The Cubic Formula: ax^3+bx^2+cx+d=0

[Bill Foster]

The quadratic formula is, as you are well aware:

$$ax^2+bx+c=0$$ $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

What is the cubic formula?

$$ax^3+bx^2+cx+d=0$$

Thanks.

 

[jhicks]

Cardano's formula will get you one root, but you have to transform the general equation a*x^3+b*x^2+cx+d=0 into e*z^3+f*z+g=0 by use of the substitution z=x-b/3. Mathworld explains a bit more: http://mathworld.wolfram.com/CubicFormula.html

 

[awvvu]

If you really want an explicit formula...

http://www.math.vanderbilt.edu/~schectex/courses/cubic/

 

[Bill Foster]

Thanks.

 

[kennysmith39]

could someone just show me with an example on how to work the cardano and Del ferro method from start to finish say using x^3-6x^2+2x-1

 

[HallsofIvy]

The first thing you do is "reduce" the equation to the form $x^3+ bx= c$ without any $x^2$ term. To do that, let x= y- a. Then [itex]x^3= (y- a)^3= y^3- 3ay^2+ 3a^2y- a^3[/math] and [itex]x^2= (y- a)^2= y^2- 2ay+ a^2[/math]

$$x^3- 6x^2+ 2x-1= y^3 -ay^2+ a^2y- a^3- 6y^2+ 12ay- 6a^2+ 2y- 2a- 1$$[tex]= y^3+ (-a- 6)y^2+ (a^2+ 2)y+ (-a^3- 2a- 1)[/math]

That will have no "$y^2$" term is a= -6 and, in that case, the polynomial is
$y^3+ 38y+ 229$ and so our equation is $y^3+ 38y+ 229= 0$.

Here's a quick review of Cardano's formula:
$(a+ b)^3= a^3+ 3a^2b+ 3ab^2+ b^3$
$-3ab(a+ b)= -3a^2b- 3ab^3$
so that $(a+ b)^3+ 3ab(a+ b)= a^3+ b^3$.

In particular, if we let x= a+b, m= 3ab, and $n= a^3+ b^3$, $x^3+ mx= n$.

Can we go the other way? That is, given m and n, can we find a and b and so find x?

Yes, we can. From $m= 3ab$, we have $b= m/(3a)$ so $a^3+ b^3= a^3+ m^3/(3^2a^3)= n$. Multiplying through by $a^3$, $(a^3)^2+ m^3/3^3= na^3$ or $(a^3)^2- na^3+ m^3/3^3= 0$.

We can think of that as a quadratic equation in $a^3$ and solve it with the quadratic formula:
[tex]a^3= \frac{n\pm\sqrt{n^2- 4\frac{m^3}{3^3}}}{2}[/itex][itex]= \frac{n}{2}\pm\sqrt{\left(\frac{n}{2}\right)^2- \left(\frac{m}{3}\right)^3}[/tex]
and a is the cube root of that.

From $n= a^3+ b^3$ we have
$$b^3= n- a^3= \frac{n}{2}\mp\sqrt{\left(\frac{n}{2}\right)^2- \left(\frac{m}{3}\right)^3}$$.


Now, in this problem, $y^3+ 38y+ 229= 0$ or $y^3+ 38y= -229$ so m= 38 and n= -229.
$$\frac{n}{2}= -\frac{229}{2}$$
and
[tex]\left(\frac{n}{2}\right)^2= \frac{52441}{4}[/itex]

$$\frac{m}{3}= \frac{38}{3}$$
and
$$\left(\frac{m}{3}\right)^3= \frac{54872}{27}$$

So
$$a^3= -\frac{229}{2}\pm\sqrt{\frac{52441}{4}- \frac{54872}{27}}$$
and
$$b^3= -\frac{229}{2}\mp\sqrt{\frac{52441}{4}- \frac{54872}{27}}$$

Calculate those numbers, take the cube roots to find a and b and then find x= a+ b.
Each of those will have 3 cube roots but in the various ways of combining them, some things will cancel so that there will be, at most, 3 roots to the equation.

 

[kennysmith39]

thank yo for taking the time to show me this but I am still confused, after plugging in y-a into the equation how does a= -6