x^3 +px+q = 0 If equation has all its roots real, find them.


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Question Number 43756 by ajfour last updated on 15/Sep/18

$x^{3} + p x + q = 0$ $I f e q u a t i o n h a s a l l i t s r o o t s$ $r e a l, f i n d t h e m .$
	Answered by ajfour last updated on 15/Sep/18

	$A p o o r s e l f a t t e m p t :$ $________________________$ $3 \sin θ - 4 \sin^{3} θ = \sin 3 θ$ $l e t \sin θ = m x$ $k s u c h t h a t i t m a k e s t h i s e q u a t i o n$ $e q u i v a l e n t t o t h a t i n q u e s t i o n .$ $\Rightarrow 3 m x - 4 m^{3} x^{3} = \sin 3 θ$ $o r x^{3} - \frac{3 x}{4 m^{2}} - \frac{\sin 3 θ}{4 m^{3}} = 0$ $\Rightarrow p = - \frac{3}{4 m^{2}} \Rightarrow m = \pm \sqrt{\frac{- 3}{4 p}}$ $q = - \frac{\sin 3 θ}{4 m^{3}}$ $o r \sin (3 θ - 2 k π) = - 4 m^{3} q$ $\sin θ = \sin [\frac{2 k π}{3} + \frac{1}{3} \sin^{- 1} (\pm \frac{3 q}{p} \sqrt{\frac{- 3}{4 p}})]$ $\Rightarrow \pm \sqrt{\frac{- 3}{4 p}} x = \sin [\frac{2 k π}{3} + \frac{1}{3} \sin^{- 1} (\pm \frac{3 q}{p} \sqrt{\frac{- 3}{4 p}})]$ $\Rightarrow x = \pm \sqrt{\frac{- 4 p}{3}} \sin [\frac{2 k π}{3} \pm \frac{1}{3} \sin^{- 1} (\frac{3 q}{p} \sqrt{\frac{- 3}{4 p}})]$ $u n l e s s i k n o w h o w t o c h o o s e$ $t h e s i g n s, i m i g h t o b t a i n 12 r o o t s .$ $p l e a s e h e l p . .$
	Commented by MJS last updated on 15/Sep/18

	$x^{3} + p x + q = 0$ $3 real roots \neq 0 \Leftrightarrow {(\frac{q}{2})}^{2} + {(\frac{p}{3})}^{3} < 0 \Rightarrow$ $\Rightarrow p < 0 \land 27 q^{2} + 4 p^{3} < 0 \Rightarrow ∣ \frac{27 q^{2}}{4 p^{3}} ∣< 1$ $we put x = a z, a = 2 \sqrt{- \frac{p}{3}}$ $a^{3} z^{3} + a p z + q = 0$ $z^{3} + \frac{p}{a^{2}} z + \frac{q}{a^{3}} = 0$ $a^{2} = - \frac{4 p}{3} a^{3} = - \frac{8 p}{3} \sqrt{- \frac{p}{3}}$ $z^{3} - \frac{3}{4} z - \frac{3 q}{8 p} \sqrt{- \frac{3}{p}} = 0$ $4 z^{3} - 3 z - \frac{3 q}{2} \sqrt{- \frac{3}{p}} = 0$ ${(\frac{3 q}{2} \sqrt{- \frac{3}{p}})}^{2} =∣ \frac{27 q^{2}}{4 p^{3}} ∣< 1 \Rightarrow - 1 < \frac{3 q}{2} \sqrt{- \frac{3}{p}} < 1 \Rightarrow$ $\Rightarrow we put - \frac{3 q}{2} \sqrt{- \frac{3}{p}} = \sin 3 α$ $z^{3} - \frac{3}{4} z + \frac{\sin 3 α}{4} = 0$ $if we put a = - 2 \sqrt{- \frac{p}{3}} we get the same equation$ $by putting \frac{3 q}{2} \sqrt{- \frac{3}{p}} = \sin 3 α$
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