h+c x^2 +(m−h)x=1−h double root ⇒ (m−h)^2 =4(h−1) ⇒ h^2 −4h(m+1)+m^2 +4=0 (h−2m−2)^2 =m(3m+8) {((h^2 −(1−c)h+c)/(h^2 −h−c))}^2 =h{((h^2 −(1−c)h+c)/(h^2 −h−c))} +1−h−c ⇒ {(((4m+3+c)h+c)/((4m+3)h−c))}^2 =h{(((4m+3+c)h+c)/((4m+3)h−c))}+1−h−c ....$
	$y = x^{3} - x - c$ $0 = p^{3} - p - c$ $y = m x$ $y = - x^{2} + h x + 1 - h$ $c = - p^{2} + h p + 1 - h$ $p^{3} = p + c$ $p + c = h (h p + 1 - h - c) + (1 - h - c) p$ $\Rightarrow {h^{2} - (h + c)} p = h^{2} - (1 - c) h + c$ $x^{2} + (m - h) x = 1 - h$ $d o u b l e r o o t \Rightarrow$ ${(m - h)}^{2} = 4 (h - 1)$ $\Rightarrow h^{2} - 4 h (m + 1) + m^{2} + 4 = 0$ ${(h - 2 m - 2)}^{2} = m (3 m + 8)$ ${\frac{h^{2} - (1 - c) h + c}{h^{2} - h - c}}^{2} = h {\frac{h^{2} - (1 - c) h + c}{h^{2} - h - c}}$ $+ 1 - h - c$ $\Rightarrow {\frac{(4 m + 3 + c) h + c}{(4 m + 3) h - c}}^{2}$ $= h {\frac{(4 m + 3 + c) h + c}{(4 m + 3) h - c}} + 1 - h - c$ $. . . .$
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