Why was the substitution x = y - b/3a used in Cardano's Cubic Formula?

[Terrell]

in Cardano's cubic formula. when reducing a cubic to its depressed form: from ax^3 + bx^2 + cx + d = 0 to a form like y^3 +py + q = 0. what is the motivation behind making x = y - b/3a first before proceeding to depress the cubic? what was cardano's motivation for that substitution or whoever showed him that technique?

 

[micromass]

What is the inflection point of the cubic? https://en.wikipedia.org/wiki/Inflection_point

 

[Terrell]

What is the inflection point of the cubic? https://en.wikipedia.org/wiki/Inflection_point

what about inflection points?

 

[micromass]

what about inflection points?

That was a hint: think about inflection points. I'll leave the rest to you

 

[Terrell]

the roots all fall on the cubic's inflection points?

That was a hint: think about inflection points. I'll leave the rest to you

okay i'll think about it first

 

[Terrell]

okay i'll think about it first

so i noticed that inflection points is equal to -b/3a but I'm not sure if that meant anything at all. still thinking and researching

 

[murshid_islam]

in Cardano's cubic formula. when reducing a cubic to its depressed form: from ax^3 + bx^2 + cx + d = 0 to a form like y^3 +py + q = 0. what is the motivation behind making x = y - b/3a first before proceeding to depress the cubic? what was cardano's motivation for that substitution or whoever showed him that technique?

I'm wondering about that, too. And from f(x) = ax^3 + bx^2 + cx + d = 0, by setting f''(x) = 0, we get the inflection point at x = \frac{-b}{3a}

Not sure where to go from there.
Thanks in advance.

 

[mathwonk]

Since Cardano learned of this formula from my thesis advisor's thesis advisor's thesis advisor's thesis advisor's thesis advisor's thesis advisor's thesis advisor's thesis advisor's thesis advisor's thesis advisor's thesis advisor's thesis advisor's thesis advisor's thesis advisor's thesis advisor's thesis advisor's thesis advisor's thesis advisor's thesis advisor's thesis advisor, namely Tartaglia, I have it on good authority that it is because the coefficient quotient -b/a equals the sum of the roots. Hence to "depress the equation", one wants to subtract -b/a from that sum, or 1/3 that amount from each root.

 

[murshid_islam]

it is because the coefficient quotient -b/a equals the sum of the roots. Hence to "depress the equation", one wants to subtract -b/a from that sum, or 1/3 that amount from each root.

I still don't get it. Can you elaborate a little?

 

[mpresic3]

I actually bought Ars Magna by Cardano (Dover Publications) which tells how to solve the cubic and quartic equations. I have not gone through it as it is heavy reading. I am told the Ars Magna calculation for solving cubic and quartic equations is even more elaborate than algorithms that calculate the roots of these equations today. It seems the proofs in the 16th century, used geometry and lengths, which could not be less than zero. This made their proofs and calculations much more complicated, and any algebra student today would have advantages in that they would be more comfortable with negative numbers.

The substitution you refer to is the first step. You ask for the motivation. The first thing I would do is divide both sides by a. This would get rid of the a-coefficient, although it would change subsequent coefficients. Now I would note what say (a-b)(a-b)(a-b) is and I can see the term just after the leading term is -3a squared b. I could then conclude if we change b to b/3 we could likely get rid of the squared term. To be honest, I think many high school algebra students would get to this point in less than an evening. I think I got this far when I tried it one evening in ninth grade. The hard part starts right after this substitution in solving the depressed cubic.

This problem supplied the very best Italian mathematicians of the 16th century years of effort. For a good recap of the history, see "Journey to Genius" by Dunham. The problem is so difficult, you can marvel at their ingenuity, but I do not think you can expect clear motivation for each step. You might consult Ars Magna though.

You know what is cool. If you use the student version of MATLAB, I think it is MATLAB 2010, you can ask it to solve the fourth degree equation symbolically in terms of the coefficients. The four roots take up pages and pages with all kinds of mysterious coefficients, but it solves for them, just like the familiar quadratic formula we all learn in high school.

 

[mathwonk]

The clearest explanation I have seen of the cubic fprmula is in Euler's algebra book, which I summarize.

CUBIC FORMULA:

To solve a cubic equation, we start with a simplified one of form X^3 -3bX - c = 0, and again assume we want to find X as a sum X = (p+q). Plugging in gives (p+q)^3 = 3b(p+q) + c, and expanding gives p^3 + 3p^2q + 3pq^2 + q^3 = p^3 + q^3 + 3pq(p+q) =

3b(p+q) + c, and for this to hold means that pq = b, and c = p^3+q^3. Cubing the first of these gives p^3q^3 = b^3, and p^3+q^3 = c. Since we know b and c, we know both the sum and the product of the cubes p^3 and q^3. Can we find p^3 and q^3 from this? If so, then we could take cube roots and find p and q, and finally add them and get our root X = p+q.

Just recall in a quadratic equation of form X^2 - BX + C, that B and C are precisely the sum and product of the desired roots, and we can find those roots from B and C. I.e. we can find any two numbers when we know their sum and product, by solving a quadratic.

Since p^3+q^3 = c and p^3q^3 = b^3, the numbers p^3 and q^3, which can be used to give a solution X = p+q of our cubic, are solutions of the quadratic equation

t^2 -ct + b^3 = 0, where X^3 = 3bX + c.

e.g. to solve X^3 = 9X + 28, we have b = 3, c = 28, and so we solve t^2 -28t + 27 = 0. Here B^2-4C = 676, whose square root is 26, so we get t = (1/2)( 28 ± 26) = {27, 1}, for p^3 and q^3, so p,q are 1 and 3, and hence X=1+3 = 4 solves the cubic. Of course if we know about complex numbers, there are two more cube roots of 1 and 27, and we get two more complex roots. (Only two more because b = pq, so we must always have q = b/p, i.e. the choice of the cube root q is determined by the choice of p.)