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Question Number 141289 by ajfour last updated on 17/May/21

Commented by ajfour last updated on 17/May/21

$x^{3} - x - c = 0 . T o f i n d e x c e s s!$
	Answered by ajfour last updated on 18/May/21
	$y=x^3 −x R^2 =x^2 +y^2 (R^2 −c^2 )(R^2 −c^2 −1)^2 =c^2 (R^2 )^3 −(3c^2 +2)(R^2 )^2 +{(c^2 +1)^2 +2c^2 (c^2 +1)}R^2 −c^2 (c^2 +1)^2 −c^2 =0 let roots of x^3 −x=c be α, −β, −γ and corresponding radii p, q, r. p^2 =α^2 +c^2 q^2 =β^2 +c^2 r^2 =γ^2 +c^2 2prcos (θ+δ)+(α−γ)^2 +4c^2 = p^2 +r^2 pcos θ=α, rcos δ=γ psin θ=c, rsin δ=c ⇒ 2pr(((αγ−c^2 )/(pr)))+(α−γ)^2 =p^2 +r^2 −4c^2 ⇒ 2(αγ−c^2 )+(α−γ)^2 = p^2 +r^2 −4c^2 p^2 +q^2 +r^2 =3c^2 +2 p^2 r^2 +q^2 (p^2 +r^2 )=(c^2 +1)(3c^2 +1) p^2 q^2 r^2 = c^2 (c^2 +1)^2 +c^2 And αγ+1=(α+γ)^2 (α−γ)^2 =1−3γα αγ(α+γ)=c ⇒ αγ+(αγ)^2 =c(α+γ) ⇒ p^2 +r^2 =2c^2 +1−αγ α^2 +γ^2 +αγ=1 ...$
	$y = x^{3} - x$ $R^{2} = x^{2} + y^{2}$ $(R^{2} - c^{2}) {(R^{2} - c^{2} - 1)}^{2} = c^{2}$ ${(R^{2})}^{3} - (3 c^{2} + 2) {(R^{2})}^{2}$ $+ {{(c^{2} + 1)}^{2} + 2 c^{2} (c^{2} + 1)} R^{2}$ $- c^{2} {(c^{2} + 1)}^{2} - c^{2} = 0$ $l e t r o o t s o f x^{3} - x = c$ $b e α, - β, - γ a n d c o r r e s p o n d i n g$ $r a d i i p, q, r .$ $p^{2} = α^{2} + c^{2}$ $q^{2} = β^{2} + c^{2}$ $r^{2} = γ^{2} + c^{2}$ $2 p r \cos (θ + δ) + {(α - γ)}^{2} + 4 c^{2}$ $= p^{2} + r^{2}$ $p \cos θ = α, r \cos δ = γ$ $p \sin θ = c, r \sin δ = c$ $\Rightarrow 2 p r (\frac{α γ - c^{2}}{p r}) + {(α - γ)}^{2}$ $= p^{2} + r^{2} - 4 c^{2}$ $\Rightarrow 2 (α γ - c^{2}) + {(α - γ)}^{2}$ $= p^{2} + r^{2} - 4 c^{2}$ $p^{2} + q^{2} + r^{2} = 3 c^{2} + 2$ $p^{2} r^{2} + q^{2} (p^{2} + r^{2}) = (c^{2} + 1) (3 c^{2} + 1)$ $p^{2} q^{2} r^{2} = c^{2} {(c^{2} + 1)}^{2} + c^{2}$ $A n d α γ + 1 = {(α + γ)}^{2}$ ${(α - γ)}^{2} = 1 - 3 γ α$ $α γ (α + γ) = c$ $\Rightarrow α γ + {(α γ)}^{2} = c (α + γ)$ $\Rightarrow p^{2} + r^{2} = 2 c^{2} + 1 - α γ$ $α^{2} + γ^{2} + α γ = 1$ $. . .$
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