y=x^3 −x−c ; find the roots. 0≤c≤(2/(3(√3)))


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Question Number 140604 by ajfour last updated on 10/May/21

$y = x^{3} - x - c; f i n d t h e r o o t s .$ $0 ⩽ c ⩽ \frac{2}{3 \sqrt{3}}$
	Answered by ajfour last updated on 11/May/21

	Commented by ajfour last updated on 11/May/21

	$A E = 1 = E H$ $E P = x = t a n θ$ $E Q = x^{2} = {t a n}^{2} θ$ $H Q = x^{2} - 1$ $B H = c (n o t = 2 c a s i n i m a g e)$ $∠ B H Q = θ$ $\Rightarrow \frac{B H}{H Q} = t a n θ = \frac{c}{x^{2} - 1} = x$ $\Rightarrow x (x^{2} - 1) = c$ $l e t s f i n d o n e x i n t e r m s o f c .$ $A l s o \Rightarrow x^{2} - 1 = \frac{c}{x}$ $L e t c i r c u m r a d i u s ρ$ $O E = \frac{c}{2}$ $\Rightarrow ρ^{2} = \frac{c^{2}}{4} + x^{4}$ $a l s o ρ = \frac{c}{2} + x$ $\Rightarrow {(ρ - \frac{c}{2})}^{4} + \frac{c^{2}}{4} = ρ^{2}$ $l e t ρ - \frac{c}{2} = s; \frac{c}{2} = k$ $\Rightarrow s^{4} + k^{2} = s^{2} + 2 k s + k^{2}$ $\Rightarrow s^{3} - s - 2 k = 0$ $\Rightarrow s^{3} - s - c = 0$ $. . . . . . . . .$
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