4
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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

4
$\begingroup$

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 : $$ x=-2\pm \sqrt{3}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.