x^3 −bx−c=0 ; b, c >0 ; ((b/3))^3 >((c/2))^2 To find the three real roots without the use of trigonometric solution to cubic polynomial...


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Question Number 102490 by ajfour last updated on 09/Jul/20

$x^{3} - b x - c = 0; b, c > 0; {(\frac{b}{3})}^{3} > {(\frac{c}{2})}^{2}$ $T o f i n d t h e t h r e e r e a l r o o t s w i t h o u t$ $t h e u s e o f t r i g o n o m e t r i c s o l u t i o n$ $t o c u b i c p o l y n o m i a l . . .$
	Answered by ajfour last updated on 10/Jul/20
	$let z=(x/(√b)) , a=(c/(b(√b))) As (c/2)<((b(√b))/(3(√3))) ⇒ a<(2/(3(√3))) ⇒ z^3 −z−a=0 Now let (z−1)(z^3 −z−a)=0 z^4 −z^3 −z^2 +(1−a)z+a=0 Now let z=((t+s)/(t+1)) ⇒ (t^4 +4st^3 +6s^2 t^2 +4s^3 t+s^4 ) −(t+1)(t^3 +3st^2 +3s^2 t+s^3 ) −(t^2 +2t+1)(t^2 +2st+s^2 ) +(1−a)(t+s)(t^3 +3t^2 +3t+1) +a(t^4 +4t^3 +6t^2 +4t+1)=0 ⇒ t^3 {4s−3s−1−2s−2+3(1−a) +s(1−a)+4a}+ t^2 {6s^2 −3s^2 −3s−s^2 −4s−1 +3(1−a)+3s(1−a)+6a}+ t{4s^3 −s^3 −3s^2 −2s^2 −2s +(1−a)+3s(1−a)+4a}+ {s^4 −s^3 −s^2 +s(1−a)+a} = 0 ⇒ t^3 {a(1−s)}+t^2 {2s^2 −s(4+3a)+3a+2} +t{s^3 −5s^2 −(3a−1)s+3a+1} +{s^4 −s^3 −s^2 +s(1−a)+a}=0 Now let 2s^2 −s(4+3a)+3a+2=0 ⇒ s=((3a+4)/2)±((√((3a+4)^2 −2(3a+2)))/2) = ((3a+4±(√((3a+4)^2 −2(3a+4)+4)))/2) =((3(a+1)+1±(√(9(a+1)^2 +3)))/2) Now t^3 +t{((s^3 −5s^2 −(3a−1)s+3a+1)/(a(1−s)))} +{((s^4 −s^3 −s^2 +s(1−a)+a)/(a(1−s)))}= 0 ⇒ t^3 +t{((s^3 −5s^2 −(3a−1)s+3a+1)/(a(1−s)))} −(1/a)(s^3 −s−a) = 0 ⇒ t^3 +(t/a){(s+3)^2 +3a+4(((2−s)/(s−1)))} −(1/a)(s^3 −s−a) = 0 lets say above expression is t^3 +((p/a))t−(q/a)=0 we can apply Cardano′s formula if p>0 ; say t=t_0 , then z_0 =((t_0 +s)/(t_0 +1)) and x=(√b)(z_0 ) ....... lets check the sign of p = (s+3)^2 +3a+4(((2−s)/(s−1))) where s =((3(a+1)+1±(√(9(a+1)^2 +3)))/2) and a<(2/(3(√3))) (someone please help here... can p be >0 for one of the two values of s ? )$
