x^3 +x−16=0


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Question Number 89687 by student work last updated on 18/Apr/20

$x^{3} + x - 16 = 0$
Commented by mr W last updated on 18/Apr/20
https://en.m.wikipedia.org/wiki/Cubic_equation
Commented by mr W last updated on 18/Apr/20

${i t}^{'} s a l r e a d y s o l v e d i n Q 89653 .$ $m e t h o d i s a l s o t o l d y o u . h a v e y o u t r i e d$ $t o s t u d y t h e m e t h o d ? i f y o u j u s t w a n t$ $t o k n o w t h e f o r m u l a, y o u c a n f i n d i t$ $i n i n t e r n e t . i f y o u w a n t t o u n d e r s t a n d$ $i t, y o u s h o u l d s t u d y i t t h r o u g h s o m e$ $r e a d i n g b y y o u r s e l f .$
Commented by MJS last updated on 18/Apr/20
$Cardano′s Method ax^3 +bx^2 +cx+d=0 (1) divide by a ⇒ x^3 +(b/a)x^2 +(c/a)x+(d/a)=0 (2) let z=x−(b/(3a)) ⇒ z^3 +((3ac−b^2 )/(3a^2 ))z+((27a^2 d−9abc+2b^3 )/(27a^3 ))=0 (3) let ((3ac−b^2 )/(3a^2 ))=p and ((27a^2 d−9abc+2b^3 )/(27a^3 ))=q ⇒ z^3 +pz+q=0 ⇔ z^3 =−pz−q (4) Cardano′s new idea: let z=u+v on lhs (u+v)^3 =−pz−q u^3 +3u^2 v+3uv^2 +v^3 =−pz−q 3uv(u+v)+u^3 +v^3 =−pz−q 3uvz+u^3 +v^3 =−pz−q by matching constants we get { ((I. uv=−(p/3) ⇔ u^3 v^3 =−(p^3 /(27)))),((II. u^3 +v^3 =−q)) :} (5) let u^3 =s and v^3 =t ⇒ { ((I. st=−(p^3 /(27)))),((II. s+t=−q)) :} this leads to a quadratic: I. ⇒ s=−(p^3 /(27t)); insert in II. ⇒ t−(p^3 /(27t))=−q ⇔ t^2 +qt−(p^3 /(27))=0 ⇒ ⇒ t=−(q/2)±(√((q^2 /4)+(p^3 /(27)))) ⇒ s=−(q/2)∓(√((q^2 /4)+(p^3 /(27)))) ⇒ we are free to choose the “+” value for t and the “−” value for s ⇒ u=((−(q/2)−(√((q^2 /4)+(p^3 /(27))))))^(1/3) and v=((−(q/2)+(√((q^2 /4)+(p^3 /(27))))))^(1/3) ⇒ z_1 =u+v z^3 +pz+q=(z−z_1 )(z^2 +(u+v)z+u^2 −uv+v^2 ) and we can solve this the only limitation is D=(q^2 /4)+(p^3 /(27))≥0 if D<0 we need the trigonometric solution$
${Cardano}^{'} s Method$ ${a x}^{3} + {b x}^{2} + c x + d = 0$ $(1) divide by a \Rightarrow$ $x^{3} + \frac{b}{a} x^{2} + \frac{c}{a} x + \frac{d}{a} = 0$ $(2) let z = x - \frac{b}{3 a} \Rightarrow$ $z^{3} + \frac{3 a c - b^{2}}{3 a^{2}} z + \frac{27 a^{2} d - 9 a b c + 2 b^{3}}{27 a^{3}} = 0$ $(3) let \frac{3 a c - b^{2}}{3 a^{2}} = p and \frac{27 a^{2} d - 9 a b c + 2 b^{3}}{27 a^{3}} = q \Rightarrow$ $z^{3} + p z + q = 0 \Leftrightarrow z^{3} = - p z - q$ $(4) {Cardano}^{'} s new idea : let z = u + v on lhs$ ${(u + v)}^{3} = - p z - q$ $u^{3} + 3 u^{2} v + 3 {u v}^{2} + v^{3} = - p z - q$ $3 u v (u + v) + u^{3} + v^{3} = - p z - q$ $3 u v z + u^{3} + v^{3} = - p z - q$ $by matching constants we get$ ${\begin{cases} I . u v = - \frac{p}{3} \Leftrightarrow u^{3} v^{3} = - \frac{p^{3}}{27} \\ II . u^{3} + v^{3} = - q \end{cases}$ $(5) let u^{3} = s and v^{3} = t \Rightarrow$ ${\begin{cases} I . s t = - \frac{p^{3}}{27} \\ II . s + t = - q \end{cases}$ $this leads to a quadratic :$ $I . \Rightarrow s = - \frac{p^{3}}{27 t}; insert in II . \Rightarrow$ $t - \frac{p^{3}}{27 t} = - q \Leftrightarrow t^{2} + q t - \frac{p^{3}}{27} = 0 \Rightarrow$ $\Rightarrow t = - \frac{q}{2} \pm \sqrt{\frac{q^{2}}{4} + \frac{p^{3}}{27}}$ $\Rightarrow s = - \frac{q}{2} \mp \sqrt{\frac{q^{2}}{4} + \frac{p^{3}}{27}}$ $\Rightarrow we are free to choose the ‘ ‘ +^{″} value$ $for t and the ‘ ‘ -^{″} value for s$ $\Rightarrow u = \sqrt[3]{- \frac{q}{2} - \sqrt{\frac{q^{2}}{4} + \frac{p^{3}}{27}}} and v = \sqrt[3]{- \frac{q}{2} + \sqrt{\frac{q^{2}}{4} + \frac{p^{3}}{27}}}$ $\Rightarrow z_{1} = u + v$ $z^{3} + p z + q = (z - z_{1}) (z^{2} + (u + v) z + u^{2} - u v + v^{2})$ $and we can solve this$ $the only limitation is$ $D = \frac{q^{2}}{4} + \frac{p^{3}}{27} ⩾ 0$ $if D < 0 we need the trigonometric solution$
	Answered by TANMAY PANACEA. last updated on 18/Apr/20

	$f (x) = x^{3} + x - 16$ $f (1) = - 14$ $f (0) = - 16$ $f (2) = - 6$ $f (3) = 14$ $f (4) = 54$ $n o w f (2) < 0 b u t f (3) > 0$ $s o o n e r o o t l i e s b e t w e e n 2 a n d 3$ $3 > x > 2$ $f (- 1) = - 18$ $f (- 2) = - 26$ $s o i t h i n k x^{3} + x - 16 = 0$ $h a s o n l y o n e s o l u t i o n w h e r e 3 > x > 2$
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