Math Is Fun Forum

wrlorimer

Member

Registered: 2014-08-06

Posts: 5

Help with Cardano's method solving cubic

Given a cubic polynomial of the form x^3+ax+b=0, I understand that Cardano solved it by substituting x=(p+q) and setting 3pq+a=0, or pq=-a/3. He then set u=p^3 and v=q^3, converting it into a quadratic equation in v.

u + v = -b
uv = -(a^3/27)

uv = -(b+v)v = -(a^3/27)

v^2 +bv = a^3/27

As long as this quadratic equation had real roots, he could find x=cuberoot(v) - a/3*(cuberoot(v)). (Yes, I know that Tartaglia is believed to be the one who came up with this method.)

Cardano became the first mathematician to make use of what he called an "imaginary number", the sqrt(-15), when he encountered a quadratic equation with roots v=5+sqrt(-15), and u=5-sqrt(-15), giving x=cuberoot(v) - a/(3*cuberoot(v))

This much, I understand.

What I can't figure out is how Cardano derived a real value for x from the difference of two cube roots of complex numbers.

Can anyone explain to me how Cardano went from v=5+sqrt(-15) to a real value for x?

Thanks.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Help with Cardano's method solving cubic

Hi;

Cardano, using the method that falsely bears his name did not know about complex numbers so he would assume that the roots are real. Check this page out and tell me exactly where you are stuck.

http://en.wikipedia.org/wiki/Cubic_function

About 1 / 3 down the page is his method.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

wrlorimer

Member

Registered: 2014-08-06

Posts: 5

Re: Help with Cardano's method solving cubic

I get stuck at the line "At this point, Cardano, who did not know complex numbers, supposed that the roots of this equation were real, that is that q^2/4 + p^3/27 >0 "

My understanding is that Cardano came across the situation where q^2/4 + p^3/27 = -15, and this was the first instance ever of an imaginary number being used in a calculation.

Did Cardano, in fact, solve a cubic equation by treating sqrt(-15) as if it were a usable number? If so, how did he solve it?

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Help with Cardano's method solving cubic

Hi;

Did you get how he got u^3 and v^3 on the next line?

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

wrlorimer

Member

Registered: 2014-08-06

Posts: 5

Re: Help with Cardano's method solving cubic

Yes. He set t=u+v and expanded the expression (u+v)^3 + p(u+v) + q = u^3 + 3uv^2 + 3(u^2)v + v^3 +pu + pv + q

Simplifying, this becomes  u^3 + v^3 + q + (3uv)v + (3uv)u +p(u+v) =   u^3 + v^3 + q + (3uv+p)(u+v)

Then he set 3uv+p = 0 and set r=u^3, s=v^3, to get the two equations:

(1) rs = (uv)^3 = (-p/3)^3 = -p^3/27

and

(2) r+s = u^3 + v^3 = -q

He then substituted r = -(q+s) into equation (1) to get

(3) -(q+s)s = -p^3/27

which is a quadratic in s (q and p are the known coefficients from the original cubic, s is the unknown variable).

If this quadratic has real roots, then the solution to the cubic is straightforward: solve for r and s, and set t = cuberoot(r)+cuberoot(s).

But it is my understanding that Cardano found an equation for which r = 5 + sqrt(-15), s = 5 - sqrt(-15), and still found real roots of the cubic function, even though he described it as "mental torture" to work with what he called an "imaginary" number.

Is this true? If so, how did he do it?

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Help with Cardano's method solving cubic

Hi;

Over here they discuss a similar problem that may give you some ideas.

http://math.stackexchange.com/questions … os-formula

I do not think any of those methods were known to Cardano.

What was the cubic if he was able to reduce r and s to numbers?

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

wrlorimer

Member

Registered: 2014-08-06

Posts: 5

Re: Help with Cardano's method solving cubic

I don't know what the original cubic was. I got interested in this by viewing this youtube video:

https://www.youtube.com/watch?v=_qvp9a1x2UM

For the original cubic, it must have been x^3 + ax + b

If my reverse engineering is correct, then b=-10 and a = -3*cuberoot(40)

The links you posted look very helpful. Thanks.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Help with Cardano's method solving cubic

Hi;

Let me know if you get anything out of it.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.