	$l e t z = \frac{x}{\sqrt{b}}, a = \frac{c}{b \sqrt{b}}$ $A s \frac{c}{2} < \frac{b \sqrt{b}}{3 \sqrt{3}} \Rightarrow a < \frac{2}{3 \sqrt{3}}$ $\Rightarrow z^{3} - z - a = 0$ $N o w l e t (z - 1) (z^{3} - z - a) = 0$ $z^{4} - z^{3} - z^{2} + (1 - a) z + a = 0$ $N o w l e t z = \frac{t + s}{t + 1} \Rightarrow$ $(t^{4} + 4 {s t}^{3} + 6 s^{2} t^{2} + 4 s^{3} t + s^{4})$ $- (t + 1) (t^{3} + 3 {s t}^{2} + 3 s^{2} t + s^{3})$ $- (t^{2} + 2 t + 1) (t^{2} + 2 s t + s^{2})$ $+ (1 - a) (t + s) (t^{3} + 3 t^{2} + 3 t + 1)$ $+ a (t^{4} + 4 t^{3} + 6 t^{2} + 4 t + 1) = 0$ $\Rightarrow$ $t^{3} {4 s - 3 s - 1 - 2 s - 2 + 3 (1 - a)$ $+ s (1 - a) + 4 a} +$ $t^{2} {6 s^{2} - 3 s^{2} - 3 s - s^{2} - 4 s - 1$ $+ 3 (1 - a) + 3 s (1 - a) + 6 a} +$ $t {4 s^{3} - s^{3} - 3 s^{2} - 2 s^{2} - 2 s$ $+ (1 - a) + 3 s (1 - a) + 4 a} +$ ${s^{4} - s^{3} - s^{2} + s (1 - a) + a} = 0$ $\Rightarrow$ $t^{3} {a (1 - s)} + t^{2} {2 s^{2} - s (4 + 3 a) + 3 a + 2}$ $+ t {s^{3} - 5 s^{2} - (3 a - 1) s + 3 a + 1}$ $+ {s^{4} - s^{3} - s^{2} + s (1 - a) + a} = 0$ $N o w l e t 2 s^{2} - s (4 + 3 a) + 3 a + 2 = 0$ $\Rightarrow s = \frac{3 a + 4}{2} \pm \frac{\sqrt{{(3 a + 4)}^{2} - 2 (3 a + 2)}}{2}$ $= \frac{3 a + 4 \pm \sqrt{{(3 a + 4)}^{2} - 2 (3 a + 4) + 4}}{2}$ $= \frac{3 (a + 1) + 1 \pm \sqrt{9 {(a + 1)}^{2} + 3}}{2}$ $N o w$ $t^{3} + t {\frac{s^{3} - 5 s^{2} - (3 a - 1) s + 3 a + 1}{a (1 - s)}}$ $+ {\frac{s^{4} - s^{3} - s^{2} + s (1 - a) + a}{a (1 - s)}} = 0$ $\Rightarrow t^{3} + t {\frac{s^{3} - 5 s^{2} - (3 a - 1) s + 3 a + 1}{a (1 - s)}}$ $- \frac{1}{a} (s^{3} - s - a) = 0$ $\Rightarrow t^{3} + \frac{t}{a} {{(s + 3)}^{2} + 3 a + 4 (\frac{2 - s}{s - 1})}$ $- \frac{1}{a} (s^{3} - s - a) = 0$ $l e t s s a y a b o v e e x p r e s s i o n i s$ $t^{3} + (\frac{p}{a}) t - \frac{q}{a} = 0$ $w e c a n a p p l y {C a r d a n o}^{'} s f o r m u l a i f$ $p > 0; s a y t = t_{0}, t h e n$ $z_{0} = \frac{t_{0} + s}{t_{0} + 1} a n d x = \sqrt{b} (z_{0})$ $. . . . . . .$ $l e t s c h e c k t h e s i g n o f$ $p = {(s + 3)}^{2} + 3 a + 4 (\frac{2 - s}{s - 1}) w h e r e$ $s = \frac{3 (a + 1) + 1 \pm \sqrt{9 {(a + 1)}^{2} + 3}}{2}$ $a n d a < \frac{2}{3 \sqrt{3}}$ $(s o m e o n e p l e a s e h e l p h e r e . . .$ $c a n p b e > 0 f o r o n e o f t h e t w o$ $v a l u e s o f s ?)$
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