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As an exercise, I have to use Cardano's formula $$ x^3 = px + q$$ $$x = \sqrt[3]{\frac{q}{2}+\sqrt{\frac{q^2}{4} - \frac{p^3}{27}}} + \sqrt[3]{\frac{q}{2}-\sqrt{\frac{q^2}{4} - \frac{p^3}{27}}} $$ to solve the equation $ x^3 = 15x+4. $ I finally get $$ x = \sqrt[3]{2 + 11i} + \sqrt[3]{2 - 11i}$$ $$ x = 4 $$ but I have no idea how to find the other two solutions to the equation, and the procedures I've found googling are not employing Cardano's formula.

Thank you for your help.

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  • $\begingroup$ How did you prove that $\sqrt[3]{2 + 11i} + \sqrt[3]{2 - 11i}=4$? $\endgroup$
    – user
    Feb 22, 2018 at 12:20

1 Answer 1

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Since the degree is 3, if you have a solution, you can make a factorisation and have a second degree equation to solve : $$\begin{align*} x^3&=15x+4 \tag{1}\\ 4^3&=15\times 4 +4\tag{2} \end{align*} $$ with (1)-(2) : $$x^3-4^3=15x-15\times 4 $$ so (with $a^3-b^3=(a-b)(a^2+ab+b^2)$ ) $$ (x-4)(x^2 + 4x +16)= 15(x-4) \quad \Leftrightarrow \quad (x-4)(x^2 + 4x +16-15)= 0$$ and you just have to solve then $$ x^2 +4x +1$$ to find the others solutions : $$ x=-2\pm \sqrt{3}.$$

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