Developed the formula for the cubic equation when it has three real roots and Cardano is just not suitable as discriminant gets negative.. Given x^3 −bx−c=0 with b>0 , c>0 (t/b)=1+((15)/2)((c^2 /b^3 ))+(√(((c^2 /b^3 ))[49−((75)/4)((c^2 /b^3 ))])) x=(√(t+((4c^2 )/b^2 ))) −((3c)/b) ★ mrW Sir, MjS Sir please check!


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Question Number 114388 by ajfour last updated on 18/Sep/20

$D e v e l o p e d t h e f o r m u l a f o r t h e c u b i c$ $e q u a t i o n w h e n i t h a s t h r e e r e a l r o o t s$ $a n d C a r d a n o i s j u s t n o t s u i t a b l e a s$ $d i s c r i m i n a n t g e t s n e g a t i v e . .$ $G i v e n x^{3} - b x - c = 0$ $w i t h b > 0, c > 0$ $\frac{t}{b} = 1 + \frac{15}{2} (\frac{c^{2}}{b^{3}}) + \sqrt{(\frac{c^{2}}{b^{3}}) [49 - \frac{75}{4} (\frac{c^{2}}{b^{3}})]}$ $x = \sqrt{t + \frac{4 c^{2}}{b^{2}}} - \frac{3 c}{b} ★$ $m r W S i r, M j S S i r p l e a s e c h e c k!$
Commented byMJS_new last updated on 18/Sep/20
$it′s not exact x^3 −((244)/3)x−((1456)/(27))=0 ⇒ x∈{−((26)/3), −(2/3), ((28)/3)} but your formula gives x≈9.33355157 instead of 9.3^•$
${it}^{'} s not exact$ $x^{3} - \frac{244}{3} x - \frac{1456}{27} = 0 \Rightarrow x \in {- \frac{26}{3}, - \frac{2}{3}, \frac{28}{3}}$ $but your formula gives$ $x \approx 9.33355157 instead of 9 . \overset{∙}{3}$
Commented byajfour last updated on 18/Sep/20

$y e s s i r, i f o u n d t h e e r r o r! t h a n k s .$
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