Find-the-cardano-s-solution-of-the-equation-28x-3-9x-2-1-0-

Question Number 200460 by faysal last updated on 19/Nov/23

Find the {cardano}^{'} s solution of the

equation {28 x}^{3} - {9 x}^{2} + 1 = 0

Answered by Frix last updated on 19/Nov/23

28 x^{3} - 9 x^{2} + 1 = 0

x^{3} - \frac{9 x^{2}}{28} + \frac{1}{28} = 0

Before using any other method we try all

factors of \pm \frac{1}{28} \Rightarrow x_{1} = - \frac{1}{4} . We now have

(x + \frac{1}{4}) (x^{2} - \frac{4 x}{7} + \frac{1}{7}) = 0

\Rightarrow x_{2, 3} = \frac{2}{7} \pm \frac{\sqrt{3}}{7} i

Cardano means more work :

28 x^{3} - 9 x^{2} + 1 = 0

x^{3} - \frac{9 x^{2}}{28} + \frac{1}{28} = 0

x = t + \frac{3}{28}

t^{3} - \frac{27 t}{784} + \frac{365}{10976} = 0 \Rightarrow p = - \frac{27}{784} \land q = \frac{365}{10976}

u = \sqrt[3]{- \frac{q}{2} + \sqrt{\frac{q^{2}}{4} + \frac{p^{3}}{27}}} \land v = \sqrt[3]{- \frac{q}{2} - \sqrt{\frac{q^{2}}{4} + \frac{p^{3}}{27}}}

t_{1} = u + v

t_{2} = ω u + ω^{2} v

t_{3} = ω^{2} u + ω v

\dots

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