How Do You Solve a Cubic Equation Using Cardano's Formula?

[cscott0001]

Homework Statement

Use Cardano's formula to find a real root for $3x^3-45x^2+243x-525=0$. [Edited to correct mistake]

Homework Equations

$$x = u - \frac{b}{3a}$$
Depressed cubic: $$u^3=3pu+2q$$
Cardano's formula: $$u=\sqrt[3]{q+\sqrt{q^2-p^3}}+\sqrt[3]{q-\sqrt{q^2-p^3}}$$

The Attempt at a Solution

I have found the depressed cubic to be $x^3=20-6x$. I found $3p= -6$ and $2q=20 \to p=-2$ and $q=10$. Using Cardano's formula, I arrived at $u=\sqrt[3]{10+\sqrt{108}}+\sqrt[3]{10-\sqrt{108}}$. However, I graphed the equation and know that it should be 7. If I change my answer to $u=\sqrt[3]{10+\sqrt{108}}-\sqrt[3]{-10+\sqrt{108}}$ and use this answer (2) to solve $x = u - \frac{b}{3a}$ I get the right answer (7), but I have no idea mathematically why I would do that, or how I'd arrive there. I have been staring at this for two days with little progress; where did I go wrong?

 

[Ray Vickson]

Homework Statement

Use Cardano's formula to find a real root for $x^3-15x^2+243x-525=0$.

Homework Equations

$$x = u - \frac{b}{3a}$$
Depressed cubic: $$u^3=3pu+2q$$
Cardano's formula: $$u=\sqrt[3]{q+\sqrt{q^2-p^3}}+\sqrt[3]{q-\sqrt{q^2-p^3}}$$

The Attempt at a Solution

I have found the depressed cubic to be $x^3=20-6x$. I found $3p= -6$ and $2q=20 \to p=-2$ and $q=10$. Using Cardano's formula, I arrived at $u=\sqrt[3]{10+\sqrt{108}}+\sqrt[3]{10-\sqrt{108}}$. However, I graphed the equation and know that it should be 7. If I change my answer to $u=\sqrt[3]{10+\sqrt{108}}-\sqrt[3]{-10+\sqrt{108}}$ and use this answer (2) to solve $x = u - \frac{b}{3a}$ I get the right answer (7), but I have no idea mathematically why I would do that, or how I'd arrive there. I have been staring at this for two days with little progress; where did I go wrong?

I don't know where you went wrong, but you did, indeed, go wrong. Using Maple, the solution to
$$ (1) \hspace{2em} x^3-15x^2+243x-525=0$$
is
$$-(220+4*14001^{1/2})^{1/3}+56/(220+4*14001^{1/2})^{1/3}+5 \doteq 2.476595025 $$
(together with two complex roots), while the solution of
$$(2) \hspace{2em} x^3 +6x - 20 = 0$$
is $x = 2$ (together with two complex roots).

 

[cscott0001]

Oh geez, that's my fault. The original equation is $3x^3-75x^2+243x-525=0$. I gave one half with the 3 factored out and one half without the 3 factored out...

 

[Ray Vickson]

Oh geez, that's my fault. The original equation is $3x^3-75x^2+243x-525=0$. I gave one half with the 3 factored out and one half without the 3 factored out...

OK, so now the real root of this is
$$1/3*(8875+3*2558073^{1/2})^{1/3}+382/3/(8875+3*2558073^{1/2})^{1/3}+25/3 \doteq 21.62912601 $$

BTW: when discussing Cardano's formula you should tell us what $a,b,c,d$ stand for. I have seen different sources use different conventions!

Also, write the formulas for $p$ and $q$ in terms of $a,b,c,d$. We can't help if we can't figure out what you are doing.

 

[cscott0001]

Thank you for your time, I'll try to add as much relevant detail to my question as I can (and finally state the given equation correctly). It is $3x^3-45x^2+243x-525=0$, making $a=3$, $b=-45$, $c=243$, $d=-525$. I quadruple-checked this time!

The method we were taught was to take an equation in the form $ax^3+bx^2+cx+d=0$ and find the depressed cubic in the form $u^3-3pu+2q=0$ by substituting $x = u-\frac{b}{3a}$ and then solving $a(u-\frac{b}{3a})^3+b(u-\frac{b}{3a})^2+c(u-\frac{b}{3a})+d=0$ for $u$. I factored a 3 out of the given equation, which gave me $a=1$, $b=-15$, $c=81$, $d=-175$, which makes the equation $x^3-15x^2+81x-175=0$. I made the substitution of $x = u -\frac{(-15)}{3(1)}=u+5$. Substituting this value in for $x$: $(u+5)^3-15(u+5)^2+81(u+5)-175=u^3+6u-20=0$. We were shown examples in class where the next step was to use this depressed cubic in the form $x^3=3px+2q$ to supply the values of $p$ and $q$: $u^3=20-6u$, so $20=2q$ and $3p=-6$. I'm not sure how to write that in terms of the original equation, I did not use $a, b, c,$ or $d$ to find $p$ and $q$. I used these values for $p$ and $q$ in Cardano's formula listed above to arrive at my thoroughly incorrect solution.

 

[PAllen]

It seems you only went wrong at the very end. The evaluation of Cardano’s formula as you computed it (before fiddling with it) should be approx

2.732 + - .732 ( note: cube root of negative number -.392...). This yields the expected answer of 7 for u + 5